HOMOTOPY THEORY OF THE SUSPENSIONS OF THE
PROJECTIVE PLANE
J. WU
Abstract. The homotopy theory of the suspensions of the real projective p*
*lane
is largely investigated. The homotopy groups are computed up to certain r*
*ange.
The decompositions of the self smashes and the loop spaces are studied wi*
*th some
applications to the Stiefel manifolds.
Contents
1. Introduction 2
2. Preliminary and the Classical Homotopy Theory 5
2.1. Action of the Steenrod Algebra 5
2.2. The James Construction and the Cohen Groups 9
2.3. An Application to the Stiefel Manifolds 14
2.4. The Strong Form of the Kervaire Invariant One Problem 16
2.5. The Exponent Problem 25
2.6. Some Lemmas on Fibrations 30
3. Decompositions of Self Smash Products 32
3.1. Functorial Decompositions of Self Smashes of Twocell Complexes 32
3.2. Decompositions of Self Smashes of RP2 36
4. Decompositions of the Loop Spaces 44
4.1. Functorial Decompositions of Loop Suspensions 44
4.2. Determination of ~Qmaxk(P n(2)) for k 9 *
*45
4.3. The Fibre of the Pinch Map 48
4.4. The EHP Sequences for mod 2 Moore Spaces 51
5. The Homotopy Groups ßn+r( nRP2) for n 2 and r 8 58
5.1. Preliminary and Notations 59
5.2. The Homotopy Groups ß*(P 4(2)) 60
5.3. The Homotopy Groups ß*(P 5(2)) 69
5.4. The Homotopy Groups ß*(P 6(2)) 77
5.5. The Homotopy Groups ß*(P 7(2)) 81
5.6. The Homotopy Groups ß*(P 8(2)) 86
5.7. The Homotopy Groups ß*(P 9(2)) 89
5.8. The Homotopy Groups ß*(P n(2)) for n 10 91
5.9. Remarks on the generators for the homotopy groups 93
6. The Homotopy Theory of RP2 94
6.1. Fibrations over RPn 95
___________
Research is supported in part by the Academic Research Fund of the National U*
*niversity of
Singapore RP3992646.
1
2 J. WU
6.2. A Decomposition of 30P 3(2) 98
6.3. The Homotopy Groups ß*( RP2) 101
Appendix A. The Table of the Homotopy Groups of nRP2 127
References 128
1. Introduction
Let RP2 be the projective plane. The ndimensional mod 2 Moore space P n(2) is
defined by P n(2) = n2RP2 for n 2. As a space, P n(2) is the homotopy cofib*
*re
of the degree 2 map [2]: Sn1 ! Sn1. In other words, P n(2) is the cell compl*
*ex
obtained by attaching an ncell to Sn1 where the attaching map is of degree 2.*
* In
geometry, P 2n(2) is the 2nskeleton of the Stiefel manifold V2n+1,2(or the ass*
*ociated
spherical bundle ø(S2n) of the tangent bundle of S2n). More precisely, let x be*
* any
point in V2n+1,2. Then the open manifold V2n+1,2\ {x} is homotopy equivalent to
P 2n(2). Also an even dimensional projective plane admits a cellular decomposit*
*ion
in terms of mod 2 Moore spaces.
In this paper, we study homotopy theoretic aspects of mod 2 Moore spaces. Par*
*tic
ularly we will compute the homotopy groups up to 8 + dimension. By knowing these
homotopy groups, we obtain some information on threecell complexes. For instan*
*ce,
by using the fact that ß10(P 6(2)) = Z=8, we show that if a simply connected co*
*mplex
X has the same mod 2 cohomology ring as that of ø(S6), then X is homotopy equiv*
*a
lent to ø(S6). (See Proposition 5.12.) In particular, if X is a simply connecte*
*d closed
11manifold such that its first nontrivial integral homology group is H5(X; Z) *
*= Z=2,
then X is homotopy equivalent to ø(S6). This statement is not true for other St*
*iefel
manifolds in general. For instance, by considering ß6(P 4(2)) = Z=4 Z=2, ther*
*e is
a simply connected complex X with property that the mod 2 cohomology ring of
X is isomorphic to that of ø(S4) but X is not homotopy equivalent to ø(S4). The
homotopy groups ß*(P n(2)) also helps to classify the complexes X which has the
cell structure of the form P n(2) [ en+r. For example, ß4(P 3(2)) = Z=4 shows *
*that
if X has the cell structure P 3(2) [ e5, then X is homotopy equivalent to one o*
*f the
following three spaces:
1) the wedge P 3(2) _ S5;
2) the homogeneous space SU(3)=SO(3);
3) P 3(2) [2~3e5, where ~3: S4 ! P 3(2) represents the generator of ß4(P 3(*
*2)).
If in addition X is a closed 5manifold, then X is homotopy equivalent to SU(3)*
*=SO(3).
(See Example 6.15.)
Our methods for computing the homotopy groups are: (1). to use the results on
the homotopy groups of spheres, (2). to investigate various fibre sequences and*
* (3).
to study splittings of spaces.
The homotopy groups of spheres have been much studied. Up to the range what
we need, we use Toda's book [42] as the main reference for the homotopy groups *
*of
spheres and so we directly use the notations for the elements in the ß*(Sn) giv*
*en in [42]
without further explanation. Relations between the homotopy groups of spheres a*
*nd
that of a space X can be set up by constructing a map from a sphere or its loops
to X or its loops, or vice versa. For instance, by knowing a spherical class i*
*n the
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 3
homology of the loop space of X, we obtain a map from a sphere to the loop space
of X. This map then induces a group homomorphism from the homotopy groups of a
sphere to that of X. In our case, the mod 2 homology of the loop space P n(2) *
*is the
tensor algebra on two generators for n 3. Many spherical classes can be obtai*
*ned by
considering the Samelson products. We should point out that the existence of ce*
*rtain
spherical classes in the mod 2 homology H*( P n(2)) is equivalent to the strong*
* form
of Kervaire invariant one conjecture. (See Section 2.4 for details.)
The canonical method to study the homotopy groups of a space X is to consider
certain fibration resolution of X or its loop spaces. The resolution produces a*
* spec
tral sequence in general. Depending on the type of fibration resolution consid*
*ered,
there are several types of the Adams spectral sequences using different homology
theories. In this paper, roughly speaking, we consider the spherical fibration*
* res
olutions of P n(2). These resolutions will be considered case by case dependin*
*g on
special properties of the considered mod 2 Moore space. For instance, there are*
* sev
eral ways to produce fibration resolutions for P 4(2): (1). Consider the pinch*
* map
P 4(2) ! S4 and then do analysis on the homotopy fibre F 4{2}. (2). Consider the
canonical inclusion P 4(2) ! ø(S4)and study its homotopy fibre. (3). Consider t*
*he
map OE: P 4(2) ! HP1 given by the composition of the pinch map P 4(2) ! S4
and the inclusion S4 ! HP1 . In the last case, the homotopy fibre of the map OE
is homotopy equivalent to the 7skeleton of the double loop space 2S5. Since *
*our
purpose is to compute low homotopy groups, we are not going to study the "entire
fibration resolutionsä nd their associated spectral sequences in this paper. O*
*n the
other hand, we freely consider different fibrations until the homotopy groups a*
*re de
termined, where the different fibrations will be used particularly for determin*
*ing the
group structure of the homotopy groups.
Although the space P n(2) itself is indecomposable up to homotopy, its loop s*
*pace
P n(2) admits various product decompositions for n 3. By applying the func
tor ß* to decompositions of loop spaces, we obtain decompositions of the homoto*
*py
groups and then the homotopy groups can be obtained by studying the factors in *
*the
decompositions. For instance, the homotopy group ß11(P 3(2)) = Z=2 14 Z=4 2 is
obtained by taking the direct sum of the corresponding homotopy groups of many
factors in a decomposition of the triple loop space of P 3(2).
Our decompositions of P n(2) for n 3 are the special case of general resul*
*ts
in [36, 37]. Let X be a path connected plocal space. Then there is a homotopy
decomposition
`1
X ' Amin(X) x ( Qmaxk(X)),
k=2
where Aminand Qmaxkare functors and Qmaxk(X) is a functorial retract of the sus*
*pen
sion of the kfold self smash product X(k). By applying the HiltonMilnor Theor*
*em
to the second factor, one gets further decompositions of X. The study of this
decomposition for the special case where X = P n1(2) is given in chapter 4.
Since the space Qmaxk(X) is a retract of X(k), the study of splittings of se*
*lf smash
products of P n(2) is important for our purpose to compute the homotopy groups.*
* In
chapter 3, we study splittings of the self smash products of RP2. The homology *
*of
indecomposable factors in (RP2)(k)has been determined in [38]. The cell structu*
*re of
4 J. WU
the indecomposable factors is the main issue that we are going to study in this*
* paper.
One of the interesting points is that these indecomposable factors admits smash
product decomposition. For instance, the smallest retract of (RP2)(7)which cont*
*ains
the bottom cell is homotopy equivalent HP2 ^ CP2 ^ RP2. (See Corollary 3.11.)
This gives a relation among different projective planes. We should point out t*
*hat
the splittings of self smash products of more general spaces play a role in Mor*
*ava
Ktheory, see [32]. Furthermore, the problem on determining functorial splittin*
*gs of
self smashes of suspensions is equivalent to the fundamental problem in the mod*
*ular
representation theory of the symmetric groups, see [38]. This problem remains o*
*pen
in both homotopy theory and the representation theory in general. However much
progress has been achieved recently in both areas. In homotopy theory, the homo*
*logy
of functorial indecomposable retracts of self smashes of a twocell suspension *
*has been
determined in [38] with applications given in [10]. The corresponding results i*
*n the
representation theory was obtained in [13, 14].
The EHP sequence is one of the main tools for computing the homotopy groups
of spheres. In Section 4.4, we study an analogue for mod 2 Moore spaces. More
precisely, we consider the reduced JamesHopf map H~2: P n+1(2) ! P 2n+1(2) for
n 2 which is given by the second JamesHopf map H2: P n+1(2) ! (P n(2))(2)
composing with the loop of the pinch map (P n(2))(2)! P 2n+1(2). We show that *
*the
atomic retract of the homotopy fibre of the map ~H2is homotopy equivalent to Sn*
*{2},
the homotopy fibre of the degree 2 map [2]: Sn ! Sn. (See Theorem 4.21.) Also a
product decomposition of the homotopy fibre of the map H~2is given in Section 4*
*.4
up to certain dimension.
The space P 3(2) = RP2 will be treated as a special case. There is a de
composition of the triple loop space of P 3(2) which is obtained as follows. L*
*et
fl2: RP2 ! BSO(3) be the adjoint map of the canonical inclusion RP2 ! SO(3).
Then the homotopy fibre of fl2 is homotopy equivalent to RP42_ P 6(2), where
RPba= RPb=RPa1 is a truncated projective space. Let 0X denote the path
connected component of X which contains the basepoint and let X denote
the nconnected cover of X. Then we have a decomposition
30P 3(2) ' 30( RP42_ P 6(2)) x 2(S3<3>),
see Theorem 6.7. This induces a decomposition of ßr(P 3(2)) for r 4. By apply*
*ing
the HiltonMilnor theorem to ( RP42_ P 6(2)), we are able to compute ßr(P 3(2))
for r 11.
The main part of this paper, particularly the calculation of the homotopy gro*
*ups,
is from my Ph.D. thesis [45] of the University of Rochester under the supervise*
* of
Professor Frederick Cohen. The table of the homotopy groups computed in this
paper have been announced without proofs in [8, pp.1202]. Some of the homotopy
groups of the mod 2 Moore spaces have been computed by Mukai and others with
applications [27, 28, 29] and so there are some overlap between our work and th*
*eirs.
An application to construct nonsuspension coHspaces is given in [48].
Some terminologies and notations used in this paper are as follows. Every spa*
*ce is
localized at 2. We always assume that a space is a pathconnected CW complex w*
*ith
a nondegenerate base point * and any map is a pointed map. Let f :X ! Y be a
map. We write Ff for the homotopy fibre of f and Cf is the homotopy cofibre of *
*f.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 5
The homotopy fibre of the pinch map P n(2) ! Sn is denoted by F n{2}. The mod 2
homology of a space X will be denoted by H*(X). On the other hand, the integral
homology is denoted by H*(X; Z) to avoid confusion. The nfold self smash produ*
*ct
of X is denoted by X(n). Let X be a coHspace. A map of degree k, denoted by
~0k `k fold 0
[k]: X ! X, means the composite X ____ X ____X, where ~k is a choice of k
j=1
fold comultiplication. (Note. If X is not homotopy associative, the map [k] dep*
*ends
on the choice of a kfold comultiplication and so may not be unique in general.)
The homotopy fibre of [k]: Sn ! Sn is denoted by Sn{k}. The homotopy cofibre
of [k]: Sn ! Sn is the Moore space P n+1(k). For an Hspace X, a power map of
k k ~k
degree k, denoted by k :X ! X, means the composite X ____X ____X, where ~k
is a choice of kfold multiplication. Let X be a complex. We write skn(X) for*
* the
nskeleton of X. The nconnected cover of X is denoted by X. The basepoint
pathconnected component of the loop space of X is denoted by 0X. In addition,*
* we
write RPbafor RPb=RPa1. Furthermore we directly use notations for the elements*
* of
the homotopy groups of spheres in [42]. Let x be a homogeneous element in a gra*
*ded
module V . We write x for the degree of x. The notation yn 2 V indicates that*
* y is
a homogeneous element in V of degree n, that is, y 2 Vn. Let ff be an element i*
*n the
set of homotopy classes [X, Y ]. We write ff :X ! Y for a map which represents *
*the
homotopy class ff.
The article is organized as follows. In chapter 2, we give some preliminaries*
* and
discuss some classical problems related to mod 2 Moore spaces. The study of se*
*lf
smashes of the projective plane RP2 is given in chapter 3. In chapter 4, we st*
*udy
the loop space of mod 2 Moore spaces. The computation of the homotopy groups
of P n(2) for n 4 is given in chapter 5. The homotopy theory of P 3(2) is stu*
*died
in chapter 6. The computation of the homotopy groups of P 3(2) is given in the *
*last
section of this chapter.
The author is indebted to Professors Fred Cohen, John Harper, John Moore,
Mamoru Mimura, Juno Mukai and Paul Selick for many fruitful conversations on
this project. These conversations defined what was most important in this pape*
*r.
The author would like to thank the referee for his/her important comments and h*
*elp
ful suggestions. The author also would like to thank many of our colleagues for*
* their
encouragement and suggestions on this project.
2. Preliminary and the Classical Homotopy Theory
2.1. Action of the Steenrod Algebra. Recall the Steenrod algebra A can be
described as all linear natural transformations of the mod 2 cohomology functor*
*. We
refer the reader to [40] for several basic properties of A. The algebra A is ge*
*nerated
by symbols Sqn, n 1, Sqn = n, subject to the Ä dem relations". That is, if
one takes the tensor algebra on these symbols and quotients by the twosided id*
*eal
generated by the Adem relations one obtains A. The fundamental properties of Sq
are as follows
1) For all integers i 0 and n 0, there is a linear natural transformati*
*on
Sqi:Hn(X, A) ____Hn+i(X, A).
6 J. WU
2) Sq0 = 1 and Sq1 is the Bockstein homomorphism.
3) Sqnx = x2 if x = n > 0.
4) Sqix = 0 if x < n.
5) The Cartan formula
Xn
Sqn(x y) = Sqix Sqniy.
i=0
6) The Adem relations
[a=2]X` '
b  1  j a+bj j
SqaSqb = Sq Sq
j=0 a  2j
for 0 < a < 2b, where [a=2] = max {k 2 Zk a=2}.
We write Sqn*:Hq(X) ! Hqn(X) for the dualaction of Sqn on the homology
functor. By the duality, the Cartan formula and the Adem relations are given by
Xn
Sqn*(x y) = Sqi*(x) Sqni*(y)
i=0
[a=2]X` '
b  1  j j a+bj
Sqb*Sqa*= Sq*Sq*
j=0 a  2j
for 0 < a < 2b, respectively.
Recall that the cohomology ring H*(RP2) is the quotient of the polynomial alg*
*ebra
on the symbol x, x = 1, by the ideal generated by x3. Thus H*(RP2) has a basis
{1, x, x2 = Sq1x} and so the reduced homology H~*(RP2) has a basis {u, v} with
v = 2 and Sq1*v = u. It follows that H~*(P n(2)) has a basis {u, v} with v*
* = n
and Sq1*v = u. Let [A, A] be the twosided ideal generated by the commutator
[ff, fi] = fffi  fiff for ff, fi 2 A. Recall that A=[A, A] is a divided polyno*
*mial algebra
on one generator, that is
` '
i + j i+j
SqiSqj = Sq mod [A, A],
i
see [40, pp.25]. Let (x1, . .,.xk) be the divided polynomial algebra generate*
*d one
x1, . .,.xk. We have A=[A, A] ~= (Sq1).
Lemma 2.1. Let X be a pathconnected space. Suppose that the commutator [Sqa*, *
*Sqb*]
acts trivially on H*(X) for a, b 1. Then the action of A on
(1).H*( X) and
(2).H~* (X)(n) for n 1
factors through the quotient A=[A, A].
Proof.It suffices to show that [Sqa*, Sqb*] acts trivially on H*( X). Let V = *
*~H*(X).
Recall that, as an algebra, H*( X) is isomorphic to the tensor algebra T (V ).*
* Thus
the action of the Steenrod algebra on H*( X) is uniquely determined by its act*
*ion
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 7
on V and the Cartan formula. By using the fact that Sqi*Sqj*x = Sqj*Sqi*x for x*
* 2 V ,
we have
X
Sqb*Sqa*(x1x2. .x.k) = Sqj1*Sqi1*x1Sqj2*Sqi2*x2. .S.qjk*S*
*qik*xk
i1 + i2 + . .+.ik = a
j1 + j2 + . .+.jk = b
X
= Sqi1*Sqj1*x1Sqi2*Sqj2*x2. .S.qik*Sqjk*xk = Sqa*Sqb*(x1x2*
*. .x.k)
i1 + i2 + . .+.ik = a
j1 + j2 + . .+.jk = b
for xi2 V and hence the result.
Corollary 2.2. The action of A on H*( P n(2)) and ~H* (P n(2))(m)factors through
A=[A, A] ~= (Sq1).
Lemma 2.3. Let f :A ! B be a map and let i: B ! Cf be the inclusion. Suppose
that there exist positive integers r and k such that
(1).Hj(B) = 0 for j r and Hr(Cf) 6= 0.
(2).there is an element v 2 Hr(Cf) with the property that Sqk*v 6= 0 and Sqj*
**v
lies in the image of i*: H*(B) ! H*(Cf) for j k.
Then the map f ^ X :A ^ X _____B ^ X is essential for any finite dimensional
complex X with H~*(X) 6= 0.
Proof.Suppose that f ^ X :A ^ X ! B ^ X were null homotopic. Then there is a
map OE: Cf ^ X ! B ^ X such that OE O (i ^ X) is homotopic to the identity map *
*of
B ^ X. Let y be a top cell of X, that is, Hj(X) = 0 for j > y and y is not ze*
*ro in
Hy(X). There exists an integer t 0 such that Sqj*y = 0 for j > t and Sqt*y *
*6= 0.
By the assumption (1), Hr+y(B ^ X) = 0 and so OE*(v y) = 0. For each j k,*
* let
uj be an element in Hrj(B) such that Sqj*v = i*(uj). It follows that
X
0 = Sqk+t*OE*(v y) = OE*(Sqk+t*(v y)) = OE*(Sqj*v Sqk+tj*y)
j k
X
= uk Sqt*y + uj Sqk+tj*y 6= 0
j>k
M
because ~Hk+t(B ^ X) = H~j(B) ~Hk+tjX. We obtain a contradiction and hence
j
the result.
Corollary 2.4. Let X be any finite dimensional complex with ~H*(X) 6= 0. Then t*
*he
degree map [2]: X ! X is essential.
Proposition 2.5. Let n 3. Then the degree map [2]: P n(2) ! P n(2) is homotop*
*ic
to the composite
pinch n '' n1 n
P n(2) ____S ____S ____P (2),
where j is a suspension of the Hopf map.
8 J. WU
Proof.It suffices to show that the statement holds for n = 3. Clearly the map *
*[2]
restricted to the bottom cell S2 is null homotopic and so there is a map ff :S3*
* ! P 3(2)
such that the diagram
pinch 3
P 3(2)_______S
 
 
 
[2] ff
 
? ?
P 3(2)===== P 3(2)
commutes up to homotopy. Since P 3(2) is a torsion space, the composite
ff 3 pinch 3
S3 ____P (2) ____S
is null homotopic and so ff lifts to the homotopy fibre F 3{2} of the pinch map
P 3(2) ! S3. By computing the low homology of F 3{2}, the bottom cell S2 is the
3skeleton of F 3{2} and so there is a map ff0:S3 ! S2 such that the map ff is
ff0 2 3
homotopic to the composite S3 _____S ____P (2). By Corollary 2.4, the map
pinch 3 ff0 2
[2]: P 3(2) ! P 3(2) is essential and so is the composite P 3(2) ____S ____S*
* . It
follows that [ff0] is a generator of ß3(S2) = Z or ff0' j. The assertion follo*
*ws.
Proposition 2.6. Let n 3. Then the group [P n(2), P n(2)] is isomorphic to Z=*
*4.
Proof.We only need to compute [P 3(2), P 3(2)]. By the cofibre sequence
[2] 2 3
S2 ____S ____P (2),
there is an exact sequence
2 3 3 3 3 2 3 2 2 3
[S3, P 3(2)]_[S , P (2)]__[P (2), P (2)]__[S , P (2)]_[S , P (2)].
Thus there is a short exact sequence
0______ß3(P 3(2))=2___[P 3(2), P 3(2)]_ß2(P 3(2)) = Z=2____0.
By using the fact that S2 is the 3skeleton of F 3{2}, the map ß3(S2) ____ß3(P*
* 3(2))
is an epimorphism and so ß3(P 3(2)) is a cyclic group Z=k for some integer k. *
*It
follows that [P 3(2), P 3(2)] is isomorphic to Z=4, Z=2 Z=2 or Z=2. By Coroll*
*ary 2.4,
we have
2 . [P 3(2), P 3(2)] 6= 0
and hence [P 3(2), P 3(2)] = Z=4, which is the result.
Corollary 2.7. Let f be a self homotopy equivalence of P n(2) with n 3. Then f
is either homotopic to the identity map or the degree map [1]: P n(2) ! P n(2).
Note. Propositions 2.5 and 2.6 can be found in [43, Corollary, p.307] and [26,
Lemma 1.1, p.271].
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 9
2.2. The James Construction and the Cohen Groups. Let X be a pointed
space with the base point *. Recall that the James construction J(X) is the fr*
*ee
monoid on X subject to the single relation * = 1, see for instance [18, 44]. T*
*he
James filtration {Jn(X)}n 0 with J0X = * and J1X = X is induced by the word
length filtration. Note that Jn(X)=Jn1(X) ~= X(n). Let qn: Xn ! Jn(X) be the
quotient map. The fundamental properties of J(X) are as follows [5]:
1) If X is pathconnected, then J(X) is (weak) homotopy equivalent to X.
2) The quotient qn: Xn ! Jn(X) has a functorial crosssection.
3) The inclusion Jn1(X) ! Jn(X) has a functorial retraction.
4) There is a functorial decomposition
`n
Jn(X) ' X(n)
j=1
for 0 n 1.
5) The JamesHopf map Hn: J(X) ____J(X(n)) is defined combinatorially by
Y
Hn(x1x2. .x.q) = (xi1^ xi2^ . .^.xin),
1 i1 i
Let
di:Xn1 ! Xn, (z1, z2, . .,.zn1) 7! (z1, . .,.zi1, *, zi, . .,.zn1)
~ßi:Xn ! Xn1, (z1, z2, . .,.zn) 7! (z1, . .,.zi1, zi+1, . .,.zn)
be the maps for 1 i n. Then there are commutative diagrams
ef n ef n1
Fn _______[X , Y ] Fn1 _____[X , Y ]
  \ 
   
di di* si ~ß*i
   
?? ?  ?
ef n1 ? ef n
Fn1 _____[X , Y ] Fn _______[X , Y ]
di
for 1 i n. It follows that the composite Fn _____Fn1 ____Kn1(C) fac
si
tors through Kn(C) and the composite Fn1 _____Fn ____Kn(C) factors through
Kn1(C). Let di:Kn(C) ! Kn1(C) and si:Kn1(C) ! Kn(C) denote the resulting
homomorphisms. Then there are commutative diagrams
ef n ef n1
Kn(C) _____[X , Y ] Kn1(C) ___[X , Y ]
  \ 
   
di di* si ß~*i
   
?? ?  
ef n1 ? ef n ?
Kn1(C) ___[X , Y ] Kn(C) _____[X , Y ]
for f 2 C and 1 i n.
By direct calculation, we have
Lemma 2.8. In the sequence of the groups {Kn(C)}n 1, the following identities h*
*old
1) djdi= didj+1 for j i.
2) sjsi= si+1sj for j i.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 11
3) 8
< si1dj for j < i
djsi= id for j = i
: s
idj1 for j > i.
Note. The third identity is different from the simplicial identity.
Lemma 2.9. The [Jn(X), Y ] is the equalizer of the homomorphisms
di*:[Xn, Y ] ! [Xn1, Y ]
for 1 i n.
Proof.The proof is given by induction on n and the assertion holds for n = 1.
Suppose that the assertion holds for n  1 with n > 1. By the suspension splitt*
*ing
theorem, q*n:[Jn(X), Y ] ! [Xn, Y ] is a monomorphism and so [Jn(X), Y ] can*
* be
regarded as a subgroup of [Xn, Y ]. Let Gn be the equalizer of the homomorphis*
*ms
di:[Xn, Y ] ! [Xn1, Y ] for 1 i n. By the definition of Jn(X), we have
[Jn(X), Y ] Gn.
Conversely let ff 2 Gn, that is, d1ff = d2ff = . .=.dnff. Then d1ff 2 Gn1 and*
* so
d1ff 2 [Jn1(X), Y ] by induction. Consider the commutative diagram
[Jn(X), Y ]____[Xn, Y ]
 
 
 
 d1
 
?? ??
[Jn1(X), Y ]__[Xn1, Y ].
There exists an element fi 2 [Jn(X), Y ] such that d1fi = d1ff. It follows th*
*at
di(fffi1) = 1 for all 1 i n. Let g :Xn ! Y be a map which represents fffi*
*1
and let g0: Xn ! Y be the adjoint of g. Then the composite
n`
(d1,...,dn)n g0
Xn1 ________ X ____Y
i=1
is null homotopic. let
(X*)n = {(z1, . .,.zn) 2 Xnzi= * for somei}.
Recall that the map
n`
(d1,d2,...,dn) n
Xn1 __________ (X*)
i=1
has a crosssection. Thus the composite
g0
(X*)n ____ Xn ____Y
is null homotopic and so g0 factors through X(n)up to homotopy or
fffi1 2 [X(n), Y ] [Jn(X), Y ] [Xn, Y ].
Thus ff 2 [Jn(X), Y ] and hence the result.
12 J. WU
The Cohen group Hn(C) is defined to the equalizer of the group homomorphisms
di:Kn(C) ! Kn1(C) for 1 i n. By Lemma 2.9, there is a homomorphism
ef: Hn(C) ! [Jn(X), Y ] such that the diagram
ef
Hn(C) ____[Jn(X), Y ]
\ \
 
 
 
 
 
? ef ?
Kn(C) _____[Xn, Y ]
commutes for f 2 C. By the definition, there is a homomorphism pn: Hn(C) ! Hn1*
*(C)
such that the diagram
Hn(C) _____Kn(C)
 
 
 
pn di
 
 
? ?
Hn1(C) ___Kn1(C)
commutes for each 1 i n.
Lemma 2.10. The map pn: Hn(C) ! Hn1(C) is an epimorphism for each n.
Proof.We show by induction that pk,n = pk+1 O . .O.pn: Hn(C) ! Hk(C) is an
epimorphism for k n. Clearly p1,nis an epimorphism. Suppose that pk1,nis an
epimorphism with k > 1 and let ff 2 Hk(C). Since pk1,n:Hn(C) ! Hk1(C) is onto,
we may assume that ff lies in the kernel of pk: Hk(C) ! Hk1(C). Let
Y
ffk,n= sinksink1. .s.i1ff 2 Kn(C)
1 i1 1. Then [2] ' 2 . (H2 O !m ): Sm ! Sm .
Proof.Let C consist of the single object Sm1 ____ Sm . Observe that the maps
2Jn(Sm1)and [2]Jn(Sm1):Jn(Sm1 ) ! Sm represent the element ef(x1. .x.n)2
and ef(x21x22. .x.2n), respectively. By commutator calculus,
Y
x21x22. .x.2n (x1x2. .x.n)2 [xi, xj] mod 3(Fn).
1 i = 0, kxk = 1, kyk = 1)}
is the spherical tangent bundle ø(S2n) of S2n. Let M be an orientable mmanifo*
*ld
and let f :M ! M be a map. The degree of f, denoted by deg(f), is defined to
the degree of the homomorphism f*: Hm (M, Z) = Z ! Hm (M, Z) = Z. A question
arising from geometry is whether there is a self homeomorphism f of M which cha*
*nges
the orientation, that is, deg(f) = 1. In the case where M = V2n+1,2with n > 1,*
* we
will show that for any self homotopy equivalence of preserves the orientation a*
*nd so
the answer of the question above is negative when M = V2n+1,2with n > 1.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 15
Recall that P 2n(2) is the (4n  2)skeleton of V2n+1,2and so there is a cofi*
*bre
sequence
~2n 2n
S4n2 ____P (2) ____V2n+1,2.
The following lemma can be found in [31], [25, Lemma 12, p.185] or [26, Proposi*
*tion
2.1, p.274].
Lemma 2.17. The order of the homotopy class [~2n] in ß4n2(P 2n(2)) is 4 if n *
* 0
mod 2 and 8 if n 1 mod 2 and n > 1.
Let ~02n:S4n3 ! P 2n(2) be the adjoint map of ~2n.
Lemma 2.18. Let n > 1. Then
[2] O ~2n ' * and [1] O ~2n ' ~2nO [3]: S4n2 ! P 2n(2)
for n > 1.
Proof.Observe that J2(P 2n1(2)) is the (6n  7)skeleton of J(P 2n1(2)) ' P *
*2n(2).
It suffices to show that the homotopy class
[ [1]J2(P2n1)O ~02n] = 3[~02n]
in ß4n3( P 2n(2)). Since the pinch map P 2n(2) ! S2n factors through V2n+1,2,
the composite [2] O ~2n is null homotopic by Proposition 2.5. Consider the Coh*
*en
representation
ef: H2(E) ! [J2(P 2n1(2)), P 2n(2)],
where f is the inclusion P 2n1____ P 2n(2). The maps
[1]J2(P2n1(2)), [2]J2(P2n1(2))and kJ2(P2n1(2))
represent elements x11x12, x21x22and (x1x2)k, respectively. Since
x21x22= (x1x2)2 . [x1, x2] = (x1x2)2 . (x11x12) . (x1x2)
in H2(E), we have
0 = [ [2]J2(P2n1(2))O ~02n] = 2[~02n] + [ [1]J2(P2n1(2))O ~02n] + [*
*~02n]
in ß4n3( P 2n(2)) and hence the result.
Theorem 2.19. Let n > 1 and let f :V2n+1,2! V2n+1,2be a map. Let
g = fP2n(2):P 2n(2) ! P 2n(2)
be the restriction of f. Then
8
>> 1 mod 4 if g ' id and n 0 mod 2
>>
>> 1 mod 8 if g ' id and n 1 mod 2
>> 1 mod 4 if g ' [1] and n 0 mod 2
><
5 mod 8 if g ' [1] and n 1 mod 2
deg(f)
>> 0 mod 4 if g ' [2] and n 0 mod 2
>>
>> 0 mod 8 if g ' [2] and n 1 mod 2
>>
>: 0 mod 4 if g ' * and n 0 mod 2
0 mod 8 if g ' * and n 1 mod 2
16 J. WU
Proof.There is a homotopy commutative diagram of fibre sequences
@ q 4n1
S4n1 _______Fq_______V2n+1,2_____S
   
   
   
[deg(f)] f~ f [deg(f)]
   
? @ ? ? q ?
S4n1 _______Fq_______V2n+1,2_____S4n1,
where q is the pinch map. By computing lowdimensional homology of Fq, P 2n(2) *
*is
the (6n  4)skeleton of Fq and so there is a homotopy commutative diagram
~2n 2n
S4n2 ______P (2)
 
 
 
[deg(f)] ' g
 
? ~ ?
S4n2 _____2nP 2n(2).
The assertion follows from Lemmas 2.17 and 2.18.
Corollary 2.20. Let n > 1. Then
(1).Any self homotopy equivalence of V2n+1,2preserves the orientation. In pa*
*rtic
ular, there is no self homeomorphism of V2n+1,2which changes the orienta*
*tion.
(2).If f is a 2local homotopy equivalence of V2n+1,2, then deg(f) 1 mod *
*4.
(3).If n 0 mod 2, then there is no self map of V2n+1,2of degree 2 or 3 m*
*od 4.
(4).If n 1 mod 2, then there is no self map of V2n+1,2is of degree 2, 3, *
*4, 6 or
7 mod 8.
(5).If n 1 mod 2 and f is a self homotopy equivalence of V2n+1,2, then f
restricted to P 2n(2) is homotopic to the identity map of P 2n(2).
(6).If n 0 mod 2, then there is a self homotopy equivalence f of V2n+1,2s*
*uch
that fP2n(2):P 2n(2) ! P 2n(2) is homotopic to [1].
2.4. The Strong Form of the Kervaire Invariant One Problem. Consider the
Whitehead square !n in ß2n1(Sn). That !n is zero precisely when n = 1, 3, 7 is
equivalent to the classical problem of the existence of elements of Hopf invari*
*ant one
which was solved completely in [1]. Restrict attention to odd integers n which*
* are
not equal to 1, 3 or 7. In these cases !n is nonzero and is of order 2. There *
*is a short
exact sequence of groups localized at 2
0 ____Z=2 ____ß2n1(Sn) ____ßs2n1(Sn) ____0.
Hence !n is divisible by 2 in ß2n1(Sn) if and only if this sequence fails to s*
*plit. It is
well known that !n is not divisible by 2 if n 6= 2k1 for some k. A proof can b*
*e found
in [5]. Of the remaining cases given by n = 2k  1, it is known that !n is divi*
*sible
by 2 if n = 1, 3, 7, 15, 31 or 63, see [3, 20, 34]. The cases for which n = 2k *
* 1 > 63
remain open.
Note. For even integers n, it is a routine exercise to show that !n is not div*
*isible
by 2 if n 6= 2, 4 or 8. According to [42], !2 is divisible by 2 but !4 and !8 a*
*re not.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 17
The Strong form of the Kervaire invariant one conjecture is that !n is divisi*
*ble
by 2 when n = 2k  1. There are many reformations of this conjecture, see for
instance [5, 8, 34]. Several reformulations in terms of mod 2 Moore spaces and *
*the
Stiefel manifolds are discussed below.
2.4.1. Spherical Classes. Let X be a space. An element a 2 H*(X) is called sphe*
*rical
if there is an element b in ß*(X) with Hurewicz image in mod 2 homology given by
a. Recall that H*( P n+1(2)) = T (u, v), where v = n and Sq1*v = u.
Proposition 2.21. The Whitehead square !2n1 is divisible by 2 in ß4n3S2n1 if
and only if u2 is spherical in H*( P 2n+1(2)).
Proof.First suppose that !2n1 : S4n3 ! S2n1 is divisible by 2. Let f :S4n3 !
S2n1 be any map and let f0: S4n4 ! S2n1 be the adjoint map of f. Then
f0 2n1 H2 4n3
the composite S4n4 ____ S ____ S is null homotopic and so, by Theo
rem 2.12, there is a homotopy commutative diagram
[2] 2n1
S2n1 ______S
6 6
 
f f
 
 [2] 
S4n3 ______S4n3.
!2n1 2n1 2n
Thus the composite S4n3 _____S ____P (2) is null homotopic. Consider the
homotopy commutative diagram
!2n1 2n1
S4n3 ______________S
\ \
 
 
 
 
 
? W ?
P 2n1(2) ^ P 2n1(2)___2P 2n(2),
where W2 : X ^X ! X is Whitehead product. It follows that there is a homotopy
commutative diagram of cofibre sequences
S4n3 ______________*_________S4n2============ S4n2
\   
   
   
   
   
   
? W ? ? pinch ?
P 2n1(2) ^ P 2n1(2)___2P 2n(2)__J2(P 2n(2))_____P 2n(2) ^ P 2n(2)
and so u2 is spherical.
Conversely suppose that u2 is spherical in H*( P 2n+1(2)). We may assume that*
* n > 1.
Since J2(P 2n(2)) is the (6n4)skeleton of P 2n+1(2) ' J(P 2n(2)), the elemen*
*t u2 is
spherical in H*J2(P 2n(2)). Thus there is a homotopy commutative diagram of fib*
*re
18 J. WU
sequences
q 2n 2n
(P 2n(2) ^ P 2n(2))____Fq______J2(P 2n(2))__P (2) ^ P (2)
6 6 6 6
   
   i
   
   
   [
S4n2 ____________* _________S4n2========= S4n2
where i is the canonical inclusion and p is the pinch map. By inspecting the Se*
*rre
spectral sequence for the fibre sequence
F ____J2(P 2n(2)) ____P 2n(2) ^ P 2n(2),
the space P 2n(2) is the (6n  5)skeleton of F . Thus the composite
S4n3 ____ P 2n1(2) ^ P 2n1(2) ____P 2n(2)
is null homotopic or the composite
!2n1 2n1 2n
S4n3 _____S ____P (2)
is null homotopic. It follows that !2n1 : S4n3 ! S2n1 lifts to the homotopy *
*fibre
E of the injection S2n1 ! P 2n(2). Consider the fibre sequence
j 2n f 2n1 j 2n
S2n1 ____ P (2) ____E ____S ____P (2).
Since j* : H*( S2n1) ! H*( P 2n(2)) is a monomorphism, the Serre spectral
sequence for the fibre sequence S2n1 ! P 2n(2) ! E collapses. By [12, Lemma
2.1], the class [u, v] is spherical in H* P 2n(2) and so is in H*(E). Hence the*
* (4n3)
skeleton sk4n3(E) of E is homotopy equivalent to S2n1 _ S4n3 and fS4n3is n*
*ull.
Now let ` : S4n3 ____E be a lifting of !2n1. Let `1 and `2 denote the compos*
*ites
` proj.2n1
S4n3 ____ sk4n3(E) ____S ____E and
` proj.4n3
S4n3 ____ sk4n3(E) ____S ____E,
respectively. Because fS2n1is of degree 2, we have
!2n1 ' f O ` ' f O `1 ' [2] O `1.
The assertion follows from the fact that
[2]* = 2 : ß4n3(S2n1) ! ß4n3(S2n1).
Recall that H*(Sn{2}) ~=P (u) E(v) as P (u)module with v = n and Sq1*v =*
* u.
Corollary 2.22. The Whitehead square !2n1 is divisible by 2 if and only if u2 *
*is
spherical in H*(S2n{2}).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 19
Proof.The homotopy commutative diagram
pinch 2n+1
S2n+1 ____F 2n+1{2}__P 2n+1(2)______S
6 6
 
 
 
 
[ [2] 
S2n ________S2n
induces a canonical inclusion j : S2n{2} ! P 2n+1(2) so that
j* : Hr(S2n{2}) ____Hr( P 2n+1(2))
is one to one for all r and onto for r 4n  2 and hence the result.
2.4.2. Exponents of Homotopy Groups.
Proposition 2.23. Let n 2. Suppose that u2 is spherical in H4n2(S2n{2}).
Let x 2 ß4n2(S2n{2}) be any element with Hurewicz image in mod 2 homology given
by u2. Then x generates a Z=8summand in ß4n2(S2n{2}).
Proof.Since u2 is spherical, we have
sk4n2(S2n{2}) ' P 2n(2) _ S4n2
and so there is a cofibre sequence
ffi 2n 4n2 j 2n 4n1
S4n2 ____P (2) _ S ____ sk4n1(S {2}) ____S .
Let OE1 and OE2 denote the composites
ffi 2n 4n2 proj. 2n 4n2 ffi 2n 4n2 proj.4n2
S4n2 ____P (2)_S ____P (2) and S ____P (2)_S ____S ,
respectively. Since ~H*(sk4n1(S2n{2})) has a basis u, v, u2 and uv with Sq1*v *
*= u and
Sq1*(uv) = u2, the map OE2 : S4n2 ! S4n2 is of degree 2. Consider the homotopy
commutative diagram of cofibre sequences
OE 2n 4n2 2n 4n1
S4n2w____P (2) _ S ___sk4n1(S {2})___S w
ww   w
ww   ww
ww   ww
ww  q ww
w   ww
OE1 2n? f 2n? 4n1
S4n2 ________P (2) ___________X _________S ,
where X2n = Cffi1. Since q*: H*(sk4n1(S2n{2})) ! H*(X2n) is onto, we have
H*(X2n) ~=E(~u) E(~v)
as coalgebras, where ~x= q*(x). By inspecting the EilenbergMoore spectral sequ*
*ence
for X2n ! * ! X2n, we have
H*( X2n) ~=P (s1~u, s1~v)
the polynomial on s1~uand s1~v. Thus
f* : H*( P 2n(2)) ~=T (s1~u, s1~v) ! H*( X2n) ~=P (s1~u, s1~v)
20 J. WU
is onto and so there is a short exact sequence of Hopf algebras
H*( Ff) ____H*( P 2n(2)) ____H*( X2n).
Observe that OE1 : S4n2 ! P 2n(2)lifts to Ff with OE1* : H4n2(S4n2) ~= H4n2*
*(Ff)
and H~r(Ff) = 0 for r < 4n  2. Let OE01: S4n3 ! P 2n(2) the adjoint map of O*
*E1.
Then OE01*('4n3) = [s1~u, s1~v] which is the only primitive element in H4n3*
*( P 2n(2)).
Since 2n should be a power of 2, the homotopy class [OE1] of OE1 is of order 4 *
*in
ß4n2(P 2n(2)) by [12, Theorem 2.2]. Now, by the cofibre sequence
ffi 2n 4n2 j 2n 4n1
S4n2 ____P (2) _ S ____ sk4n1(S {2}) ____S ,
there is a short exact sequence
ffi* 2n j* 2n
0 ____Z = ß4n2(S4n2) ____ß4n2(P (2)) Z ____ß4n2(S {2}) ____0
with OE*('4n2) = [OE1] + 2['4n2]. Thus
8x = 8j*(['4n2]) = 4j*(2['4n2]) = 4j*([OE1]) = 0
and
4x = 4j*(['4n2]) = j*(2[OE1]) 6= 0
since 2[OE1] 62 Im(OE*). Observe that x is not divisible by 2. Thus x generates*
* a Z=8
summand and hence the result.
Proposition 2.24. If !2n1is not divisible by 2, then 4ßr(S2n{2}) = 0 for r 4*
*n2.
Proof.Consider the homotopy commutative diagram of fibre sequences
2[4] 2 4n1 4n1 4n1 [4] 4n1
2S4n1 _____ S ___ (S {4}) ____ S ____ S
6 6 6 6 6
    
  ~H2 H2 H2
    
    
 2[2]  f   [2] 
2S2n _______ 2S2n _____ (S2n{2}) _____ S2n ______ S2n
6 6 6 6 6
    
  ~E E E
    
    
 [2]    [2] 
S2n1 ______ S2n1 _____S2n1{2} ______S2n1______S2n1.
The map
~E*:ßr(S2n1{2}) ____ßr( (S2n{2}))
is onto for r 4n  4 and there is an exact sequence
ß4n3(S2n1{2}) ____ß4n3( (S2n{2})) ____ß4n3( (S4n1{4})).
Since u2 is not spherical, ß4n3( (S2n{2})) ! ß4n3( (S4n1{4})) = Z=4 is not an
epimorphism. Recall that
[ H2 O !02n] = 2['4n3]
in ß4n3( 2S4n1), where !02n:S4n3 ! 2S2n is the adjoint map of Whitehead squ*
*are !2n.
Thus
Im (H~2*: ß4n3( (S2n{2})) ____ß4n3( (S4n1{4}))) = Z=2
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 21
which is generated by [H~2O f O !02n]. By Theorem 2.12, we have
2[2] O !02n= 2!02n+ 2!2nO H2 O !02n= 4!02n.
Thus 4 f*(!02n) = 0 and so
ß4n3(S2n1{2}) Z=4 ! ß4n3( (S2n{2}))
is onto. According to [12], 4ßr(S2n1{2}) = 0 for r 4n3 and hence the result.
Corollary 2.25. Let n 2. Then the Whitehead square !2n1 is not divisible by 2
if and only if 4ßr(S2n{2}) = 0 for r 4n  2.
Proposition 2.26. Let n > 1. Suppose that u2 is spherical in H*( P 2n+1(2)). Let
x be any element in ß4n2( P 2n+1(2)) with the Hurewicz image in mod 2 homology
given by u2. Then x is of order 4.
Proof.Suppose that 2x = 0. Then there is a map OE : P 4n+1(2) ! P 2n+1(2) such
that OES4n2is a representative of x in ß4n2( P 2n+1(2)). Let {U, V } be a ba*
*sis for
~H*(P 4n1(2)) with Sq1*V = U. Since OE*(U) = u2, we have Sq1*OE*(V ) = OE*(U) *
*= u2
and so OE*(V ) = uv or vu. This is a contradiction because uv and vu are not pr*
*imitive.
Thus 2x 6= 0.
Now we show that 4ß4n2( P 2n+1(2)) = 0 from which we obtain 4x = 0. In the
Cohen group K2(EZ=4), we have
(x1x2)4 = [x1, x2]2
and so
4y = 2( W2 O H2)*(y)
for any y 2 ßr P 2n+1(2) with r 6n  4, where W2 : (P 2n(2))(2)! P 2n+1(2) i*
*s the
Whitehead product. Since 2 . ß4n2( (P 2n(2))(2)) = 0, we have 4y = 0 and hence
the result.
2.4.3. Partial Hspaces. A map f :X ! Y is called a partial Hspace if there i*
*s a
map ~f:J2(X) ! Y such that ~fX = f, or equivalently there is a map ~: X xX ! Y
such that ~Xx* = ~*xX = f. The identity map f = idX: X ! X is a partial H
space if and only if X is an Hspace. If X is an Hspace, then any map from X !*
* Y
is a partial Hspace.
It is well known that S2n1 is an Hspace if and only if !2n1 is null homoto*
*pic
(and so if and only if 2n  1 = 1, 3 or 7). Consider the spherical fibration
i 2n
S2n1 ____V2n+1,2____S .
One may ask whether the inclusion i: S2n1 ! V2n+1,2is a partial Hspace. This
question is equivalent to the strong form of the Kervaire invariant one problem.
Proposition 2.27. The following are equivalent.
(1).The Whitehead square !2n1 is divisible by 2.
(2).the inclusion i: S2n1 ! P 2n(2) is a partial Hspace.
(3).the inclusion i: S2n1 ! V2n+1,2is a partial Hspace.
22 J. WU
Proof.(2) ) (3) is obvious. (1) ) (2). Since !2n1 is divisible by 2, the compo*
*site
!2n1 2n1 i 2n
S4n3 _____S ____P (2)
is null homotopic and so the inclusion i: S2n1 ! P 2n(2) extends to J2(S2n1) *
*or i
is a partial Hspace.
(3) ) (1). Since the inclusion i: S2n1 ! V2n+1,2extends to J2(S2n1), the comp*
*osite
!2n1 2n1 i
S4n3 _____S ____V2n+1,2
is null homotopic and so there is a homotopy commutative diagram
! 2n1
S4n3 ______Sw
 ww
 w
 ww
g ww
 ww
? @ i
S2n ______S2n1_____wV2n+1,2_____ S2n
6 www
 ww
E ww
 ww
 [2]
S2n1 ______S2n1,
where the middle row is the fibre sequence. Recall that E*: ßr(S2n1) ! ßr( S2n*
*) is
onto for r 4n  3. Thus there is a map ~g:S4n3 ! S2n1 such that !2n1 ' [2]*
* O ~g
and hence the result.
The question whether the inclusion i: S2n1 ! V2n+1,2is a partial Hspace adm*
*its
some geometric means, namely whether there is a map
(ffi0,`0,ffi1,`1)2n 2n
OE: S2n1 x S2n1 _________R x R x R x R
such that
1) OE0(x, *) = OE0(*, x) = x;
2) OE1(x, *) = OE1(*, x) = 0
3) `0(x, *) = `0(*, x) = 0;
4) `1(x, *) = `1(*, x) = 1;
5) (OE0, `0) and (OE1, `1) are nowhere zero;
6) (OE1, `1) is orthogonal to (OE0, `0).
The map OE0 looks like a multiplication on R2n with possible zero divisors, whe*
*re 2n is
assumed to be a power of 2. The CayleyDickson multiplication might be a candid*
*ate
for the map OE0. Condition (5) requires that `0(x, y) 6= 0 whenever OE0(x, y) *
*= 0
for x, y 2 S2n1. The map (OE1, `1) looks like certain `derivative' of (OE0, `0*
*).
2.4.4. Spherical Fibrations.
Proposition 2.28. Suppose that the Whitehead square !2n1 is divisible by 2. Th*
*en
there is a spherical fibration
S2n1 ____E4n1 ____P 2n(2)
such that the boundary map @ : P 2n(2) ! S2n1 is onto in mod 2 homology.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 23
Proof.When 2n  1 = 1, 3 or 7, the space E4n1 can be chosen as the homotopy
q 2n pinch 2n
pull back of the diagram S4n1 _____S oe__ P (2) and so we may assume that
n 6= 1, 3, 7.
Let q :P 2n1(2) ! S2n1 be the pinch map. Consider the homotopy commutative
diagram
pinch 2n1 2n2 4n5
S4n5 ____P 2n1(2) ^ P 2n2(2)_____P_(2) ^ P (2)=S
  
  
  
!2n2 W2 W~2
  
? ? q ?
S2n2 _________P 2n1(2)___________________S2n1
\ \ \
  
  
  
  
  
? ? ?
J2(S2n2)______J2(P 2n1(2))_________J2(P 2n1(2))=J2(S2n2),
where W~2is the composite
pinch 4n3 pinch 4n3 !2n1 2n1
P 2n1(2) ^ P 2n2(2)=S4n5 ____P (2) ____S _____S .
Since !2n1is divisible by 2, the map ~W2is null homotopic and so there is a re*
*traction
r :J2(P 2n1(2))=J2(S2n2) ! S2n1. Let ~q:J2(P 2n1(2)) ! S2n1 be the composi*
*te
pinch 2n1 2n2 r 2n1
J2(P 2n1(2)) ____J2(P (2))=J2(S ) ____S .
Then ~qP2n1(2)= q. Let {u, v} be a basis for ~H*(P 2n1(2)) with Sq1*v = u. C*
*onsider
the Serre spectral sequence for the fibre sequence
~q 2n1
S2n1 ____F~q____J2(P 2n1(2)) ____S .
Then
Hr(F~q) ! Hr(J2(P 2n1(2)))
is a monomorphism for r 6n  7. A basis for Hr(F~q) for r 6n  7 is given *
*by
{u, u2, [u, v], v2]}. Since ~qJ2(S2n2)is null homotopic. The inclusion
J2(S2n2) ! J2(P 2n1(2))
lifts to F~qand so sk4n4(F~q) ' J2(S2n2) with a cofibre sequence
P 4n3(2) ____J2(S2n2) ____ sk4n2(F~q).
Let s: X(2)! J2(X) be a functorial crosssection of the pinch map J2(X) ! X*
*(2)
and let E4n1 be the homotopy cofibre of the composite
s 2n2
S4n3 ____ J2(S ) ____ sk4n2F~q.
24 J. WU
Then there is a homotopy commutative diagram
S4n3 _____P 2n1^ P 2n(2)
 
 
 
 s
 
 
? ?
sk4n2(F~q)__ J2(P 2n1(2))
 
 
 
OE 
 
? ` ?
E4n1 _________P 2n(2).
We show that the homotopy fibre of the map ` is S2n1. By considering the commu
tative diagram
H*( sk4n2(F~~q))__H*( J2(P 2n1(2)))
 
 
 
 
 
 
? ` ?
H*( E4n1) _________*H*( P 2n(2)),
the elements u, [u, v] and v2 of H*( P 2n(2)) = T (u, v) lie in the image of `**
* and so
the Poincar'e series
Ø(H*( E4n1)) Ø(T (u, [u, v], v2)).
By considering the EilenbergMoore spectral sequence for E4n1 ! * ! E4n1, we
have
Ø(H*( E4n1)) Ø(T (u, [u, v], v2)).
Thus Ø(H*( E4n1)) = Ø(T (u, [u, v], v2)) or H*( E4n1) ~=T (u, [u, v], v2) and*
* there
fore `*: H*( E4n1) ! H*( P 2n(2)) is a monomorphism. Hence the Serre spectral
@
sequence for the fibre sequence E4n1 ____ P 2n(2) ____F` collapses and so
Ø(T (u, v)) 2n1
Ø(H*(F`)) = ________________= 1 + t
Ø(T (u, [u, v], v2))
It follows that F` is homotopy equivalent to S2n1 and hence the result.
Proposition 2.29. Let n 2. Suppose that there is a spherical fibration
S2n1 ____E4n1 ____P 2n(2)
such that the boundary map @ : P 2n(2) ! S2n1 is onto in mod 2 homology. Then
1) the homotopy cofibre of the map E4n1 ! P 2n(2) has a nontrivial cup pr*
*od
uct.
2) The Whitehead square !2n1 is divisible by 2.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 25
Proof.Consider the Serre spectral sequence for S2n1 ! E4n1 ! P 2n(2). The
homology H~*(E4n1) has a basis x2n1, x4n2 and x4n1, where xi = i. Since
@*: H*( P 2n(2)) ! S2n1 is onto, the Serre spectral sequence for
E4n1 ! P 2n(2) ! S2n1
collapses and so H*( E4n1) ~=T (u, [u, v], v2). By expecting the Serre spectr*
*al se
quence for E4n1 ! * ! E4n1, the homology suspension
oe :Q(H*( E4n1)) ! ~H*(E4n1)
is an isomorphism, where Q(A) is the set of indecomposable elements of an algeb*
*ra A.
It follows that Sq1*x4n1 = x4n2.
Let X be the homotopy fibre of E4n1 ! P 2n(2). Then ~H*(X) has a basis y2n, *
*y4n1
and y4n with Sq1*y4n = y4n1. Consider the homotopy commutative diagram
P 2n(2)_____S2n1______E4n1 _____P 2n(2)
   
   
   
 f   f
   
   
? ? ? ?
X ======= X _________* _________X.
Let ø denote the transgression. We have
ø(y2n)2 = f*(v2) = 0.
By expecting the Serre spectral sequence for X ! * ! X, we have
y*22n= y*4n
in the cohomology H*(X) and hence assertion (1). By [5, Theorem 9.1], assertion*
* (2)
follows. This finishes the proof.
2.5. The Exponent Problem.
2.5.1. The Exponent Problems in the Homotopy Theory. Let p be a prime integer
and let G be an abelian group. We write Torp(G) for the ptorsion component of
G. We call pr an exponent of G if pr . Torp(G) = 0. Let X be a space. The integ*
*er
pr is called an exponent of ß*(X) if pr . Torp(ßn(X)) = 0 for all n 2. If X *
*is a
simply connected CW of finite type, then each ßn(X) has a bounded exponent which
depends on n in general. For instance ß*(S2_S2) does not have a bounded exponen*
*t.
On the other hand, one can check that the homotopy groups of the mapping space
from a simply connected finite torsion space to a space has a bounded exponent.
It was first known by James [19] that ß*(S2n+1) has an exponent bounded by 22*
*n.
The improvements given in [5, 35] state that the exponent of ß*(S2n+1) is bound*
*ed
by 22n[n=2], where [a] is the maximal integer a. In the cases where p > 2, T*
*oda [41]
showed that ß*(S2n+1) has an exponent bounded by p2n. Selick [33] then showed t*
*hat
ß*(S3) has an exponent bounded by p for p > 2. Later CohenMooreNeisendorfer [*
*9]
proved that ß*(S2n+1) has an exponent bounded by pn for p > 2. This is the best
exponent because a theorem of Gray [15] gives that there are Z=pnsummands in
ß*(S2n+1) for p > 2. The best 2exponent of ß*(S2n+1) remains open.
The Moore conjecture says that the homotopy groups of a simply connected fini*
*te
complex X has a bounded pexponent if the rational cohomology ring H*( X; Q) is
26 J. WU
finitely generated. For any given simply connected finite complex X, it was kno*
*wn
by McGibbon and Wilkerson [21] that the Moore conjecture holds when the prime p
is sufficiently large. By the arguments above, the Moore conjecture holds for s*
*pheres
(for all primes p) and so for any space which admits a finite fibration resolut*
*ion in
terms of spheres, for instance the classical Lie groups.
The Barratt conjecture states that if the degree map [pr]: X ! X is null
homotopic, then pr+1ß*( X) = 0. It was known by Neisendorfer [30] that the Barr*
*att
conjecture holds for the Moore spaces P n(pr) with p > 2. By [9], there is a fa*
*mily of
Z=pr+1summands in ß*(P n(pr)) and so pr+1 is the best exponent.
In [6], Cohen showed that ß*(P n(2r)) has a bounded exponent when r 2 and
so the Moore conjecture holds for these spaces. It is unknown whether the Barra*
*tt
conjecture holds for P n(2r) with r 2. But there is a family of Z=2r+1summan*
*ds
in ß*(P n(2r)), see [4], and so 2r+1 will be the best exponent if the Barratt c*
*onjecture
holds.
It is still unknown whether ß*(P n(2)) has a bounded exponent. In other word*
*s,
the conjectures of Moore and Barratt both remain open for mod 2 Moore spaces.
In [12], we showed that there is a family of Z=8summands in ß*(P n(2)) and so *
*8 will
be the best exponent for ß*(P n(2)) if the Barratt conjecture holds.
2.5.2. Exponents of ß*(Sn{2}). Consider the homotopy commutative diagram of fib*
*re
sequences
[4] 2n+1
2S2n+1 ___ (S2n+1{4}) ____ S2n+1 ____ S
6 6 6 6
   
 H~2 H2 H2
   
   
   [2] 
2Sn+1 ____ (Sn+1{2}) ____ Sn+1 _____ Sn+1
6 6 6 6
   
   
 E~ E E
   
   
   [2] 
Sn ________Sn{2} _________Sn ________Sn.
We obtain the fibre sequence
~E n+1 H~2 2n+1
Sn{2} ____ (S {2}) ____ (S {4}).
Recall that Sn{2} is an Sn = J(S2n1)space. By the above diagram, the map E~
can be chosen as an Snmap.
Lemma 2.30. Let n 3. Then the map
4 . ~E: Sn{2} ____ (Sn+1{2})
is null homotopic.
Proof.Filter Sn{2} by setting FkSn{2} to be the ((n  1)k + 1)skeleton. Let
~: J(Sn1)xP n(2) ! Sn{2} be the J(Sn1)action on Sn{2}. Then ~k = ~Jk(Sn1)x*
*Pn(2)
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 27
lifts to Fk+1Sn{2}. Recall that
`k `k
JkSn1 x P n(2) ' S(n1)j^ P n(2) _ S(n1)j_ P n(2) and
j=1 j=1
`k
Fk+1Sn{2} ' S(n1)j^ P n(2).
j=0
The map ~k is a retraction for 0 k +1. Let
qk : (Sn1)k x P n(2) ____Jk(Sn1) x P n(2)
denote the projection. Then qk is a retraction and therefore
q*kO ~*k: [Fk+1Sn{2}, Y ] ! [(Sn1)k x P n(2), Y ]
is a monomorphism for any Y . Since ~Eis an Snmap, there is a homotopy commu
tative diagram
(E~Sn1)k x ~EPn(2) n+1 k+1
(Sn1)k x P n(2)______________________( (S {2}))
 
 
 
qk ~x id
 
? E~J (Sn1)x ~EPn(2) ?
JkSn1 x P n(2)__________________k (Sn+1{2}) x (Sn+1{2})
 
 
 
~k ~
 
 
? E~ ?
Fk+1Sn{2} ___________________________ (Sn+1{2}).
Let xi and y denote the homotopy classes of
ii n1 ~ESn1 n+1
(Sn1)k x P n(2) ____S ______ (S {2}) and
ik+1 n ~EPn(2) n+1
(Sn1)k x P n(2) ____P (2) ______ (S {2}),
respectively, where ßi the ith coordinate projection. Then
q*kO ~*k([E~Fk+1Sn{2}]) = x1x2. .x.ky.
Now let Gk be the subgroup of [(Sn1)k x P n(2), (Sn+1{2})] generated by xi,
1 i k, and y. Then the following identities hold in Gk:
1) x2i= 1 for 1 i k;
2) y4 = 1;
3) the commutators [xi, xj] = 1;
4) The (iterated) commutators
[[a1, a2], . .,.as] = 1
if ai= aj for some i < j, where ai= y or xj for some j.
28 J. WU
Identities (1) and (2) are obvious. Identity (4) follows from Theorem 2.15 and*
* (3)
follows from the homotopy commutative diagram
2Sn+1 ____ (S2n+1{2})
6 6
 
 
E E~
 
 
 
Sn ________Sn{2}.
Now, by using identities (1) to (4),
(x1. .x.ky)2 = x1. .x.ky . x1. .x.ky = (x1. .x.k)2 . y2 . ffi = y2 . f*
*fi
______ ______
for some ffi 2 < y_>,_where_< y > is the subgroup of Gk generated by [y, ff] fo*
*r ff 2 Gk.
By identity (4), < y > is abelian. Thus (x1. .x.ky)4 = y4.ffi2 = 1 and so 4.E~*
*Fk+1Sn{2}
is null homotopic for each k. Since
Y1
[Sn{2}, Y ] ~= [P n+(n1)j(2), Y ]
j=0
as sets for any Y , the map 4 . ~Eis null homotopic and hence the result.
An Hspace X is said to have an exponent pr if the power map pr: X ! X is null
homotopic.
Proposition 2.31. Let n 3. Then the space 2(Sn{2}) has an exponent 16.
Proof.By Lemma 2.30, the power map 4 : (Sn{2}) ! (Sn{2}) lifts to the fi
bre 3(S2n+1{4}). Since 2[4] ' 4 : 2S2n+1 ! 2S2n+1, 3(S2n+1{4}) is homotopy
equivalent to the homotopy fibre ( 2S2n+1){4} of the power map 4: 2S2n+1 ! 2S*
*2n+1.
Thus 4(S2n+1{4}) has exponent 4 and so 2(Sn{2}) has exponent 16, which is the
assertion.
Corollary 2.32. Let n 3. Then 16 . ß*(Sn{2}) = 0.
For the case n = 2, we have
Proposition 2.33. There is a homotopy decomposition
2(S2{2}<3>) ' 2(S3<3>) x 30S3.
In particular, 4 . (ß*(S2{2})) = 0.
Proof.By the homotopy commutative diagram
[4] ' 4 3
S3 ________ S
6 6
 
H2 H2
 
 [2] 
S2 ________ S2,
we have 0(S2{2}) ' ( S3){4}. According to [4], 2(S3<3>) has an exponent 4 and
hence the result.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 29
By Proposition 2.23, there are Z=8summands in ß*(Sn{2}) for n = 4, 8, 16, 32*
* or
64. We do not know any element in ß*(Sn{2}) which order is 16 and so we wonder
whether ß*(Sn{2}) has an exponent bounded by 8. This is true when n is odd.
Proposition 2.34. The space 4(S2n+1{2}) has an exponent 8.
Proof.By using the fact that 2 [2] ' 4 : 2S2n+1 ! 2S2n+1, there is a homotopy
commutative diagram of fibre sequences
ffi 3 2n+1 2 3 2n+1
( 3S2n+1){2}w_____ S _______ S
ww  
ww  
ww k 
ww  
g 2 2n+1? f 2 2n+1?
( 3S2n+1){2} ___ (S {2})___( S ){4}
  
  
  
  
  
 ? ?
? 2 2n+1 2 2n+1
* ___________ S ======= S
  
  
  
  2[2] 4
  
  
? ? 2 ?
( 2S2n+1){2} _____ 2S2n+1 _______ 2S2n+1.
Observe that
2 g ' 2 k O ffi ' k O 2 O ffi ' *.
There is a homotopy commutative diagram of fibre sequences
g 3 2n+1
4(S2n+1{2}) ___( 4S2n+1){4}___( 4S2n+1){2} ___ (S {2})
  
   
   
2   2
   
? ? ? ?
4(S2n+1{2})== 4(S2n+1{2}) ________* _________ 3(S2n+1{2})
and so a homotopy commutative diagram
4(S2n+1{2}) ___( 4S2n+1){4}
 
 
 
4 4' *
 
? ?
4(S2n+1{2}) ___( 4S2n+1){4}
 
 
 
2 
 
? ?
4(S2n+1{2})== 4(S2n+1{2})
30 J. WU
Thus 8: 4(S2n+1{2}) ! 4(S2n+1{2}) is null homotopic and hence the result.
Corollary 2.35. The homotopy groups ß*(S2n+1{2}) = 0 have an exponent bounded
by 8.
So far we do not find any element in ß*(S2n+1{2}) which order is 8. Below is*
* a
conjecture due to F. R. Cohen.
Conjecture 2.36 (Cohen). The homotopy groups ß*(S2n+1{2}) have an exponent
bounded by 4.
2.6. Some Lemmas on Fibrations. We refer the reader to [22] for basic properties
of Hopf algebras. A Hopf algebra means an (graded) algebra A together with a
comultiplication _ :A ! A A such that A is a coalgebra under _ and _ is an
algebraic map. Let M be an augmented module with the augmentation ffl: M ! k,
where k is the ground field. We write IM for the kernel of ffl. Let A be a Ho*
*pf
algebra. Let InA denote the nfold self product of IA in A and let Q(A) = IA=I2A
be the set of indecomposable elements.
In the case where A is a tensor algebra (as an algebra), the indecomposable e*
*lements
of subHopf algebras of A can be understood in the following sense.
Lemma 2.37. Let A be a connected Hopf algebra of finite type and let B be a sub*
*Hopf
algebra of A. Suppose that A is a tensor algebra as an algebra. Then there is a*
* short
exact sequence
0 ____Q(B) ____(k B A) Q(A) ____I(k B A) ____0.
Proof.Let C = k B A. Then C is a right Amodule. Let V be a submodule of IA
such that V ~= Q(A) and let J be the left ideal of A generated by IB. Consider *
*the
commutative diagram
J V======= J V
 
 
 
mult 
 
 
? ?
J ______A V ~= IA _____ICw
  w
  w
  ww
  w
  ww
  w
?? ?? w
J=(J . V )_____C V _______IC.
we obtain the short exact sequence
0 ____J=(J . V ) ____C V ___IC.
Now the composite
IB ____J ____J=(J . V ) = J=(J . IA)
is epimorphism and it factors through Q(B) = IB=I2B. Let OE: Q(B) ! J=(J . V )
be the resulting map. Since OE is an epimorphism, the Poincar'e series
Ø(Q(B)) Ø(J=(J . V )) = Ø(C V )  Ø(IC) = Ø(C)Ø(V )  Ø(C) + 1.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 31
Recall that any sub algebra of tensor algebra is free. By [22, Theorem 4.4], th*
*ere is
an isomorphism of Bmodules
A ~=B C.
In particular, we have
____1____ 1
= ____________. Ø(C)
1  Ø(V ) 1  Ø(Q(B))
or Ø(Q(B)) = Ø(C)Ø(V )  Ø(C) + 1. Thus the map OE: Q(B) ! J=(J . V ) is an
isomorphism and hence the result.
This lemma helps to understand the homology of certain spaces.
i q
Lemma 2.38. Let F ____E ____ X be a fibration. Suppose that
1) X is pathconnected;
2) F is simply connected and
3) the boundary H*( X) ! H*(F ) is an epimorphism.
Then
1) There is a short exact sequence
0 ____Q(H*( E)) ____H*(F ) ~H*(X) ____H~*(F ) ____0.
2) The homology suspension
oe :Q(H*( E)) ! ~H*+1(E)
is an isomorphism.
Proof.Since H*( X) ! H*(F ) is an epimorphism, the Serre spectral sequence for
the fibre sequence E ____ X ____F collapses and so assertion (1) follows from
Lemma 2.37.
Consider the cofibre sequence
i `
F ____E ____ X x F=(* x F ).
Since H*( X) ! H*(F ) is onto, the map i*: ~H*(F ) ! ~H*(E) is zero or
`*: ~H*(E) ! ~H*( X x F=(* x F ))
is a monomorphism. It follows that
Ø(s1H~*(E)) = Ø(Q(H*( E))),
where s1 is the desuspension of graded modules. Since
1
Ø(H*( E)) = ________________,
1  Ø(s1H~*(E))
the cobar spectral sequence for E collapses and hence assertion (2). We finish*
* the
proof.
i q
Lemma 2.39. Let F _____E _____ X be a fibration and let f :X ! F be the
@
composite X ____ X ____F. Then there is a cofibre sequence
Cf ____E ____ X ^ F.
32 J. WU
Proof.By [44, Theorem 1.1, chapter VII], there is a cofibre sequence
i
F ____E ____( X x F )=(* x F ).
Since i O f :X ! E is null homotopic, there is a homotopy commutative diagram of
cofibre sequences
Fw_________Cf _____________ X
ww  
ww  
ww  
ww  
w  
i ? ?
F _________E ______( X x F )=(* x F )

  
  
  
  
  
  
? ? ?
* _______ X ^ F ======== X ^ F
and hence the result.
3.Decompositions of Self Smash Products
In this chapter, we investigate splittings of self smash products of the proj*
*ective
plane RP2. An instructional example is the 3fold self smashes (RP2)(3). Firs*
*t we
can decompose this space as wedge of 3 factors
(RP2)(3)' X1 _ X2 _ X3,
where Xiis an indecomposable space. Since there is only one bottom cell in (RP2*
*)(3),
we assume that X1 contains the bottom cell, that is, H3(X1) = Z=2. Then we
show that X2 ' X3 ' P 5(2). By analysis on the cell structure, we show that
X1 ' CP2 ^ RP2. In other words, the atomic space X1 admits a smash product
decomposition. This gives an interesting relation between different projective *
*planes.
We will largely investigate this kind of decompositions in this chapter.
There is a connection between decompositions of self smashes of RP2 and the
modular representation theory. The homology of the (functorial) indecomposable
factors of self smashes of a twocell suspension has been determined recently i*
*n [38].
We will go over these results and connections in the first section. In the sec*
*ond
section, we focus on decompositions of self smash products of RP2. Since RP2 is*
* not
a suspension, we first desuspension the decomposition formulas. Then we study t*
*he
smash product decompositions of indecomposable factors.
3.1. Functorial Decompositions of Self Smashes of Twocell Complexes.
Let X be a pathconnected plocal CW complex. We start with some observations.
Consider the nfold self smash product X(n). Let the symmetric group Sn act on *
*X(n)
by permuting positions. Thus, for each oeX2 Sn, we have a map oe :X(n)! X(n). L*
*et
Z(p)denote plocal integers and let ff = kffoebe an element in the group al*
*gebra
ff2Sn
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 33
Z(p)(Sn). By using the group structure in [ X(n), X(n)], we obtain a map
X
ff = kffoe : X(n)! X(n).
ff2Sn
(Note. [ X(n), X(n)] is not abelian in general and the product above is given *
*by a
fixed choice of order.) The map ff has the following properties:
1) The map ff is functorial with respect to X.
2) Let S*(Y ) be the singular chain complex of Y with coefficients in Z(p).*
* Then
the induced chain map
ff*: S*( X(n)) ' S*(X) n ! S*( X(n)) ' S*(X) n
X
is homotopic to the linear combination kff oe, where Sn acts on S*(X*
*) n
ff2Sn
by permuting factors in graded sense. In other words, we have the commu
tative diagram
`1 n n
Z(p)(Sn)___[ S*(X) , S*(X) ]
ww 
ww 6
ww 
ww 
w 
`2 (n) (n)
Z(p)(Sn)_______[ X , X ],
where `1 is a map of rings and `2 is only a function.
For each ff 2 Z(p)(Sn), let ff(X) = hocolimff X(n)be the homotopy colimit of the
sequence
ff (n) ff (n) ff
X(n)____ X ____ X ____ . ...
Now let
X
(1) 1 = eff
ff
be an orthogonal decomposition of the identity in Z(p)(Sn) in terms of primitive
idempotents. For each eff, we have a eff(X) = hocolimeff X(n). The composite
comult` (n) `
(2) X(n)_____ X ____ eff(X)
ff ff
is a homotopy equivalence because its induced map on the singular chains over p*
*local
integers is a homotopy equivalence. Furthermore, this decomposition is functor*
*ial
with respect to X. In other words, each effis a functor from pathconnected spa*
*ces
to coHspaces. (Note: for each ff, we can fix a choice of the representative m*
*ap eff.
This gives a particular choice of space eff(X) such that eff(X) is strictly fun*
*ctorial
with respect to Y . Furthermore, there is a choice of the map
`
OE: X(n)! eff(X)
ff
such that OE is strictly functorial on X and is a homotopy equivalence for each*
* X.)
34 J. WU
Recall that primitive idempotents, up to conjugates, in Z(p)(Sn) are onetoo*
*ne
corresponds to pregular partitions of n. A (proper) partition of n is a seque*
*nce of
positive integers ~ = (~1, . .,.~s) such that
Xs
~1 ~2 . . .~s and = n.
j=1
s is called the length of ~, which is denoted by len(~). A partition ~ = (~1, .*
* .,.~s)
is called pregular if there is NO subscript i with 1 i s such that ~i = ~i*
*+1 =
. .=.~i+p1. For each pregular partition ~, there is an idempotent Q~ 2 Z(p)(S*
*n)
such that the (left) ideal generated by Q~ is a projective cover of the Specht *
*mod
ule corresponding to ~ (See [18] for details). Any primitive idempotent in Z(p)*
*(Sn)
is conjugate to one and only one Q~. Let d~ be the number of idempotents in de
composition 1 which are conjugate to Q~. We called d~ the multiplicity of primi*
*tive
idempotent Q~ in Z(p)(Sn). Let Q~(X) be the homotopy colimit
Q~(X) = hocolime~ X(n).
Then decomposition 2 can be rewritten as
` d~`
(3) X(n)' Q~(X),
~
where ~ runs over all pregular partitions of n.
Let H*(X) denote the mod p homology of X.
Theorem 3.1. [38, Theorem 1.1] Let X = Y be a plocal suspension. Suppose
that p = 2 or H~*(X)odd = 0. Then Q~(X) is contractible if and only if len(~) >
dimH~*(X).
Note: If p > 2 and H~*(X)odd 6= 0, then Q~(X) may not be contractible even if
len(~) > dim ~H*(X). For instance, let X be an odd sphere, then Q(1,1)(X) ' X(*
*2).
Let X be a pathconnected twocell complex localized at 2. By Theorem 3.1, we
have
` c(a,na)`
(4) X(n)' Q(a,na)(X).
n=2>2 ifm > 0 and m is even,
>>
><
(5) ffl(m) = 1 if m is odd,
>>
>>
>:
0 if m = 0.
Let I = (i1, . .,.is) be a sequence of nonnegative integers. Let
ffl(I) = (ffl(i1), . .,.ffl(is)).
36 J. WU
Theorem 3.3. [38, Theorem 1.3] For each n=2 < a n, the multiplicity d(a,na)*
*of
Q(a,na)is given by the formula
X
(6) d(a,na)= kI,
I = (i0, . .,.is), i0 = n
ffl(I) = !(a)
0 il il1ffl(il1)_2
il1> ffl(il1) for0 l s
0 is ffl(is)
where
` i0ffl(i0)'` ' ` '
______ i1ffl(i1)_ is1ffl(is1)_(n(ffl(i )+i +ffl(i )+...+i *
* +ffl(i )+i ))=2
kI = 2 . 2 . . . 2 2 0 1 1 s*
*1. s1 s
i1 i2 is
3.2. Decompositions of Self Smashes of RP2. When X is a projective plane, the
decompositions in Theorem 3.2 are desuspensionable.
Proposition 3.4. Let F = R, C, H or K and let X = FP2. Then there is a sequence
of spaces FQk for k 1 such that
[n1_2]`ci`
1) X(n)' FQn _ (u+v)iFQn2i.
i=1j=1
2) FQn ' Q(n)(X).
3) FQ2n ' FQ2n1^ FP2.
Proof.Suppose that there is a sequence of space FQn such that assertions (1) an*
*d (2)
hold. Then the composite
FQ2n1^ FP2 ____(FP2)(2n)____FQ2n
is a homotopy equivalence by checking the homology. Now we show that there exis*
*ts
a sequence of spaces FQn such that assertions (1) and (2) hold by induction on *
*n.
Let FQ1 = FP2. Suppose that the spaces FQk defined for k < n with n > 1 such
that assertions (1) and (2) holds for all k < n. Let d = dim F over R. Then Q(n*
*)(X)
is (dn)connected (2dn + 1)dimensional complex and so there is a space Y such *
*that
Y ' Q(n)(X). Since X(n)is (dn  1)connected (2dn)dimensional complex, there
is a decomposition
sk2dn1(X(n)) ' sk2dn1(Y ) _ B,
where
[n1_2]`ci`
B = (u+v)iFQn2i(X).
i=1j=1
Consider the cofibre sequence
@ (n)
S2dn1 ____ sk2dn1(Y ) _ B ____X .
Since X(n)' Y _ B and B is (dn)connected, the composite
@ proj.
S2dn1 ____ sk2dn1(Y ) _ B ____B
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 37
is null homotopic. Observe that sk2dn1(Y ) is the (2dn  1)skeleton of the ho*
*motopy
fibre of the projection map sk2dn1(Y ) _ B ! B. There is a homotopy commutative
diagram of cofibre sequences
S2dn1w______sk2dn1(Y )______FQn
ww \ 
ww  
ww  
ww  
w  ?
@ ? (n)
S2dn1 ____sk2dn1(Y ) _ B____X .
By expecting the homology, the map FQn _ B ! X(n)is a homotopy equivalence,
which is assertion (1). By assertion 3 of Theorem 3.2, FQn ' Q(n)(X). The
induction is finished and hence the result.
Now we investigate the atomic spaces RQn for low n. By expecting the homology,
RQ1 = RP2 and RQ2 = RP2^ FP2. The first proper factor is RQ3. By expecting the
homology, this is a four cell complex. This factor of X(3)can obtained in an ex*
*plicit
way after suspension.
Lemma 3.5. Let X be a 2local pathconnected space and let oe = (123) be the cy*
*clic
element in S3 of order 3. Then
Q(3)(X) = hocolim1_3(ff+ff2+1) X(3).
1 2
The proof follows immediately by observing that __(oe + oe +i1)s a primitive*
* idem
3
potent in Z=2(S3) which acts on x x x trivially.
Proposition 3.6. The space RQ3 is homotopy equivalent to RP2 ^ CP2.
Proof.By assertion 1 of Theorem 3.2, H~*(RQ3) has a basis x3, x4, x5 and x6 with
the nontrivial Steenrod operations Sq1*x6 = x5, Sq2*x6 = x4, Sq3*x6 = x3, Sq2*x*
*5 = x3
and Sq1*x4 = x3. Clearly sk4(RQ3) ' RP43= P 4(2). Since ß4(P 4(2)) = Z=2, we ha*
*ve
sk5(RQ3) ' RP53. Now we compute ß5(RP53). Since P 4(2) is the 6skeleton of t*
*he
homotopy fibre of the pinch map q :RP53! S5, there is an exact sequence
ß5(S4) = Z=2 ____ß5(P 4(2)) = Z=4 ____ß5(RP53) ____ß5(S5) ____ß4(P 4(2)) __*
*__0
and so a short exact sequence
q* 5
0 ____ß5(P 4)=2 ____ß5(RP53) ____2 . ß5(S ).
Let ff 2 ß5(RP53) such that q*(ff) = 2'5 and let ffi be the nontrivial element *
*of
Im(ß5(P 4(2)) ! ß5(RP53)) = Z=2. Let fi be any element in ß5(RP53) such that
q*(fi) = 2 . '5. Then fi is one of the four elements { ff, (ff + ffi)}. It *
*follows
that there are exactly two different, up to homotopy, 4cell complexes Y such *
*that
sk5(Y ) ' RP53and Sq1*:H6(Y ) ! H5(Y ) is an isomorphism. Observe that RP63and
RP2 ^ CP2 are the complexes which satisfy these conditions with different Steen*
*rod
operations. By expecting the Steenrod operations, RQ3 is homotopy equivalent to
RP2 ^ CP2 and hence the result.
Corollary 3.7. There is homotopy decomposition
(RP2)(3)' RP2 ^ CP2 _ P 5(2) _ P 5(2).
38 J. WU
By using this corollary, we can directly work out decompositions of (RP)(n)fo*
*r n < 6.
Example 3.8. . When n = 4, then
(RP2)(4)' (RP2^CP2_P 5(2)_P 5(2))^RP2 ' (RP2)(2)_P 5(2)^RP2_P 5(2)^RP2
is a complete decomposition by considering the homology and so RQ4 ' (RP2)(2)^ *
*CP2 .
When n = 5, then
(RP2)(5)' (RP2)(3)^ CP2 _ 3(RP2)(3)_ 3(RP2)(3)
`4 `4
' RP2 ^ (CP2)(2)_ P 5(2) ^ CP2 _ P 8(2).
j=1 j=1
Thus RQ5 ' RP2 ^ CP2 ^ CP2.
The next example is RQ7. Observe that RQ7 is a retract of RQ6 ^ RP2 and so,
by Corollary 3.7, RQ7 is a retract of RP2 ^ (CP2)(3). The space (CP2)(3)has fur*
*ther
decomposition by Proposition 3.4.
Let f, g 2 ßn(Y ). We write f ~ g if there is a homotopy commutative diagram
f
Sn _________Y
 
 
 
OE1 OE2
 
 
? g ?
Sn _________Y
for some self homotopy equivalences OE1 and OE2. Clearly, if f ~ g, then the ho*
*motopy
cofibre Cf ' Cg.
Lemma 3.9. Let Z = S4 [ e8 be a complex such that Sq4: H4(Z) ! H8(Z) is an
isomorphism. Then Z is homotopy equivalent to HP2 localized at 2.
Proof.Let f :S7 ! S4 be the attaching map for Z. Recall that ß7(S4) ~=Z Z=4 with
generators 4 and E( 0), see [42]. Since Sq4: H4(Z) ! H8(Z) is an isomorphism, *
*we
may assume that f = 4 + aE( 0), where a = 0, 1 or 2. Now we apply the Cohen
group to show that 4 + aE( 0) ~ 4 for each a. The result will follow from th*
*is
statement.
Observe that [k]: S4 ! S4 restricted to J2(S3) is represented by the ele
ment xk1xk2in the Cohen group for the inclusion E :S3 ___J(S3). Let ffk denote*
* the
element xk1xk2(x1x2)k. Then ffk 2 2K2(E). By Example 2.11, we have 3K2(E) = 1
and so
ffk+1 = xk+11xk+12(x1x2)k1 = x1 . xk1. x2 . xk2. (x1x2)k . (x1x2)1
= x1[xk1, x2]x2(xk1.xk2.(x1x2)k).(x1x2)1 = [x1, x2]kffk.x1.x2.(x1x2)1 = [x1,*
* x2]kffk.
Since ff1 = 1, we have
ffk = [x1, x2](k1)+(k2)+...+1= [x1, x2]k(k1)=2.
Note that [x1, x2] is represented by the composite
pinch 6 !4 3 4
J2(S3) ____S ____J(S ) ' S .
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 39
Thus [x1, x2]*( 4) = !4 = 2 4  E( 0) and so
k(k  1) 0 2 k(k  *
*1) 0
([k])*( 4) = k 4+[x1, x2]k(k1)=2*( 4) = k 4+________(2 4E( )) = k 4______*
*__E( ).
2 2
By taking k = 3, we have
1 0 0
4 ~ 9 4  3E( 0) ~ 4  __E( ) = 4 + E( )
3
and
4  E( 0) ~ 9 4  3E( 0)  3E( 0) ~ 4 + 2E( 0).
By taking k = 5, we have
10 0 0
4 ~ 25 4  10E( 0) ~ 4  ___E( ) = 4 + 2E( )
25
and hence the result.
A homology class x 2 Hn(Y ; Z=2) is called (mod 2) cospherical if there is a
map f :Y ! Sn such that f*(x) = 'n 2 Hn(Sn; Z=2)
Lemma 3.10. There is a homotopy decomposition
(CP2)(3)' HP2 ^ CP2 _ 6CP2 _ 6CP2
localized at 2 and so CQ3 ' HP2 ^ CP2.
Proof.Observe that the homology class [u, v] = u vv u is spherical in ~H*(CP2 *
*^ CP2).
There is a cofibre sequence
f 2 2
S6 ____CP ^ CP ____Cf,
where f*('6) = [u, v]. Now ~H*(Cf) has a basis {x4, x6, x8} with the nontrivial*
* Steen
rod operations Sq2*x6 = x4 and Sq4*x8 = x4. Since Sq2*x8 = 0, the homology clas*
*s x6
is cospherical. Let g :Cf ! S6 such that g*(x6) = '6. Let Z be the 8skeleton o*
*f the
homotopy fibre of g :Cf ! S6. By expecting the Serre spectral sequence for the *
*fibre
sequence Fg ! Cf ! S6, the map Z ! Cf induces a monomorphism in homology
and so there is a cofibre sequence
g 6
Z ____Cf ____S .
Since Z is a twocell complex with nontrivial Sq4, we have Z ' HP2 by Lemma 3.9.
By Proposition 3.4, we have
(CP2)(3)' S6 ^ CP2 _ Cf ^ CP2 ' S6 ^ CP2 _ S6 ^ CP6 _ HP2 ^ CP2
and hence the result.
Corollary 3.11. There is a homotopy decomposition
`6 14` `8
(RP2)(7)' HP2 ^ CP2 ^ RP2 _ 3(CP2)(2)^ RP2 _ 6CP2 ^ RP2 _ P 11(2).
j=1 j=1 j=1
and so RQ7 ' HP2 ^ CP2 ^ RP2.
In general, we have the following theorem for RQn with n 7.
Theorem 3.12. Let k 1. Then
40 J. WU
1) RQ8k1' HQ2k1^ CP2 ^ RP2;
2) RQ8k ' HQ2k1^ CP2 ^ (RP2)(2);
3) RQ8k+1' HQ2k1^ (CP2)(2)^ RP2;
4) RQ8k+2' HQ2k1^ (CP2)(2)^ (RP2)(2);
5) RQ8k+3' HQ2k1^ HP2 ^ CP2 ^ RP2;
6) RQ8k+4' HQ2k1^ HP2 ^ CP2 ^ (RP2)(2);
7) RQ8k+5' HQ2k1^ HP2 ^ (CP2)(2)^ RP2;
8) RQ8k+6' HQ2k1^ HP2 ^ (CP2)(2)^ (RP2)(2);
The proof follows by induction and by using Theorem 3.2, Propositions 3.4 and*
* 3.6
and Lemma 3.9.
Example 3.13. The spaces RQ2n+1 for 9 n 23 admit the following smash
product decompositions:
RQ9 ' HP2 ^ (CP2)(2)^ RP2; RQ11' (HP2)(2)^ CP2 ^ RP2;
RQ13' (HP2)(2)^ (CP2)(2)^ RP2; RQ15' HQ3 ^ CP2 ^ RP2;
RQ17' HQ3 ^ (CP2)(2)^ RP2; RQ19' HQ3 ^ HP2 ^ CP2 ^ RP2;
RQ21' HQ3 ^ HP2 ^ (CP2)(2)^ RP2; RQ23' HQ5 ^ CP2 ^ RP2.
One may wonder whether we can get the geometric analogue of the algebraic
theorem 3.2, namely, whether HQn admits further smash product decompositions.
The first case is HQ3 which appears in RQ15. The space HQ3 has the same homology
as KP2 ^ HP2. However, by analysis on the cell structure, HQ3 is not homotopy
equivalent to KP2 ^ HP2. Below we are going to determine the cell structure of
Q(3,0)(X) for a general twocell complex in the stable category. In particular*
*, this
will give the stable cell structure of HQ3.
For convenience, we write ~Qn(X) for 1Q(n)(X) in the stable category. Thus *
*~Qn(X)
is the functorially atomic factor of X(n)which contains the bottom cell. Let {u*
*, v}
be a basis for ~H*(X). We assume that u = 0 and v > 1. (So mod 2 Moore spac*
*e is
excluded.) As a stable complex, X = S0 [f ev, where f runs over all elements *
*in the
2torsion of the stable homotopy groups ßs*(S0). Observe that the integral homo*
*logy
of X is torsion free in these cases. We write ab for a b in a tensor product *
*V W
and write snx for the nfold suspension of an element x in a graded module.
Lemma 3.14. Let X = S0[fevbe a 2cell complex with v > 1. Suppose that v *
* 0
mod 2. Then there exists a map OEn: (n1)vX ! X(n)= sk(n2)v(X(n))Xsuch th*
*at in
the integral homology OEn*(s(n1)vv) = vn and OEn*(s(n1)vu) = viuvj.
i+j=n1
Proof.The proof is given by induction on n. The statement is obvious for n = 1.
Now we show that the statement holds for n = 2. Consider the cofibre sequence
@ (2) (2)pinch 2v @ (2)
1S2v____ skv(X ) ____X ____S ____ skv(X ).
Let g1, g2: Sv! skv(X(2))=S0 be the maps which sends 'vto uv and vu, resp*
*ec
tively. By using the cofibre sequence
 idX^( f) 1
X ^ S0 ____X(2)____X ^ Sv_________X ^ S ,
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 41
we obtain the commutative diagram
@ (2) 0 (2) 0
1S2v_______________skv(Xw )=S ___X =S
ww 
ww  
ww  
ww p1 
w ? ?
1( f ^ idSv) 0 v v
1S2v_________________S ^ S _____X ^ S .
By computing the homology, we have
(p1 O g1)*('v) = 'v and (p1 O g2)*('v) = 0.
Similarly, we have a map p2: skv(X(2))=S0 ! Svsuch that
p2 O @ = 1(idSv^ f): 1S2v____Sv,
(p2 O g1)*('v) = 0, and (p2 O g2)*('v) = 'v.
Since 1(idSv^ f) ' (1)uv 1( f ^ idSv): S2v1___Sv, there is a *
*homo
topy commutative diagram
1( f ^ idSv) v v
1S2v____________________Sw _________ X
ww  
ww  
ww  
ww g OE2
w ? ?
@ (2) 0 (2) 0
1S2v_______________skv(X )=S ___X =S ,
where in the integral homology g* sends 'vto uv + (1)uvvu = uv + vu. Thu*
*s the
statement holds for n = 2.
Suppose that the statement holds for n  1 with n > 2. There is a homotopy
commutative diagram
OEn1 ^ idX (n1) (n1)
(n2)vX ^ X___________(X = sk(n3)v(X )) ^ X

 
 
pinch  pinch
 
? ~OE ?
(n2)vX(2)=S0______________X(n)= sk(n2)v(X(n)).
Let OEn: (n1)vX ! X(n)= sk(n2)v)(X(n)) be the map defined by the composi*
*te
(n2)vffi2(n2)v(2)0~ffi(n) (n)
(n1)vX ________ X =S ____X = sk(n2)v(X ).
By induction, we have OEn*(s(n1)vv) = vn and
X X
OEn*(s(n1)vu) = viuvjv + vn1u = viuvj.
i+j=n2 i+j=n1
Hence the result.
42 J. WU
Theorem 3.15. Let X = S0 [f evbe a stable 2local twocell complex with v >*
* 0
k1
and v 0 mod 2. Then the cell structure of ~Q2 (X) is given by
Q~2k1(X) = S0 [f ev[2fe2v[3fe3v[ . .[.(2k1)fe(2k1)v
for each k 0.
k1) 2k1
Proof.Let V = H~*(X) and let q :X(2 ! ~Q (X) be the projection map. By
k1 2k1
Theorem 3.2, ~H*(Q2 (X)) is the symmetric quotient S2k1(V ) of V . Let
k1n) (n) (2k1)
jn: X(n)= (S0)(2 ^ X ____X
be the injection for n 2k  1. Let ` be the composite
ffin (n) (n) jn 2k1 2k1
(n1)vX ____X = sk(n2)v(X ) ____ sknv(Q~ (X))= sk(n2)v(Q~ *
* (X)),
where OEn is given in Lemma 3.14. Then
k1
`*: Hj( (n1)vX) ____Hj(sknv(Q2 (X))= sk(n2)v(X))
is an isomorphism for j 6= (n  1)v and of degree n for j = (n  1)v. Since
(n1)vX is attached by f, the attaching map for the twocomplex
k1 2k1
sknv(Q2 (X))= sk(n2)v(Q (X))
is nf and hence the result.
Corollary 3.16. The space HQ3 has the stable cell structure
S12[ e16[2 e20[3 e24.
Thus HQ3 is not homotopy equivalent to KP2 ^ HP2.
Proposition 3.17. Let X = S0 [f evbe a stable twocell complex with v > 0 a*
*nd
v 0 mod 2. Then
k1 2t1
skn(Q~2 (X)) ' skn(Q~ (X))
for n min{2k  1, 2t 1}.
Proof.We may assume that 2k  1 2t 1. Then the composite
Q~2k1(X) ____X(2k1)____j2k1X(2t1)____pinch~Q2t1(X)
k1 2t1
induces a homotopy equivalence from ~Q2 (X) to sk2k1(Q~ (X)).
k1 k
Let ~Jn(X) denote the nvskeleton of ~Q2 (X) for some k such that n 2 *
* 1.
By the above proposition, the homotopy type of ~Jn(X) is independent on the cho*
*ice
k1
of ~Q2 (X). By Theorem 3.15, the cell structure of ~Jn(X) is given by
J~n(X) = S0 [f ev[2fe2v[3fe2v[ . .[.nfenv.
k1) 2k1
The projection q :X(2 ! ~Q (X) induces a map
k1) qX(n) 2k1
qn: X(n)= skn(X(2 ) _____J~n(X) = skn(Q~ (X)).
Let D denote the SpanierWhitehead dual and let cn be the composite
Dqn 2nv (n) (n) qn
2nvD(J~n(X)) ____ D(X ) ' X ____J~n(X).
The homology information on the map cn is given as follows.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 43
Proposition 3.18. Let X = S0 [f evwith v > 0 and v 0 mod 2. Then
cn*: ~Hjv( 2nvD(J~n(X)); Z(2)) = Z(2)! ~Hjv(J~n(X); Z(2)) = Z(2)
` '
n
is of degree .
j
Proof.Let V = ~H*(X). Observe that
qn*: ~H*(X(n)) = V n ____H~*(J~n(X)) = Sn(V )
is the canonical quotient from the self tensor product to symmetric product. Th*
*us
Dqn*: ~H*( 2nvD(J~n(X))) = Sn(V )* ____H~*(X(n)) = V n
is the canonical inclusion of the nth homogeneous of the commutative divided a*
*lgebra
into nfold self tensor product. It follows that
cn*: ~Hjv( 2nvD(J~n(X))) ! ~Hjv(J~n(X))
n
is of degree dim(Vj n) = j , which is the assertion.
k1
Corollary 3.19. Let X = S0[fevwith v > 0 and v 0 mod 2. Then ~Q2 (X)
is selfdual for each k 0.
As an application, we give a new proof of the classical result that the White*
*head
square !15 is divisible by 2.
Theorem 3.20 (Toda). The Whitehead square !152 ß29(S15) is divisible by 2.
Proof.Let X = 8KP2 and let f be the composite
c2 0 0
KP2 ' 32D(J~2(X))=S0 ____J~2(X)=S ____J~3(X)=S .
Then f*: H*(KP2; Z(2)) ! H*(J~3(X)=S0; Z(2)) is of degree 2 and 1 in dimensions*
* 8
and 16, respectively. Thus H~*(Cf) has a basis {x8, x9, x24} with Sq1*x9 = x8 *
*and
Sq16*x24= x8. It is wellknown that the existence of such a complex if and only*
* if the
Whitehead square !15 is divisible by 2 and hence the result.
Example. Let X = 8KP2 and let g be th composite
c5
16~J3(X) ' ( 80D(J~5(X)))= sk8( 80D(J~5(X))) ____J~5(X)=X ____J~6(X)=X.
Then in Z(2)homology, g* is of degree 1, 5, 10 and 10 in dimensions 40, 32, 24*
* and
16, respectively. Thus H~*(Cg) has a basis {x16, x17, x24, x25, x48} with Sq1*x*
*17 = x16,
Sq1*x25 = x24, Sq8*x24 = x16, Sq8*x25 = x17 and Sq32*x48 = x16. Clearly the att*
*aching
map between the 48cell and 25cell is trivial and so we obtain a stable comple*
*x Y
such that ~H*(Y ) has a basis {x16, x17, x24, x48} with Sq1*x17= x16, Sq8*x24= *
*x16 and
Sq32*x48= x16.
44 J. WU
4. Decompositions of the Loop Spaces
This chapter is the preliminary for computing the homotopy groups of mod 2
Moore spaces. In the first section, we briefly go over some results in [12, 36,*
* 37, 39].
In section 2, we will apply these general results to our special space and dete*
*rmine the
factors of P n(2) up to certain dimension. This will help computing the homoto*
*py
groups up to certain range. We give some properties on the spaces F n{2} in sec*
*tion 3.
In section 4, we study the EHP sequences for mod 2 Moore spaces.
4.1. Functorial Decompositions of Loop Suspensions. We start with some al
gebraic results. Let V be a vector space over a field k and let T (V ) be the*
* tensor
algebra generated by V . T (V ) is a Hopf algebra by saying V primitive. Consid*
*er T
as a functor from (ungraded) modules to coalgebras. Let Amin is a smallest retr*
*act
of T which contains V in the following sense
1) Amin(V ) is a functorial coalgebra retract of T (V );
2) V Amin(V );
3) If A(V ) is any functorial coalgebra retract of T (V ) with V A(V ), *
*then
Amin(V ) is a functorial coalgebra retract of A(V ).
It was shown in [36] that the functor Aminexists and unique up to natural equiv*
*alence.
The coalgebra Amin(V ) can described as follows.
Recall that Lie(n) is module over the symmetric group k(Sn), see 2.15. Let
Liemax(n) be the maximal projective k(Sn)submodule of Lie(n). (If Lie(n) itse*
*lf
is projective, then Liemax(n) = Lie(n).) For any kmodule V , define the functor
Lmaxn(V ) = V n k(Sn)Liemax(n).
Then Lmaxn(V ) is a functorial submodule of Ln(V ). Let Bmax(V ) be the subHopf
algebra of T (V ) generated by Lmaxn(V ) for n 2. Let Qmaxn(V ) denote the s*
*et of
indecomposable elements of Bmax(V ) of tensor length n.
Theorem 4.1. [36, Theorem 6.5, Corollary 8.9] There are functorial coalgebra d*
*e
compositions
M1
T (V ) ~=Bmax(V ) Amin(V ) ~=T ( Qmaxn(V )) Amin(V ).
n=2
Let p be the characteristic of k.
Theorem 4.2. [36, Theorem 1.5] If n is not a power of p, then Ln(V ) Bmax(V *
*).
Let Aminn(V ) be the submodule of Amin(V ) consisting of homogeneous elements*
* of
tensor length n.
Proposition 4.3. [36, Proposition 8.10] There is a functorial isomorphism of k
modules
Aminn1(V ) V ~= Qmaxn(V ) Aminn(V )
for each n 1.
These algebraic results admit geometric realization in the following sense. *
*A ho
motopy functor from CW complexes to spaces means a functor from the homotopy
category of CW complexes to the homotopy category of spaces.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 45
Theorem 4.4. [37, Theorem 1.2] Let X be any pathconnected plocal CW complex.
Then there are homotopy functors ~Qmaxn, n 2, and Aminfrom pathconnected pl*
*ocal
CW complexes to spaces with the following properties
1) Q~maxn(X) is a functorial retract of X(n);
2) there is a functorial fibre sequence
`1
jX max iX
Amin(X) ____ Q~n (X) ____ X
n=2
with jX ' * functorially;
3) there is a functorial decomposition
`1
X ' Amin(X) x ( Q~maxn(X));
n=2
4) there is a functorial coalgebra filtration on the mod p homology H*(Amin*
*(X))
such that there is a functorial isomorphism of coalgebras
Gr.H*(Amin(X)) ~=Amin(V ),
where V = ~H*(X) by forgetting the comultiplication.
The following lemmas will be useful.
Lemma 4.5. [12, Lemma 3.3] The symmetric group module Lie(n) is projective if
and only if n 6 0 mod p. Thus Liemax(n) = Lie(n) if and only if n 6 0 mod p.
Recall that the set of finitely generated indecomposable projective modules o*
*f the
symmetric group k(Sn) is onetoone correspondent to the set of pregular parti*
*tions
of n. The indecomposable projective k(Sn)module corresponding to ~ is denoted
by Q~.
Lemma 4.6. [39] The symmetric group module Liemax(8) has the decomposition
Liemax(8) ~=[L3, L5]+Q(6,2)+4Q(5,2,1)+Q(4,3,1)~=2Q(6,2)+Q(5,3)+8Q(5,2,1)+4Q(4,3*
*,1),
where kM is the kfold self direct sum of a module M and M + N = M N.
The following lemma is wellknown.
Lemma 4.7. Let G be a finite group and let Sylp(G) be the Sylow psubgroup. Sup*
*pose
that P is a finitely generated projective k(G)module. Then P is a free k(Sylp(*
*G))
module, where p is the characteristic of k. In particular, the dimension of P *
* is
divisible by the order of the Sylow psubgroup of G.
4.2. Determination of Q~maxk(P n(2)) for k 9. In this section, we determine
~Qmaxk(P n(2)) for low k. Since ~Qmaxk(P n(2)) is a retract of (P n(2))(k), th*
*ere is a corre
sponding retract Qmaxk(P n(2)) of (P n(2))(k)such that ~Qmaxk(P n(2)) ' Qmaxk(*
*P n(2))
by Proposition 3.4. We will determine Qmaxk(P n(2)) for n 2 and k 9. We a*
*s
sume that the ground field k is of characteristic 2. Let V be a twodimension*
*al
module with a basis {u, v} and v = u + 1 = n. Let ad (v)(u) = [u, v] and
adn(v)(u) = [adn1(v)(u), v] for n 2. For a 2regular partition ~ of k. Let
Q~(V ) = V k k(Sk)Q~. By Theorem 3.1, we have Q~(V ) = 0 if len(~) > 2.
By Proposition 3.4, each module Q~(V ) is geometric realizable by a space Q~(P *
*n(2))
46 J. WU
which is a retract of (P n(2))(k). For simplicity, we write kX for the kfold s*
*elf wedge
of X.
Our algorithm for determining decompositions of Qmaxk(P n(2)) for low k is as*
* fol
lows:
1) Determine the module Qmaxk(V ).
2) Count the dimension t1 = dim Qmaxk(V )l, where Qmaxk(V )j = 0 for j > l
and Qmaxk(V )l 6= 0. In other words, count the dimension of the top deg*
*ree
submodule of Qmaxk(V ). By Theorem 3.2, there is one and only one indeco*
*m
posable factor Q(a1,ka1)(P n(2)) of (P n(2))(k)such that (1) Q(a1,ka1)*
*(P n(2))
has exactly one top cell in dimension l and (2) the t1fold self wedge of
Q(a1,ka1)(P n(2)) is a retract Qmaxk(P n(2)). In this step, we obtain t*
*he factor
Q(a1,ka1)(P n(2)) with the multiplicity t1.
3) By subtracting the factor t1Q(a1,ka1)(V ) from Qmaxk(V ), we obtain the*
* proper
summand Qmaxk(V )=(t1Q(a,ka)(V ) of Qmaxk(V ). By repeating step 2, we *
*ob
tain a factor Q(a2,ka2)(P n(2)) of Qmaxk(P n(2)) with a multiplicity t2.
4) After a finite steps, we obtain a (finite) sequence of indecomposable fa*
*ctors
of Qmaxk(P n(2)) together with their multiplicities. This will give a co*
*mplete
decomposition of Qmaxk(P n(2)) in terms of indecomposable factors.
4.2.1. Qmaxk(P n(2)) for k 4. Since dim Lie(2) = 1 and dim Lie(4) = 6 , we ob*
*tain
that Liemax(2) = Liemax(4) = 0 by Lemma 4.7. It follows that
Qmax2(V ) = Lmax2(V ) = Qmax4(V ) = Lmax4(V ) = 0
and so
Qmax2(P n(2)) ' Qmax4(P n(2)) ' *.
By Proposition 4.5, Liemax(3) = Lie(3) and so
Qmax3(V ) = Lmax3(V ) = L3(V ).
This is a twodimensional module with a basis {[[u, v], v], [[u, v], u]} and so
Proposition 4.8. There is a homotopy equivalence
Qmax3(P n(2)) ' P 3n1(2).
4.2.2. Qmax5(P n(2)). By Proposition 4.5, Liemax(5) = Lie(5) and so Lmax5(V ) =*
* L5(V ).
Observe that Lmax5(V ) ! Q(Bmax(V )) is a monomorphism. We have
Qmax5(V ) = L5(V ).
This is 6dimensional module.
Observe that the only top dimensional element in L5(V ) is given by [[u, v], *
*v, v, v].
Thus Q(4,1)(V ) is a summand of L5(V ) with the multiplicity one. By Example 3.*
*8,
Q(4,1)(P n(2)) ' CP2 ^ P 5n5(2).
Now L5(V )=Q(4,1)(V ) is a twodimensional module. Thus we have
Proposition 4.9. There is a homotopy decomposition
Qmax5(P n(2)) ' CP 2^ P 5n5(2) _ P 5n2(2).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 47
4.2.3. Qmax6(P n(2)). By Theorem 4.2, we have
L6(V ) Bmax(V ) ~=T (Qmax3(V ) Qmax5(V ) Qmax6(V ) . .).
and so Qmax6(V ) = L6(V )=[L3(V ), L3(V )]. Observe that the only top dimensio*
*nal
element in L6(V )=[L3(V ), L3(V )] is represented by ad5(v)(u). Thus Q(5,1)(V *
*) is a
summand of Qmax6(V ). Since
1 6 3 2 (5,1)
dim L6(V )=[L3(V ), L3(V )] = __(2  2  2 + 2)  1 = 8 = dim Q (V )
6
by the Witt formula and Example 3.8, we have
Proposition 4.10. There is a homotopy equivalence
Qmax6(P n(2)) ' CP2 ^ RP2 ^ P 6n7(2).
4.2.4. Qmax7(P n(2)). By Theorem 4.2, we have Qmax7(V ) = L7(V ). Since the top
dimensional element of L7(V ) is ad6(v)(u), Q(6,1)(V ) is a summand of L7(V ). *
* By
expecting the next top dimensional elements in L7(V ) and Q(6,1)(V ), Q(5,2)(V *
*) is a
summand of V with multiplicity 2. By extracting these three factors from L7(V )*
*, we
obtain a 2dimensional factor which is Q(4,3)(V ). By Corollary 3.11, we have
Proposition 4.11. There is a homotopy decomposition
Qmax7(P n(2)) ' (CP2)(2)^ P 7n9(2) _ 2(CP2 ^ P 7n6(2)) _ P 7n3(2).
4.2.5. Qmax8(V ). Since dim V = 2, we have
Lmax8(V ) ~=[L3(V ), L5(V )] Q(6,2)(V ) ~=Q(6,2)(V ) Q(6,2)(V ) Q(5,3*
*)(V )
by Lemma 4.6. Observe that the factor [L3(V ), L5(V )] lies in the set of decom*
*posable
elements of Bmax(V ) and in dimension 8, there are no more possible decomposable
elements. We obtain
Qmax8(V ) ~=Q(6,2)(V ).
Thus we have
Proposition 4.12. There is a homotopy equivalence
Qmax8(P n(2)) ' CP2 ^ RP2 ^ P 8n8(2).
This proposition also gives information for primitive elements of tensor leng*
*th 8 in
Amin(V ). Let Lminn(V ) = Ln(V ) \ Amin(V ). Observe that there is a decomposit*
*ion
L8(V ) ~=[L3(V ), L5(V )] Qmax8(V ) Lmin8(V ).
Thus
1 8 4
dimLmin8(V ) = dimL8(V )dim [L3(V ), L5(V )]dim Qmax8(V ) = _(2 2 )2x68 = *
*10.
8
Let L be the sub Lie algebra of L(V ) generated by L2(V ), L3(V ), L4(V ), L5(V*
* ) and
L6(V ). Then L is a free Lie algebra generated by
{L2(V ), L3(V ), L4(V )=L2(L2(V )) = L4(V ), L5(V ), L06(V )},
where L06(V ) ~=L6(V )=(L3(L2(V )) L2(L3(V ))) ~=L6(V )=[L3(V ), L3(V )]. This *
*gives
another (nonfunctorial) decomposition of L8(V ):
L8(V ) ~=L4(L2(V )) [L4(V ), L4(V )] [L2(V ), L06(V )] [L3(V ), L5(V )]*
* M8,
48 J. WU
where dim M8 = 7. According to [39], Qmax8(V ) is a summand of [L2(V ), L6(V )]*
* and
so Qmax8(V ) ~=[L2(V ), L06(V )]. Observe that L4(L2(V )) = 0. Thus
Lmin8(V ) ~=[L4(V ), L4(V )] M8
as modules over the Steenrod algebra. A basis for M8 is given by {Sqi*v8  1 *
*i 7}
and dim[L4(V ), L4(V )] is 3.
Note. If dim W 3, then [L4(W ), L4(W )] contains factors which lie in Bmax(V*
* ).
These factors come from some of the factors Q~ with len(~) 3 in Lemma 4.6.
4.2.6. Qmax9(P n(2)). By expecting the Lie elements of tensor length 9 in the f*
*ree Lie
algebra
L(L3(V ), L5(V ), Qmax6(V ), L7(V ), Qmax8(V )),
we have the short exact sequence
0 ____L3(L3(V )) [L3(V ), Qmax6(V )] ____L9(V ) ____Qmax9(V ) ____0.
Thus
1 9
dim Qmax9(V ) = __(2  3)  2  2 x 8 = 38.
9
The top element in Qmax9(V ) is ad8(v)(u) which occurs once. Thus
Q(8,1)(V ) ~=L2(V ) Q(7)(V )
is a summand of Qmax9(V ) with multiplicity one. Now consider the second top
cells, that is, in the submodule of V 9 spanned by monomials in which u occurs
twice and v occurs 7 times. The modules L9(V ), L3(L3(V )) and [L3, Qmax6(V )]*
* re
stricted to this submodule are of dimension 4, 0 and 1, respectively, by the Wi*
*tt
formula. Thus Qmax9(V ) restricted to this submodule is of dimension 3. By Th*
*eo
rem 3.2, Q(8,1)(V ) restricted to this submodule is of dimension 1. Thus Q(7,2*
*)(V )
is a summand of Qmax9(V ) with multiplicity 2. Similarly, by expecting the sub
module of V 9 spanned by monomials in which u occurs 3 times and v occurs 6
times, Q(6,3)(V ) is a summand of Qmax9(V ) with multiplicity 3. By subtracting*
* the fac
tors obtained, there is a twodimensional submodule of Qmax9(V ) left. Thus Q(5*
*,4)(V )
is a summand of Qmax9(V ) with multiplicity 1 and so we have
Proposition 4.13. There is a homotopy decomposition
Qmax9(P n(2)) ' HP2^CP2^P 9n13(2)_2(CP2)(2)^P 9n10(2)_3CP2^P 9n7(2)_P 9n4(2*
*).
Note. By using this algorithm, one may determine Qmaxk(V ) for k 15 with the
help of computer program.
4.3. The Fibre of the Pinch Map. Let F n+1{2} be the homotopy fibre of the
pinch map P n+1(2) ____Sn+1. There is a principal fibre sequence
sn+1 n+1 n+1
Sn+1 ____F {2} ____P (2)
and so Sn+1 acts on F n+1{2}.
Lemma 4.14. [9, Proposition 8.1, Corollary 8.2] If n 2, then
1) the reduced integral homology H~*(F n+1{2}; Z) is a free Zmodule on gen*
*era
tors yk with k 1 and degree yk = kn;
2) The map s(n+1)*:Hkn( Sn+1; Z) = Z ! Hkn(F n+1{2}; Z) = Z is of degree 2;
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 49
3) H~*(F n+1{2}) is a free H*( Sn+1) module on a single generator y1.
The following proposition can be found in [16, Corollary 5.8. p.513] and [27,
Lemma 1.4].
Proposition 4.15. The (2n)skeleton sk2n(F n+1{2}) of F n+1{2} is the homotopy
cofibre of 2!n: S2n1 ! Sn. Thus sk4n2(F 2n{2}) ' S2n1 _ S4n2.
Proof.Consider the homotopy commutative diagram of fibre sequences
[2] n+1 n+1 n+1
Sn+1w ______ S ____S {2} _____Sw
ww 6  w
ww  6 ww
ww   ww
ww g  ww
w   ww
sn+1 n+1 n+1 n+1
Sn+1 _____F {2} ____P (2)_____S .
By assertion 2 of Lemma 4.14, we have
2k'kn= [2]*('kn) = g*(s(n+1)*('kn)) = 2g*(yk)
and so g*(yk) = 2k1'kn
g*: Hkn(F n+1{2}; Z) ____Hkn( Sn+1; Z)
is of degree 2k1 for k 1. Thus there is a homotopy commutative diagram of co*
*fibre
sequences
@ n n+1 pinch 2n
S2n1 _______Sw _____sk2n(F {2}) _______S
 ww  
 w  
 ww  
[2] ww g [2]
 ww  
? ! ? pinch ?
S2n1 _______nSn _________J2(Sn)__________S2n
and hence the result.
By assertion 2 of Theorem 4.4, there is a homotopy commutative diagram of fib*
*re
sequences
jPn(2)1` max n ßPn(2) n+1
Amin(P n(2))______ ~Qk (P (2))______P (2)
 k=2 
  
  
  
 ? 
? j n1` ß n ?
Amin(Sn) _______S ~Qmaxk(Sn) ' *______SSn+1.
k=2
50 J. WU
1`
Thus the map Qmaxn(P n(2)) ! P n+1(2)lifts to F n+1{2} and so there is a hom*
*otopy
k=1
commutative diagram of fibre sequences
A~n+1============== ~An+1_______________*
  
  
  
 ~j 
  
 ? 
? jPn(2) 1` ßPn(2) ?
Amin(P n(2))_____ Qmaxk(P n(2))_____P n+1(2)w
 k=2 ww
  ww
  w
  ww
  w
? ?
Sn+1 ____________F n+1{2} __________P n+1(2).
Since jPn(2)is null homotopic, so is ~j. Thus we have
Proposition 4.16. If n 2, then there is a homotopy decomposition
`1
F n+1{2} ' ~An+1x ( Qmaxk(P n(2)))
k=2
with a fibre sequence A~n+1____Amin(P n(2)) ____ Sn+1.
1`
Let ~Bn+1be the pull back of the diagram Sn ____F n+1{2} oe___ Qmaxk(P*
* n(2)).
k=2
There is a principal fibre sequence Sn ____A~n+1 ____B~n+1. When n is odd, t*
*his
fibre sequence splits off.
Lemma 4.17. There is a homotopy decomposition
A~2n' S2n1 x ~B2n.
for n > 1.
Proof.Consider the homotopy commutative diagram of fibre sequences
Fw_______F 2n{2}_____S2n1
ww  
ww  
ww  
ww  
w ? ?
F _______P 2n(2)_____ø(S2n)
  
  
  
  
 pinch 
  
 ? ?
? 2n 2n
* _________S ======= S .
The bottom cell S2n1 is a retract of F 2n{2} and so S2n1 is a retract F 2n{*
*2}. It
follows that S2n1 is a retract of ~A2nand hence the result.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 51
4.4. The EHP Sequences for mod 2 Moore Spaces. Let H~2denote the com
posite
~H2: P n+1' J(P n(2)) ____H2J(P n(2) ^ P n(2)) ____J(q)J(P 2n(2)) ' P 2n+*
*1(2),
where q is the pinch map (P n(2))(2)____(P n(2))(2)=(Sn1 ^ P n(2)) = P 2n(2).*
* There
is a homotopy commutative diagram of fibre sequences
P n E n+1 H2 2n+1
2S2n+1 _______S _______ S ______ S
6 6 6 6
   
   
   
   
 P  E  ~H 
2P 2n+1(2)_____T n+1____ P n+1(2)___2 P 2n+1(2).
We will determine the atomic factor of T n+1which contains the bottom cell. We
need some terminologies on simplicial sets. Let X and Y be pointed simplicial s*
*ets
and let f :X ! Y be a simplicial map. Let F (Y ) be the Milnor construction on *
*Y .
Define the simplicial set Ef to be the pullback of the diagram
J(f)
J(X) ____J(Y ) F (Y ) oe___ P F (Y ),
where P Z is the Moore pathspace of a fibrant simplicial set Z. As a space, Ef*
* is
homotopy equivalent to the loop space of the homotopy fibre of f : X ! Y .
Consider the strictly commutative diagram
J(f)
J(X) ____________J(Y ) ______F (Y )oe_____P F (Y )
   
   
   
H2 H2 ~H2 P~H2
   
   
? J(f ^ f) ? ? ?
J(X ^ X) ________J(Y ^ Y ) __F (Y ^ Y )oe_P F (Y ^ Y ),
where H~2:F (Y ) ! F (Y ^ Y ) is a functorial extension of the map H2. There i*
*s a
unique simplicial map
H2: Ef ____Ef^f
such that the diagram
Ef _________J(X) x P F (Y )
 
 
 
H2 H2x P ~H2
 
? ?
Ef^f ____J(X ^ X) x P F (Y ^ Y )
commutes strictly.
Lemma 4.18. Let f :X ! Y be a simplicial map and let Z be a simplicial subset
of X such that fZ :Z ! Y is the trivial map. Then there exist simplicial homom*
*or
phisms OE1: J(Z) ! Ef and OE2: J(Z ^ X) ! Ef^f such that
1) OE1 is a lifting of J(Z) ____J(X);
52 J. WU
2) OE2 is a lifting of J(X ^ Z) ____J(X ^ X) and
3) the diagram
OE1 f
J(Z) ___________E
 
 
 
H2 H2
 
 
? OE2J(Z^Z) ?
J(Z ^ Z) _________Ef^f
commutes strictly.
Proof.Since J(f)J(Z)and J(f ^f)J(Z^X)are trivial maps, the simplicial submono*
*ids
J(Z) of J(X) and J(X ^ Z) of J(X ^ X) admits the crosssections, x 7! (x, 1), i*
*nto
the pullback Ef and Ef^f, respectively. The assertions follow immediately.
Lemma 4.19. There exists a map OE: T n+1! Sn{2} such that
OE*: Hn1(T n+1) ____Hn1(Sn{2})
is an isomorphism.
Proof.Let f :X ! Y be a fibration with the fibre F such that
1) X ' Y ' Sn;
2) f is a map of degree 2.
Then there is a map g :P n(2) ! F such that the composite P n(2) ! F ! X is
homotopic to the pinch map P n(2) ! Sn. By Lemma 4.18, there is a homotopy
commutative diagram of fibre sequences
~H2
T n+1_____ P n+1(2)___ (P n(2) ^ Sn)
  
  
  
OE OE1 OE2
  
  
? ? H2 ?
Sn{2} ____ Sn+1{2} _____ S2n+1{4}.
Clearly OE induces an isomorphism on Hn1 and hence the result.
Our next step is to construct a crosssection map for T n+1! Sn{2}. Let
i
A ____X ____X=A
be a cofibration such that A ' X ' Sn1 and i is of degree 2. Then P n+1(2) ' *
*F (X=A)
is the quotient simplicial group of F (X) by the normal subgroup generated by F*
* (A).
Clearly the simplicial coset of F (X) by the action of the simplicial subgroup *
*F (A) is
homotopy equivalent to Sn{2}. Thus there is a map of F (X)spaces
s: Sn{2} ! P n+1(2)
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 53
such that the diagram
F (X)====== F (X)
 
 
 
 
 
 
? s ?
Sn{2} ____ P n+1(2)
commutes. Let c: J(X) ^ J(Y ) ____J(X ^ Y ) be the map defined by
c((x1x2. .x.n) ^ (y1y2. .y.m)) =
(x1 ^ y1) . (x2 ^ y1) . .(.xn ^ y1) . (x1 ^ y2) . .(.xn ^ y2) . .(.x1 ^ ym ) *
*. .(.xn ^ ym )
for xi2 X and yj 2 Y , see [47]. Let K~k,lbe the subgroup of
[(Sn1)k x (P n(2))l, P n+1(2)]
generated by the elements xi, 1 i k and yj, 1 j l, represented by the m*
*aps:
ii n1 n+1
xi:(Sn1)k x (P n(2))l ____S ____ P (2),
ik+j n n+1
yj: (Sn1)k x (P n(2))l ____P (2) ____ P (2),
respectively, where ßi the ith coordinate projection. By the proof of Lemma 2.*
*30,
we have
Lemma 4.20. If n 3, then the following identities hold in the group K~k,l:
1) x2i= 1 for 1 i k;
2) y4j= 1;
3) The (iterated) commutators
[[a1, a2], . .,.as] = 1
if ai= aj for some i < j, where ai= ys or xt for some s, t.
Theorem 4.21. If n 3, then the map OE: T n+1! Sn{2} has a crosssection and
so Sn{2} is a retract of T n+1.
Proof.Since Sn{2} is atomic, it suffices to show that the composite
s n+1 H~2 2n+1
Sn{2} ____ P (2) ____ P (2)
is null homotopic. We use the filtration {FkSn{2}} is the proof of Lemma 2.30.
Observe that the product element x1x2. .x.ky1 2 ~Kk,1is represented by the comp*
*osite
s n+1
(Sn1)k x P n(2) ____Fk+1Sn{2} ____ P (2).
Consider the function
H2*:[(Sn1)k x P n(2), P n+1(2)] ! [(Sn1)k x P n(2), (P n(2))(2)].
According to [47, Lemma 3.4], The element H2*(x1x2. .x.ky) is represented by the
product of the maps ff and fi given by
proj. n1 k proj. n1 H2 2n2 n n
ff :(Sn1)kxP n(2) ____(S ) ____Jk(S ) ____J(S ) ___J(P (2)^P (2*
*)),
proj. n1 n c n1 n n n
fi :(Sn1)kxP n(2) ____Jk(S )^P (2) ____J(S ^P (2)) ___J(P (2)^P (2*
*)).
54 J. WU
Since both ff and fi are null homotopic after composing with the map
J((P n(2))(2)) ____J((P n(2))(2)=(Sn1 ^ P n(2))) = J(P 2n(2)),
the element
H~2(x1. .x.ky1) = 1
in [(Sn1)k x P n(2), P 2n+1(2)] and hence the result.
The homology of the space T n+1can be described as follows.
Proposition 4.22. If n 3, then the Serre spectral sequence for the fibre sequ*
*ence
H~2 2n+1
T n+1____ P n+1(2) ____ P (2)
collapses. Furthermore, there is a submodule W of the free Lie algebra L(u, v) *
*such
that H*(T n+1) is isomorphic to T (W ) H*(Sn{2}) as coalgebras.
Proof.Let ~V= ~H*(P 2n(2)) with a basis {~u, ~v} and Sq1*~v= ~u. Consider
H~2*:H*( P n+1(2)) = T (V ) ____H*( P 2n+1(2)) = T (V~).
Let A be the cotensor product k ___E(u,v)T (u, v). Then A is a subHopf algebr*
*a of T (u, v)
generated by {u2, [u, v], v2, [[u, v], u], [[u, v], v]}. Let j :A ! T (u, v) be*
* the inclusion
map and let s: E(u, v) ! T (u, v) be a klinear map defined by s(u) = u, s(v) =*
* v
and s(uv) = uv. Then s is a coalgebra map and the composite
j s ~
A E(u, v) ____T (u, v) T (u, v) ____T (u, v)
is an isomorphism of coalgebras over A. According to [11, Proposition 5.1] and *
*[47,
Proposition 3.12], the map
H2*:T (V ) ! T (V V )
is a morphism of coalgebras over A and so is H~2*, where the map H~2*restricted*
* to
A is a morphism of Hopf algebras given by ~H2*([u, v]) = ~u, ~H2*(v2) = ~vand
H~2*(u2) = ~H2*([[u, v], u]) = ~H2([[u, v], v]) = 0.
Let B = k ___T(~V)A. By [22, Theorem 4.4], there is an isomorphism of coalgeb*
*ras
A ~=B T (V~),
where the coaction of T (V~) on B is trivial. Observe that the composite
s ~H2*
E(u, v) ____T (u, v) ____T (V~)
is trivial map. Thus
T (V ) ~=E(u, v) B T (V~)
as comodules over T (V~), where the coactions of T (V~) on the factors E(u, v) *
*and B
are trivial. Thus H*( P n+1(2)) is a free H*( P 2n+1(2))module. By expecting t*
*he
EilenbergMoore spectral sequence for the fibre sequence
T n+1____ P n+1(2) ____ P 2n+1(2),
H*(T n+1) is a subquotient of E(u, v) B. In particular, the Poincar'e series
Ø(H*(T n+1)) Ø(E(u, v))Ø(B).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 55
By expecting the Serre spectral sequence for the fibre sequence above, we have
Ø(E(u, v))Ø(B)Ø(T (V~)) = Ø(T (V )) Ø(H*(T n+1))Ø(T (V~)).
Thus
Ø(H*(T n+1)) = Ø(E(u, v))Ø(B)
and so the Serre spectral sequence collapses. Furthermore, there is an isomorph*
*ism
of coalgebra
H*(T n+1) ~=E(u, v) B.
Consider the commutative diagrams of short exact sequences of Hopf algebras
B ___P(u2)___B _______P (u2)
\ \ ww
  ww
  w
  ww
  w
  w
? ?
A ___P(u2)___A _______P (u2)
 
 
 
 
 
?? ??
T (V~)====== T (V~).
There is a submodule W of L(V ) such that B ___P(u2)~=T (W ). It follows that
H*(T n+1) ~=E(u, v) P (u2) T (W ) ~=H*(Sn{2}) T (W )
as coalgebras. The proof is finished.
Now we are going to give a product decomposition of T n+1up to certain dimens*
*ion.
We need to do some lemmas on the JamesHopf maps. Let E be the collection of
spaces consisting of all suspensions. For each suspension X, let Kn(k)(X) be t*
*he
subgroup of [Xn, J(X(k))] generated by the elements labelled by {xi1xi2 . ..*
*xik},
1 i1, i2, . .,.is n, represented by the map
ii1,i2,...,ikkpinch(k) (k)
Xn _______X ____X ____J(X ),
where ßI is the coordinate projection. By letting X run over all objects in E, *
*one gets
a universal group Kn(k). The generators for Kn(k) are labelled by {xi1xi2 . .*
*.xik}
subject to the relations given in [47, Lemma 2.2]. Observe that Kn(1) = Kn(E) *
*is
the Cohen group given in Section 2.2. The JamesHopf map Hk: J(X) ! J(X(k))
induces a function
Hk*:[Xn, J(X)] ! [Xn, J(X(k))].
56 J. WU
When X runs over all objects in E, this function induces a unique function Hk: *
*Kn(1) !
Kn(k) such that the diagram
eX n
Kn(1) _____[X , J(X)]
 
 
 
Hk Hk*
 
 
? eX ?
Kn(k) ____[Xn, J(X(k))]
commutes for all X 2 E. Let fin: X(n) ! J(X) be the nfold Samelson product.
Observe that the element eX ([[x1, x2], . .,.xn]) in [Xn, J(X)] is represented *
*by the
pinch (n) Wn
composite Xn ____X ____J(X). Thus the element
eX (Hk([[x1, x2], . .,.xn]))
in [Xn, J(X(k))] is represented by composite
pinch (n) fin Hk (k)
Xn ____X ____J(X) ____J(X ).
Lemma 4.23. If n is odd, then the composite
fin H2 (2)
X(n)____J(X) ____J(X )
is null homotopic for any suspension X.
Proof.The homotopy commutative diagram
J([1])
J(X) ________J(X)
 
 
 
H2 H2
 
? id ?
J(X(2))_______J(X(2))
induces a group homomorphism J([1]): Kn(1) ! Kn(1) such that the diagram
J([1])
Kn(1) _______Kn(1)
 
 
 
H2 H2
 
 
? ?
Kn(2)======= Kn(2)
commutes. Observe that J([1])(xi) = xi. Thus we have
*
* n
H2([[x1, x2], . .,.xn]) = H2 O J([1])([[x1, x2], . .,.xn]) = H2([[x1, x2], . .*
*,.xn](1)).
By [47, Theorem 3.8], H2 restricted to the commutator subgroup 2Kn(2) is a gro*
*up
homomorphism. Thus the element
(H2([[x1, x2], . .,.xn]))2 = 1
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 57
in the group Kn(2). Since Kn(2) is a torsion free group, the element
H2([[x1, x2], . .,.xn]) = 1
and so the composite
q (n) fin H2 (2)
Xn ____X ____J(X) ____J(X )
is null homotopic. The assertion follows from the fact that q*: [X(n), Y ] ! [*
*Xn, Y ]
is a monomorphism for any Y .
Note. This lemma can be generalized as follows: The composite
Wl Hk (k)
X(l)____J(X) ____J(X )
is null homotopic if l 6 0 mod k.
Corollary 4.24. If n 3 and k is odd, then the composite
fik n+1 H~2 2n+1
(P n(2))(k)____ P ____ P (2)
is null homotopic.
Lemma 4.25. In the group K4(2), the element H2([[x1, x2], x3, x4]) is equal to *
*the
product
[{x4x3}, {x1x2}] . [{x4x1}, {x3x2}] . [{x3x1}, {x4x2}] . [{x2x1}, {x*
*4x3}].
The proof follows from a direct calculation by using the methods in [47].
Lemma 4.26. If n 3, then the composite
fi4 n+1 H~2 2n+1
(P n(2))(3)^ Sn1 ____(P n(2))(4)____ P (2) ____ P (2)
is null homotopic.
Proof.Let
q :(P n(2))(2)____P 2n(2) = (P n(2))(2)=Sn1 ^ P n(2)
be the pinch map. Observe that the element {x4xi} restricted to (P n(2))3x Sn*
*1 is
represented by the composite
i4,i n1 n n (2)
(P n(2))3 x Sn1 ____S ^ P (2) ____J((P (2)) ).
Thus J(q)*({x4xi}) = 1 for 1 i 3. The assertion follows from Lemma 4.25.
Let f :X ! J(Y ) be a map. We write J(f): J(X) ! J(Y ) be the homomorphism
of monoids induced by f.
Lemma 4.27. Let OE: Y ! (P n(2))(k)be a map such that the composite
ffi n (k) fik n+1 H~2 2n+1
Y ____(P (2)) ____ P (2) ____ P (2)
is null homotopic. Suppose that n 3 and k 2. Then there is a map
~ffi: Y x T n+1! T n+1
58 J. WU
such that the diagram
~ffi n+1
Y x T n+1 ______________________T
 
 
 
 OE x E E
 
? J(fi ?
k) . id Pn+1(2)n+1
(P n(2))(k)x P n+1________________ P (2)
commutes up to homotopy.
Proof.By Theorem [47, Theorem 3.10], there is a map fik,2: X(k)____ X(2)such
that fik,2is homotopic to the composite
J(fik) H2 (2)
X(k)____ X ____ X
for any suspension X and any integer k 2. According to [47, Proposition 3.12*
*],
there is a homotopy commutative diagram
J(fik) . id X
X(k)x X ______________ X
 
 
 
fik,2xH2 H2
 
? ~ ?
X(2)x X(2) ____________ X(2)
for any suspension X and any integer k 3. It follows that the composite
ffixE n (k) n+1 J(fik).id Pn+1(2)n+1
Y x T n+1______ (P (2)) x P (2) ____________ P (2)
lifts to the fibre T n+1and hence the result.
Theorem 4.28. If n 3, there is a map
` :Sn{2} x P 3n(2) x P 4n1(2) x (CP2 ^ P 5n4(2) _ P 5n1(2)) ____T n+1
such that Hj(`) is an isomorphism for j 6n  6.
Proof.Let s: Sn{2} ! T n+1be a crosssection of the map OE defined in Lemma 4.1*
*9.
By Lemma 4.26 and Corollary 4.24, the maps
fi3: Qmax3(P n(2)) = P 3n1(2) ____ P n+1(2),
fi5: Qmax5(P n(2)) = CP2 ^ P 5n5(2) _ P 5n2(2) ____ P n+1(2),
~fi4:Qmax3(P n(2)) ^ Sn1 = P 4n2(2) ____Qmax3(P n(2)) ^ P n(2) ____fi4 P*
* n+1(2)
are null homotopic after composing with ~H2. The assertion follows from Lemma 4*
*.27
and by expecting the homology of T n+1described in Proposition 4.22.
5. The Homotopy Groups ßn+r( nRP2) for n 2 and r 8
In this chapter, we compute specific homotopy groups. Thus this chapter will
be divided into parts A: ß*(P 4(2)), B: ß*(P 5(2)), C: ß*(P 6(2)), D: ß*(P 7(2)*
*), E:
ß*(P 8(2)), F: ß*(P 9(2)) and G: ß*(P n(2)) for n 10.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 59
5.1. Preliminary and Notations.
Proposition 5.1. The first three homotopy groups are as follows.
(1).ßn1(P n(2)) = Z=2 ~=ßs1(RP2) for n 2;
(2).ßn(P n(2)) = Z=2 ~=ßs2(RP2) for n 4;
(3).ßn+1(P n(2)) = Z=4 ~=ßs3(RP2) for n 4.
The proof is immediate.
Recall that there is a decomposition
`1
P n+1(2) ' ( Qmaxk(P n(2))) x Amin(P n(2)).
k=3
By applying the HiltonMilnor theorem, we have a decomposition
`1 1Y
( Qmaxk(P n(2))) ' Lk.
k=3 k=3
Let V = H~*(P n(2)) with a basis {u, v} with Sq1*v = u. When k is odd, We wri*
*te
Lk for Lk(P n(2)) as a geometric realization of module Lk(V ). When k = 6, then
Lk is a geometric realization of L6(V )=[L3(V ), L3(V )] = Qmax6(V ). Observe t*
*hat Lk
is (kn  k)connected. For computing ßn+r( P n+1(2)) with n 3 and r 8, we
only need factors L3 ' P 3n1(2) and L5 ' CP2 ^ P 5n5_ P 5n2(2). We rewrite t*
*he
decomposition
P n+1(2) ' L3 x L5 x Yn+1,
where ßn+r(Yn+1) = ßn+r(Amin(P n(2))) for n 3 and r 8. The decomposition in
Proposition 4.16 is rewritten as
F n+1{2} ' L3 x L5 x An+1,
where ßn+r(An+1) = ßn+r(A~n+1) for n 3 and r 8. Recall that there is a fib*
*re
sequence
An+1 ____Yn+1 ____ Sn+1.
By Lemma 4.17, A2n has a further decomposition
A2n ' S2n1 x B2n.
We will need some information on the stable homotopy groups of RP2.
Lemma 5.2. The stable homotopy groups ßsr(RP2) for 4 r 10 are as follows
(1).ßs4(RP2) = Z=2 Z=2,
(2).ßs5(RP2) = Z=2,
(3).ßs6(RP2) = 0,
(4).ßs7(RP2) = Z=2,
(5).ßs8(RP2) = Z=2 Z=2,
(6).ßs9(RP2) = Z=2 3,
(7).ßs10(RP2) = Z=2 Z=4 Z=4.
Proof.The proof follows from the long exact sequence of stable homotopy groups
associated with the cofibre sequence
[2] 1 2
S1 ____S ____RP
60 J. WU
together with calculations from [42] for the homotopy groups of spheres and the*
* fact
that the degree 2 map [2] : RP2 ! RP2, in the stable category, is the composite
pinch 2 '' 1 2
RP2 ____S ____S ____RP .
5.2. The Homotopy Groups ß*(P 4(2)). Now we start to compute ß*P 4(2) for * 1*
*2.
5.2.1. The Space B4.
Lemma 5.3. There exists a map OE : S6 x P 10(2) ! B4 so that
OE* : H*( S6 x P 10(2)) ! H*B4
is an isomorphism for * 10.
Proof.The primitive elements P Hr(B4) for r 10 has a basis
{[u, v], [u, v]2, [[[u, v], u], u], [[[u, v], v], u]}.
Let OE1 : S6 ! F 4{2} be a map so that (OE1)*('5) = [u, v]. Since
L3(P n(2)) = P 3n1(2),
there exists a map _ :P 8(2) ! F 4{2} such that
_*(v8) = [[u, v], v] and _*(u7) = [[u, v], u].
Let j :S2 ! F 4{2} be the injection so that j*('2) = u. Define OE2 to be Samel*
*son
product [_, j] : P 10(2) = P 8(2) ^ S2 ! F 4{2}. Let OE be the composite
ffi1xffi24 4 4
S6 x P 10(2) _____ F {2} x F {2} ____ F {2} ____B4.
Then the map OE has the desired property and hence the result.
Lemma 5.4. There is a cofibre sequence
_ 6 10
S10 ____ sk15( S x P (2)) ____ sk15(B4),
which satisfies:
1) the composite
_ 6 10 proj. 10
S10 ____ S x P (2) ____P (2)
is ~jwhich is a representative of the generator for ß10P 10(2) = Z=2 and
2) the composite
_ 6 10 6
S10 ____ S x P (2) ____ S
is 2!6.
Proof.Note that P Hr(B4) for r 16 has a basis
[u, v], [[u, v, u2], [u, v]2, [[u, v, v, u], [[u, v, v, v]
with Sq1*[[u, v, v, u] = [[u, v, u2] and Sq2*[[u, v, v, v] = [[u, v, u2]. By [9*
*, Lemma 10.4],
fi2[[u, v, v, v] = [u, v]2. Thus sk15(B4)= sk15( S6xP 10(2)) = S11 and there is*
* a cofibre
sequence
_ 6 10
S10 ____ sk15( S x P (2)) ____ sk15(B4).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 61
_ 6 10 10
Since Sq2*[[u, v, v, v] = [[u, v, u2] 6= 0, S10 ____ S x P (2) ____P (2) i*
*s essential
_ 6 10 6
and hence condition (1). Let _2 be the composite S10 ____ S x P (2) ____ S .
There is a homotopy commutative diagram of cofibre sequences
_ 6 10 11
S10w_____sk15( S x P (2))___sk15(B4)______Sw
ww   w
ww   ww
ww   ww
ww   ww
w   ww
_2 ?5 ? 11
S10 ___________J3(S )____________X ________S .
Since H*(sk15(B4)) ! H*X is onto, the Bockstein fi2 : H11X ! H10X is an iso
morphism. Observe that OE2: S10 ! S6 maps into the 10skeleton J2(S5) up to
homotopy. The composite
_2 5 pinch 10
S10 ____J2(S ) ____S ,
is of degree 4. Since ß11(S6) = Z which is generated by !6, we have _2 = 2!6 and
hence the result.
5.2.2. The Homotopy Groups ß*(F 4{2}).
Lemma 5.5. The homotopy groups ßk(B4) for 5 k 12 are as follows:
8
>> Z for k = 5
>>
>> Z=2 for k = 6
>> Z=2 for k = 7
><
Z=8 for k = 8
ßk(B4) =
>> Z=2 for k = 9
>>
>> Z=4 for k = 10
>>
>: Z=2 Z=2 for k = 11
Z=4 Z=2 Z=2 for k = 12.
Proof.By Lemma 5.4, ßr(B4) ~=ßr( S6) for r 8 and ß9(B4) = ß9( S6)
ß9(P 10(2)) = Z=2.
1. ß10(B4): There is an exact sequence
_* 6 10
ß10(S10) ____ß10( S ) ß10(P (2)) ____ß10B4 ____0 or
(2,''*)
Z ____Z Z=2 ____ß10(B4) ____0.
Thus ß10( S6) ____ß10(B4) is onto and ß10(B4) = Z=4.
2. ß11(B4): There is an exact sequence
_* 6 10
ß11(S10) ____ß11( S ) ß11(P (2)) ____ß11(B4) ____0 or
_*
Z=2 ____Z=2 Z=4 ____ß11(B4) ____0.
Observe that
_*(j) = 2!6 O j + ~j2= ~jO j = 2ff,
where ff is the generator of ß11(P 10(2)) = Z=4. Thus
ß11(B4) ~=ß11( S6) ß11(P 10(2))=2 = Z=2 Z=2.
62 J. WU
3. ß12(B4): Consider the exact sequence
_* 6 10
ß12(S10) ____ß12( S ) ß12(P (2)) ____ß12(B4) ____0.
Note that
_*(j2) = 2!6 O j2 + ~jO j2 = 0.
Thus
ß12(B4) ~=ß12( S6) ß12(P 10(2)) = Z=4 Z=2 Z=2.
The calculation is finished now.
By Lemmas 5.2 and 5.5, we have
Proposition 5.6. The homotopy groups ßr = ßr(F 4{2}) for r 13 are as follows.
____________________________________________________________________________*
*
3 9 14 2
ßr_=___ßr(S_)_____ßr1(B4)____ßr(P_(2))___ßr(X)____ßr(P__(2))___*
*
 r=3  Z     
 4  Z=2     
 5  Z=2     
 6  Z=4  Z    
 7  Z=2  Z=2    
 8  Z=2  Z=2  Z=2   
 9  0  Z=8  Z=2   
 10  0  Z=2  Z=4 2   
 11  Z=2 2  Z=4 2  Z=2   
 12  Z=2  2Z=2  Z=2  Z=2  2 
__13__Z=2___Z=4__Z=2_____Z=4__________0________0_______(Z=2)____
where X = CP2^ P 11(2), one copy of P 14(2) comes from L5 and another copy of *
*P 14(2)
is the bottom twocells for L6.
5.2.3. The Fibre of P 4(2) ! BS3. Now consider the homotopy commutative diagram
of fibre sequences
4 4 3
S3w_________S7_________S ________BSw
ww 6 6 w
ww   ww
ww   ww
ww   ww
w   ww
i 4 q 4 ffi 3
S3 _________X _______P (2) _______BS
6 6
 
 j
 
 
 
F 4{2}===== F 4{2}
where 4: S7 ! S4 is the Hopf fibration.
Lemma 5.7. There is a cofibre sequence
( 0,2)3 6 4
S6 ____S _ S ____X
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 63
Proof.By Lemma 2.38, the mod 2 homology H~r(X4) = Z=2 for r = 3, 6, 7 and 0
otherwise.
Consider the Serre spectral sequence for the fibre sequence F 4{2} ! X4 ! S7.
There is a cofibre sequence
sk6(F 4{2}) ____X4 ____S7.
Since F 4{2}(6)= S3 _ S6, there is a cofibre sequence
` 3 6 f 4
S6 ____S _ S ____X .
Now consider the homotopy commutative diagram of fibre sequences
S7 ______F 4{2} _______X4
 ww 
 w 
 ww 
 w 
 ww 
 w 
? ?
S4 ______F 4{2} _____P 4(2).
Observe that H6( S4; Z) ! H6(F 4{2}; Z) is of degree 2. The composite
` 3 6 proj.6
S6 ____S _ S ____S
is of degree 2. Recall that ß6(S3) = Z=4 which is generated by 0. The compos*
*ite
` 3 6 proj.3 0
S6 ____S _ S ____S is k . We claim that k = 1.
Suppose that k 0 mod 2. Let ~~4denote the composite
S6 ____S3 _ S6 = sk6(F 4{2}) ____X4.
Then, in ß6(X4), 2[~~4] = kf*( 0). Since i : S3 ! X4 is of degree 2 and S3 is *
*an
Hspace, we have
i*(k=2 0) = f*([2]*(k=2 0)) = f*(k 0) = 2[~~4].
Hence 2q*[~~4] = 0 in ß6(P 4(2)). Let ~4: S5 ! P 4(2) denote the adjoint map
of q O ~~4:S6 ! P 4(2). Then (~4)*('5) = [u, v] in H5( P 4(2)). By [12, Theorem*
* 2.2],
2q*[~~4] = 2[~4] 6= 0.
This is a contradiction. Thus k = 1 and hence the result.
5.2.4. The Homotopy Groups ß*(P 4(2)).
Theorem 5.8. The homotopy groups ßr(P 4(2)) for 6 r 12 are as follows.
_____________________________________________________________
4 9 2 11
ßr(P_(2))___ßr1(Y4)____ßr(P_(2))___ßr(CP__^_P__(2))_
 r = 6  Z=2 Z=4   
 7  Z=2 Z=2   
 8  Z=2 Z=2  Z=2  
 9  Z=2 Z=2  Z=2  
 10  Z=2 Z=2  Z=4  
 11  Z=2 3 Z=4  Z=2 Z=2  
______12___Z=2_____Z=4________Z=2_______________Z=2__
64 J. WU
Proof.1. ß6(P 4(2)): By expecting the exact sequence
( 0,2) 3 6 4
ß6(S6) ____ß6(S ) ß6(S ) ____ß6(X ) ____0,
we have ß6(X4) = Z=8 generated by [~~4]. Since
i*( 0) = 2 0= 4[~~4],
there is a short exact sequence
0 ____Z=4 ____ß6(P 4(2)) ____ß5(S3) ____0.
Since q*([~~4]) = [~4] in ß6(P 4(2)) ~= ß5( P 4(2)) and h([~4]) = [u, v] in H* *
*P 4(2),
the element [~4] is not divisible by 2. Thus the above short exact sequence spl*
*its and
so ß6(P 4(2)) = Z=4 Z=2.
2. ß7(P 4(2)): Consider the fibre sequence
ffi 3 4 7
2S7 ____ S x B4 x Q ____ X ! S ,
Y
where Q = Lk. Since P 4(2) ' Y4 x Q, the composite
k 3
ffi 3 proj.
2S7 ____ S x B4 x Q ____Q
is null homotopic. By Lemma 5.7, the diagram
E 2 7 ffi 3
S6 _______ S _____ S x B4 x Q
 
 
 
[2] proj.
 
? OE ?
S6 _______________________1B4
commutes up to homotopy commutes and the composite
E 2 7 ffi 3 proj. 3
S6 ____ S ____ S x B4 x Q ____ S
is 0. Consider the exact sequence
ffi* 3 4
ß6( 2S7) ____ß6( S ) ß6(B4) ß6(Q) ____ß6( X ) ____0.
Observe that
ffi*(j) = 0O j + (OE1)* O [2]*(j) = 0O j.
Thus ß6( X4) ~= ß6(B4) ß6(Q) = Z=2 and ffi* : ß6( 2S7) ! ß6( F 4{2}) is 11.
Since i*( 0O j) = f*(2 0O j) = f*(0) = 0, there is a short exact sequence
q* 4 3
0 ____ß7(X4) ____ß7(P (2)) ____2 . ß6(S ) ____0 or
0 ____Z=2 ____ß7(P 4(2)) ____Z=2 ____0
and therefore (~4)* : ß6(S5) ! ß7(P 4(2)) is a monomorphism. Consider the second
reduced JamesHopf map ~H2: P 4(2) ____ P 7(2). Then
(H~2O ~4)*('5) = (H~2)*([u, v]) 6= 0
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 65
in H5( P 7(2)) and so ~H2O~4 : S5 ! P 7(2) is the inclusion of the bottom cell*
*. Since
~H2O ~4 induces an isomorphism on ß6, the subgroup ß6(S5) = Z=2 is a summand of
ß6( P 4(2)) or the above short exact sequence for ß7(P 4(2)) splits. Thus
ß7(P 4(2)) = Z=2 Z=2
___
generated by ~4 O j and 2 0.
3. ß8(P 4(2)): By the commutative diagram
ß8(S7) _____________ß8(S4)
6 6
 
0 
 
i*  4 q* 4 ffi0* 3
ß8(S3) _____________ß8(X )____________ß8(P (2))___ß7(S )
6 6
 
 
 
 
 
ß7( S3) ß7(B4) ß7(Q)== ß8(F 4{2})
6 6
 
ffi* 
 
 
ß9(S7) _____________ß9(S4),
we have
ffi*(j2) = 0O j2 + (OE1)* O [2]*(j2) = 0O j2 6= 0.
Thus
ß8(X4) ~=ß7(B4) ß7(Q) = Z=2 Z=2
and there is an exact sequence
i* q* ffi0* 3
ß8(S3) ____ß7(B4) ____ß7(Y4) ____ß7(S ) ____0.
Since i*( 0O j2) = f*(2( 0O j2)) = 0, there is a short exact sequence
q* ffi0* 3
0 ____ß7(B4) ____ß7(Y4) ____ß7(S ) ____0.
Recall that (~4)*('5) = [u, v]. There is a commutative diagram
q* 4
ß7(B4)___________ß7(Y4) _______________ß8(P (2))
6 
~= H
 2*
 
 ~ g ?
Z=2 = ß7(S5)___*ß7( (P 6(2) _ S6))__*ß7( (P 3(2) ^ P 3(2)),
E 6 6 6
where ~ : S5 ! (P 6(2) _ S6) is the composite S5 ____ S ___ (P (2) _ S ) and
g 3 3 7
P 6(2) _ S6 ____ P (2) ^ P (2) ____S
66 J. WU
is a cofibre sequence. Observe that
g*: ß8(P 6(2) _ S6) = Z=4 Z=2 ____ß8(P 3(2) ^ P 4(2)) = Z=4 Z=2
is an isomorphism. There exists a homomorphism _ :ß7( (P 3(2) ^ P 3(2))) ! Z=2
so that
ß7(Y4) ~=ß7(B4) ß7(S3)
_____
generated by ~4 O j2 and 0O j. Thus
ß8(P 4(2)) = ß7(Y4) ß7(Q) = Z=2 Z=2 Z=2.
4. ß9(P 4(2)): Consider the commutative diagram
ß9(S7)_____________ß9(S4)
6 6
 
0 
 
i* 4 q* 4 ffi0* 3
ß9(S3) = 0___________ß9(X ) ____________ß9(P (2))____ß8(S ) ________0
6 6
 
 
 
 
 
ß8( S3) ß8(B4) ß8(Q)== ß9(F 4{2})
6 6
 
ffi* 
 
 
ß10(S7) _____________ß10(S4) .
Since ß8(B4) ~=ß8( S6) = Z=8, we have ffi*( 7) = 2 6 and so
ß9(X4) ~=ß8(B4)=2 ß8(Q) = Z=2 Z=2
with a short exact sequence
0 ____ß8( S6)=2 ____ß8(Y4) ____ß8(S3) ____0.
Now consider the commutative diagram
ß8( S6)=2 ____ß8(Y4)___________ß8( P 4(2))
6 
 
 H~2*
 
 
 g ?
ß8( S6) _________________*ß8( P 7(2)) = Z=2 Z=2.
Since g*( 6) 6= 0, the above short exact sequence for ß8(Y4) splits and theref*
*ore
ß8(Y4) ~=ß8 S6=2 ß8S3.
Thus ß9(P 4(2)) ~=ß8( S6)=2 ß8(S3) ß8(Q) = Z=2 Z=2 Z=2.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 67
5. ß10(P 4(2)): Consider the commutative diagram of exact sequences
ß10(S7)_____________ß10(S4)
6 6
 
 
 
 
i*  4 q*  4 ffi0* 3
ß10(S3) = 0___________ß10(X )___________ ß10(P (2))___ß9(S ) = 0
6 6
 
 
 
 
 
ß9( S3) ß9(B4) ß9(Q)== ß10(F 4{2})
6 6
 
ffi* 
 
 
ß11(S7) = 0___________ß11(S4).
Since Ker(ffi*: ß10(S7) ! ß9(F 4{2})) = 4 . ß10(S7), we have ß10(X4) ~= ß10(P 4*
*(2))
with short exact sequences
0 ____ß9(B4) ß9(Q) ____ß10(P 4(2)) ____4 . ß10(S7) ____0 and
0 ____ß9(B4) ____ß9(Y4) ____4 . ß10(S7) ____0.
Recall that ß9(B4) ~=ß9(P 10(2)) = Z=2. Let ff be the nonzero element in ß9(B*
*4).
Then the Hurewicz image H(ff) = [[u, v], u], u] in H*(B4) and so in H* P 4(2). *
*Thus
ff is not divisible by 2 in ß10(P 4(2)) and
ß9(Y4) ~=ß9(B4) 4 . ß10(S7)
____
generated by Samelson product <<~4, u>, u> and 4 24. Hence
ß10(P 4(2)) ~=ß9(B4) 4 . ß10(S7) ß9(Q) = Z=2 Z=2 Z=4.
68 J. WU
6. ß11(P 4(2)): Consider the commutative diagram of exact sequences
ß11(S7) = 0____________ß11(S4)
6 6
 
 
 
 
i*  4 q*  4 ffi0* 3
ß11(S3) = Z=2 ____________ß11(X )____________ ß11(P (2))___ß10(S ) = 0
6 6
 
 
 
 
 
ß10( S3) ß10(B4) ß10(Q)== ß11(F 4{2})
6 6
 
ffi* 
 
 
ß12(S7) = 0____________ß12(S4).
Recall that ß11(S3) = Z=2 generated by ffl3 and i*(ffl3) = f* O [2]*(ffl3) = f**
*(2ffl3) = 0.
Thus
ß11(P 4(2)) ~=ß11(X4) ~=ß11(F 4{2}) = Z=2 Z=4 Z=2 Z=2.
7. ß12(P 4(2)): Consider the commutative diagram of exact sequences
~=
ß12(S7) = 0____________ß12(S4)_____ß11(S3)w
6 6 ww
  ww
 p* w
  ww
  w
i*  4 q* 4 ffi0* 3
ß12(S3) _____________ß12(X )____________ß12(P (2)) ____ß11(S )_______0
6 6
 
 
 
 
 
ß11( S3) ß11(B4) ß11(Q)== ß12(F 4{2})
6 6
 
ffi* 
 
 
ß13(S7) = Z=2___________ß13(S4).
Observe that
ffi*( 2) = 0O 26+ (OE1)* [2]*( 26) = 0O 26= 0
and 2ß12S3 = 0. Thus
ß12(X4) ~=ß11( S3) ß11(B4) ß11(Q)
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 69
with a short exact sequence
ffi0* 3
0 ____ß11( S3) ß11(B4) ____ß11(Y4) ____ß11(S ) = Z=2 ____0.
Now let ff 2 ß11(Y4) ß11( P 4(2)) so that ffi0*(ff) = ffl3. Then p*(ff) = ffl*
*4 2 ß11( S4).
Consider the commutative diagram
p* 4
ß11( F 4{2})____ß11( P 4(2))___ ß11( S )
  
  
  
 E1 E1
  
  
? ? p ?
ß11(Q(S2)) ___ß11(Q(P 3(2)))__*ß11(Q(S3))
Then ffi0*(E1 ff) = E1 ffl3 = ffl and E1 (2ff) = 2E1 ff = ffl O j 6= 0 in ß11Q(*
*P 3(2)). It
follows that 2ff 6= 0. Since 2ß11 S3 = 2ß11B4 = 0, we have
1 3 3
ß11(Y4) ~=__ß11(S ) ß11(B4) ß11( S )=(ffl3 O j) ~=Z=4 Z=2 Z=2 Z=2
2
generated by 1=2ffl3, ~3, ~4O 26, fi, where fi is the image of the generator fo*
*r ß11(P 10(2))
with 2fi = 0 in ß11( P 4(2)). Hence
ß12(P 4(2)) ~=ß11(Y4) ß12(P 9(2)) ß12(CP2 ^ P 11(2)) ~=(Z=2) 5 Z=4.
We finish the calculation.
5.3. The Homotopy Groups ß*(P 5(2)). Now we compute ß*P 5(2). By Proposi
tion 4.15, there is a cofibre sequence
2!4 4 5
S7 ____S ____ sk8(F {2})
and so there is a homotopy commutative diagram of fibre sequences
F 5{}==== F 5{2} _______* ________F 5{2}
6 6 6
  
j OE1 j
  
  2! 
S4 ________X5 ________4S4.
Let OE2 = [S3, j] : L3(P 4(2)) ^ S3 = P 14(2) ! F 5{2} be Samelson product and*
* let OE
denote the composite
ffi1xffi25 5 multi. 5
X5 x P 14(2) _____ F {2} x F {2} ____ F {2} ____A5.
Lemma 5.9. The map OE* : Hr(X5xP 14(2)) ! Hr(A5) is an isomorphism for r 13
and therefore OE* : ßr(X5 x P 14(2)) ! ßr(A5) is an isomorphism for r 12.
Proof.The proof follows by observing that the primitive elements P ~Hr(A5) for *
*r 14
has a basis u, u2, [u, v], u4, [[[u, v], u], u], [u, v]2, [[[u, v, ], v, ], u] *
*with fi2[u, v] = u2 and
Sq1*[[[u, v], v], u] = [[[u, v], u], u].
There is an overlap between the following theorem and [28, Theorem 1.2, p.64].
70 J. WU
Theorem 5.10. The homotopy groups ßr(P 5(2)) for 7 r 13 are as follows.
________________________________________________
 ß (P 5(2))  ß (Y )  ß (P 12(2)) 
_r______________r1__5_________r__________
 r = 7  Z=2 Z=4  
   
 8  Z=2 Z=2  
 9  Z=2 Z=2  
 10  Z=2 Z=4  
 11  Z=2 Z=4  Z=2 
 12  Z=2 Z=2  Z=2 
______13____Z=2___Z=2___Z=4_________Z=4___
Proof.1. ß7(P 5(2)): Consider the exact sequence
(2!4)* 4 j* 5
ß7(S7) = Z _____ß7(S ) = Z Z=4 ____ß6(X ) ____0.
Since
(2!4)*('7) = !4 O (2'7) = 2(2 4  E 0) = 4 4  2E 0,
we have
ß6(X5) ~=Z=8 Z=2
generated by j*( 4) and j*(!4). Now consider the exact sequence
ffi* 5 5 i* 5 p* 5
ß8(S5) ____ß7(F {2}) ~=ß6(X ) ____ß7(P (2)) ____ß7(S ) ____0
and the commutative diagram
ffi* 5
ß8(S5) ____ß7(F {2})
6 6
 
E j*
 
 [2] 
ß7(S4) ______*ß7(S4).
Since [2]*(v4) = 2 4 + !4, we have
ffi*( 5) = ffi* O E( 4) = 2j*( 4) + j*(!4).
Thus there is a short exact sequence
0 ____Z=4(i*j*( 4)) ____ß7(P 5(2)) ____Z=2 ____0
The homotopy commutative diagram of fibre sequences
F 5{2} ______P 5(2)______S5
  
  
  
  
  
  
? ? ?
Q(S4) ____Q(P 5(2))____Q(S5)
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 71
induces a commutative diagram of exact sequences
0 _________Z=4 _______ß7(P 5(2))______Z=2 _________0
  ww
  ww
 1 w
 E w
  ww
  w
? ?
0 _________Z=2 _____ß7(Q(P 5(2)))_____Z=2 ________0.
Since E1 (i*j*( 4)) 6= 0 and ß7(Q(P 5(2))) = Z=2 Z=2, the element i*j*( 4) is*
* not
divisible by 2 and so
ß7(P 5(2)) ~=Z=4 Z=2
with the epimorphism E1 :ß7(P 5(2)) ____ßs7(P 5(2)).
2. ß8(P 5(2)): By expecting the exact sequence
(2!4)*=0 4 5
ß8(S7) _______ß8(S ) ____ß7(X ) ____0,
we have
ß8(S4) ~=ß7(X5) ~=ß7(A5) ~=ß8(F 5{2}).
Now consider the commutative diagram
ffi* 5 5 5
ß9(S5) ____ß8(F {2}) ___ß8(P (2))___4 . ß8(S_)____ 0
66 6
 
E ~=
 
 [2] 
ß8(S4) ______*ß8(S4),
where the top arrow is exact. Since [2]*( 4 O j) = 2 4 O j + !4 O j = E 0O j a*
*nd
ß8(S4) = Z=2( 4 O j) Z=2(E 0O j), there is a short exact sequence
Z=2( 4 O j) ____ß8(P 5(2)) ____4 . ß8(S5) ____0
with the monomorphism ffi* : ß9(S5) ____ß8(F 5{2}). By the homotopy commutative
diagram of fibre sequences
F 5{2} ______P 5(2)______S5
  
  
  
 E E
  
  
? ? ?
F 6{2} ____ P 6(2) ______ S6,
there is a commutative diagram of exact sequences
ffi* 5 5 5 5
ß9(S5)_____ß8(F {2}) ___ß8(P (2))_____ß8S _____ß7(F {2})
    
    
    
   ~= 
    
    
? ffi ? ? ? ?
0 = ß10(S6) ___*ß9(F 6{2})_ß9(P 6(2))___ß9(S6)____ß8(F 6{2}).
72 J. WU
By Proposition 4.16 and Lemma 4.17, there is decomposition
F 6{2} ' S5 x B6 x Q6,
where both B6 and Q6 are 8connected. Thus ßr(S5) ~=ßr(F 6{2}) for r 9 and so
there is a commutative diagram of exact sequences
0 ______Z=2( 4 O j)___ß8(P 5(2))__4 . ß8(S5)____0
  
  
~= E ~=
  
  
? ? ?
0 ______Z=2( 5 O j)___ß9(P 6(2))__4 . ß9(S6)____0.
Hence ß8(P 5(2)) ~=ß9(P 6(2)). We show that 2.ß9(P 6(2)) = 0. Recall that the d*
*egree
2 map [2] : P 6(2) ! P 6(2) is the composite
pinch 6 ''5 5 6
P 6(2) ____S ____S ____P (2)
and j5 O 6 = 0 ( see [42, pp.44]). Thus [2]* : ß9(P 6(2)) ! ß9(P 6(2)) is zero*
*. By the
distributivity law [2],
[2]* = 2 + ( W2 O H2)* : ß8( P 6(2)) ____ß8( P 6(2)).
(Note the dimension restrictions in this last equation.) Since the composite
E 6 H2 5 5 W2 6
P 5(2) ____ P (2) ____ (P (2) ^ P (2)) ____ P (2)
is null homotopic, the composite
E 6 ( W2OH2)* 6
ß8(P 5(2)) ____~=ß8( P (2)) ________ß8( P (2))
is zero and so is ( W2 O H2)* : ß8( P 6(2)) ! ß8( P 6(2)). Thus
2 . ß8( P 6(2)) = 0
and hence ß8(P 5(2)) ~=ß9(P 6(2)) ~=Z=2 Z=2.
3. ß9(P 5(2)): From the exact sequence
(2!4)* 4 5 7
ß9(S7) _____ß9(S ) ____ß8(X ) ____ß8(S ) ____0,
we obtain a short exact sequence
0 ____ß9(S4) ____ß8(X5) ____ß8(S7) ____0.
Consider the commutative diagram
ffi* 5 5
ß10(S5) ________________ß9(F {2})___ß9(P (2))______0
6 6
 
E ~=
 
 [2] 
ß9(S4)_____*ß9(S4)_____ß8(X5)
j*
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 73
Since [2]*( 4 O j2) = 2 4 O j2 + !4 O j2, we have ffi*( 5 O j2) = j*(!4 O j2) 6*
*= 0 and so
there is a short exact sequence
0 ______ß9(S4)=(!4wO j2)__ß8(X5)=j*(!4 O j2)___ß8(S7)________0
ww 
ww 
ww ~=
ww 
?
Z=2( 4 O j2) ß9(P 5(2))
The commutative diagram of fibre sequences
S5 ______S5{2} _______S5w
6 6 www
  ww
  w
  ww
  w
 
F 5{2} ______P 5(2)______S5
induces a commutative diagram
ß9(S4) _______ß9(F 5{2})
 
 
 
E 
 
? ?
Z=2 = ß9( S5) === ß9( S5),
where E is onto. Thus j*( 4 O j2) is not divisible by 2 in ß9(F 5{2}) and ther*
*efore
in ß9(P 5(2)). Hence ß9(P 5(2)) ~=Z=2 Z=2.
4. ß10(P 5(2)): Consider the exact sequence
(2!4)* 4 5 7
ß10(S7) _____ß10(S ) ____ß9(X ) ____ß9(S ) ____0.
H 9 P 4 E 5
By the EHP sequence ß12(S5) ____ß12(S ) ____ß10(S ) ____ß11(S ) ____0, we
have P ( 9) = 2 24and so (2!4)*( 7) = 4 24. By the commutative diagram
ß10(S4) = Z=8___ß9(X5) ~=ß10(F 5{2})
 
 
E 
 
?? ?
Z=2 = ß11(S5)======== ß11(S5),
the element 24is not divisible by 2 in ß9(X5) and so
ß9(X5) ~=ß10(S4)=4 ß9(S7) = Z=4 Z=2.
74 J. WU
Consider the exact sequence
ffi* 5 5
ß11(S5)____ß10(F {2})___ß10(P (2))______0
66 6
 
E 
 
 [2] 
ß10(S4)_____*ß10(S4).
Since [2]*( 24) = 2 24+ !4 O 7 = 4 24 0 in ß10(F 5{2}), we have ffi*( 25) = 0*
* and
so ß10(F 5{2}) ~=ß10(P 5(2)). Thus ß10(P 5(2)) ~=ß10(S4)=4 ß9(S7) ~=Z=4 Z=2.
5 ß11(P 5(2)): Consider the exact sequences
ffi* 5 5 5 0 5
ß12(S5) ____ß11(F {2}) ____ß11(P (2)) ____ß11(S ) ____ß10(F {2})
(2!4)* 4
ß11(S4) = 0 ____ß10(X5) ____ß10(S7) _____ß10(S ).
Since (2!4)*( 7) = 4 24, we have ß10(X5) ~= 2 . ß10(S7) ~= Z=4. By Lemma 5.9,
ß10(X5) ~=ß10(A5) and
ß11(F 5{2}) ~=ß10(A5) ß11(P 12(2)) ~=ß10(X5) ß11(P 12(2)) = Z=4 Z=2,
where P 12(2) = L3 is a common factor of F 5{2} and P 5(2). We need to det*
*er
mine the boundary ffi*: ß12(S5) ! ß11(F 5{2}). Consider the homotopy commutative
diagram of EHP sequences
P 4 E 5 H 9
2S9 ________S ________ S _______ S
6 6 6 6
   
   
   
   
 P  E  ~H 
2P 9(2) ______T 5_______ P 5(2)___2 P 9(2)
6 6 6 6
   
   
   
   
 P  E~  H~ 
2F 9{2} ______E5 ______ F 5{2} ____2 F 9{2}.
Observe that the inclusion S4 ____ F 5{2} lifts to E5 and the composite
H~2 9
L3(P 4(2)) ____ F 5{2} ____ F {2}
is null homotopic. Thus sk12(E5) ' sk12( S4 x P 12(2)) and so
ß10(E5) = ß11(S4) ß11(P 12(2)) = 0 Z=2 = Z=2.
It follows that the composite
Z=4 = ß10(X5) ____ß10( F 5{2}) ____ß10( F 9{2}) ~=ß11(S8) = Z=8
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 75
is a monomorphism. Now consider the commutative diagram of exact sequences
ffi* 5 5
ß12(S5) ____ß10( F {2})___ß10( P (2))
  
  
  
H H~2* H~2*
  
? ffi0 ? ?
ß12(S9) ____ß10( F 9{2})__ß10( P 9{2}).
Recall that ß12(S5) is generated by oe000with H(oe000) = 4 9, see [42, Lemma 5.*
*13]. We
obtain
~H2*O ffi*(oe000) = ffi0*O H(oe000) = ffi0*(4 9) = 8 7 = 0
in ß10(S7) ~=ß10( F 9{2}). Thus ffi*(oe000) = 0 and so there are short exact se*
*quences
0 ______ß10( F 5{2}) = Z=4 Z=2___ß10( P 5(2))___ß11(S5)w= Z=2______0
6 6 ww
  w
  ww
  w
  ww
[ [
0 ____________ß10(X5) = Z=4__________ß10(Y5)_____ß11(S5) = Z=2 ______0.
Now we need to solve the extension problem. By the distributivity law [2],
X
[2]* = 2+( W2)*O(H~2)*+ nff W3*O oe*OH3*: ß10( P 5(2)) ____ß10( P 5(2))
ff2 3
for some nff2 Z. Note that the diagram
(P 4(2))(3)___ L3 = P 12(2)
 
 
 
W3 
 
? ?
P 5(2)========= P 5(2)
commutes up to homotopy and ß10( P 12(2)) = Z=2. We have
2 W3*O oe* O H3*(ff) = 0
for any ff 2 ß10( P 5(2)). Recall that [2] : P 5(2) ! P 5(2) is the composite
P 5(2) ____ S5 ____ S4 ____ P 5(2).
Since ß10( S4) = 0, [2]* : ß10( P 5(2)) ____ß10( P 5(2)) is zero. Since
ß11(P 5(2) ^ P 4(2)) = Z=2 Z=2,
we have 2( W2 O H2)*(ff) = 0 for any ff 2 ß11(P 5(2)). Thus
4ff = 0
for all ff 2 ß11P 5(2) and so ß10(Y5) = Z=4 Z=2. Hence
ß11(P 5(2)) = ß10(Y5) Z=2 = Z=4 Z=2 Z=2.
6. ß12(P 5(2)): Consider the exact sequence
ß12(S7) = 0 ____ß12(S4) ____ß11(X5) ____0.
76 J. WU
We have ß12(S4) ~=ß11(X5) ~=ß11(A5). Now consider the commutative diagram of
exact sequences
ffi* j* 5
ß13(S5) ____ß11(A5)_____ß11(Y5)_____ß12(S )_______0
   
   
   
 E1 E1 E1
   
   
? 2 ? j ? ?
ßs13(S5)____ßs12(S4)___*ßs12(P 5(2))_ßs12(S5),
where
E1 :ß11(A5) ~=ß12(S4) ____ßs13S5 and E1 :ß12(S5) ____ßs12(S5)
are monomorphisms, see [42, Proposition 5.15, Theorem 7.1]. Observe that
E1 j*(ffl4) = j*(ffl) 6= 0
in ßs12(P 5(2)) = Z=2 3. Thus there is a short exact sequence
0 ____ß11(A5) ____ß11(Y5) ____ß12(S5) ____0.
Since E1 ß11(A5) is a summand of ßs12(P 5(2)), we have
ß11(Y5) = ß11(A5) ß12(S5)
and E1 :ß11(Y5) ____ßs12(P 5(2)) is a monomorphism. Thus ß12(P 5(2)) = ß11(Y5)
ß12(P 12(2)) = Z=2 3.
7. ß13(P 5(2)): Consider the exact sequence
(2!4)* 4 5 7
ß13(S7) _____ß13(S ) ____ß12(X ) ____ß12(S ) = 0.
Since 2 . ß13(S7) = 0, (2!2)*: ß13(S7) ____ß13(S4) is zero and so
ß13(S4) ~=ß12(X5) ~=ß12(A5).
Consider the commutative diagram of exact sequences
ffi* 5
ß14(S5) ____ß12(A5)_____ß12(Y5)_____ß13(S )_______0
   \
   
   
   
   
   
? 2 ? ? ?
ßs14(S5)____ßs13(S4)___ßs13(P 5(2))__ßs13(S5)_____0
By Theorem [42, Theorem 7.2], ß13(S4) ~=ßs13(S4) ~=Z= 3 generated by 34, ~4 and
j4 O ffl5. There is a commutative diagram of short exact sequences
0_______ß13(S4) _____ß12(Y5)____ß13(S5) = Z=2______0
  \
  
~=  
  
  
? ? ?
0_______ßs13(S4)____ßs13(P 5(2))___ßs13(S5)_________0
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 77
Thus ß12(Y5) ! ßs13(P 5(2)) is a monomorphism and
1 5 4
ß12(Y5) ~=__ß13(S ) ß13(S )=(j4 O ffl5) ~=Z=4 Z=2 Z=2.
2
Thus ß13(P 5(2)) = ß12(Y5) ß13(P 12(2)) = Z=2 Z=2 Z=4 Z=4. This complet*
*es
the calculation.
5.4. The Homotopy Groups ß*(P 6(2)).
Theorem 5.11. The homotopy groups ßr(P 6(2)) for r 14 are as follows.
1) ßr(P 6(2)) ~=ßsr(P 6(2)) for r 8.
2) ß9(P 6(2)) ~=ß8(P 5(2)) = Z=2 Z=2 maps onto ßs9(P 6(2)).
3) ß10(P 6(2)) = Z=8 maps onto ßs10(P 6(2)) = 0.
4) ß11(P 6(2)) = Z=2 Z=2 maps onto ßs11(P 6(2)) = Z=2.
5) ß12(P 6(2)) = Z=2 Z=2 with the stable image Z=2 in ßs12(P 6(2)) = Z=2 Z=*
*2.
6) ß13(P 6(2)) = Z=2 Z=2 Z=2 with the stable image Z=2 2 in ßs13(P 6(2)*
*) =
Z=2 3.
7) ß14(P 6(2)) = Z=2 4 Z=4 with the stable image Z=2 2 Z=4 in ßs14(P 6(2)) =
Z=2 Z=4 Z=4.
An application is as follows.
Proposition 5.12. Let X be a simply connected CW complex. Suppose that the
integral homotopy H5(X; Z) = Z=2 and the mod 2 cohomology H*(X) is an exterior
algebra E(u, v) with u = 5 and v = 6. Then X is homotopy equivalent to ø(S6*
*).
In other words, the Stiefel manifold ø(S6) is uniquely determined, up to homoto*
*py,
by its fundamental group and its cohomology ring.
Proof.By the assumption, X = P 6(2) [f e11. Since H*(X) ~=E(u, v), the homology
H*( X) is isomorphic to the polynomial algebra S(u, v). Let f0: S9 ! P 6(2) be
the adjoint of the attaching map f :S10 ! P 6(2). Then f*('9) = [u4, v5] and t*
*he
homotopy class [f] is not divisible by 2 in ß10(P 6(2)). Let ~6: S10 ! P 6(2) *
*be
the attaching map for ø(S6). By assertion 3, f ' k~6 for k = 1 or 3. Clearly
X ' ø(S6) if k = 1. By Lemma 2.18, there is a homotopy commutative diagram
~6 6
S10_______P (2)
 
 
 
[3] [1]
 
? ~ ?
S10_______6P 6(2).
Thus X ' ø(P 6(2)) if k = 3. If k = 3, then X ' P 6(2) [f e11' ø(S6) and hence
the result.
Proof of Theorem 5.11.1. ßrP 6(2) for r 9: Since S5 is the 9skeleton of F 6*
*{2},
assertion (1) follows. Assertion (2) was proved in the Step 2 of Theorem 5.10.
2. ß10(P 6(2)): By Lemma 4.17, A6 ' S5 x B6. Since S9 is the 16skeleton of
B6, ßr( S5 x S9) ~= ßr(A6) for r 15. Observe that L3(P 5(2)) = P 15(2). The
78 J. WU
homotopy groups ßr(P 6(2)) ~=ßr1(Y6) for r 13 and ß14(P 6(2)) = ß13(Y4) *
* Z=2.
Now consider the commutative diagram of exact sequences
ffi* 6 6 6
ß11(S6)____ß10(F {2}) ___ß10(P (2))____ß10(S ) = 0
   
   
   
H ~H2* H~2* H
   
? 2 ? ? ?
ß11(S11)___ß10(F 11{2})__ß10(P 11(2))__ß10(S11) = 0.
Since H :ß11(S6) ! ß11(S11) is of degree 2, the composite
ffi* 6 5 9 proj. 9
ß11(S6) ____ß10(F {2}) ~=ß9( S x S ) ____ß9(S )
is of degree 4. Consider the exact sequence
ffi* 6
0 ____Z ____Z=2 Z ____ß10(P (2)) ____0,
where ffi*(!6) = k 5 O j2 + 4'9 for some k = 0, 1. By [12, Theorem 2.2], th*
*ere is a
Z=8summand in ß10(P 6(2)). Thus k = 1 and ß10(P 6(2)) = Z=8.
3. ß11(P 6(2)): Consider the commutative diagram
ß12(S6)____ß11(F 6{2})__ß11(P 6(2))_____0
6 6
~= 
 
 
 [2] 
ß11(S5)______*ß11(S5)
Recall that [2] ' 2 + !5 O H2: S5 ! S5. Thus [2]*( 25) = 2 25= 0 and
ß11(P 6(2)) ~=ß11(F 6{2}) = Z=2 Z=2.
Since ß11(S5) ~=ßs11(S5), the map ß11(P 6(2)) ! ßs11(P 6(2)) is onto.
4. ß12(P 6(2)): Consider the commutative diagram
ffi* 6 5 9 6 6
Z=4 = ß13(S6) ___ß12(F {2}) = ß12(S ) ß11(S )__ß12(P (2))____ß12(S )______*
*0
   
   
   
 E1  ~=
   
   
? 2 ? ? ?
ßs12S5 = Z=16___________ßs12S5 = Z=16___________ßs12(P 6(2))__ßs12(S6)_____*
*0
where ß13(S6) ____ßs12(S5) is a monomorphism by [42, Proposition 5.15]. Thus
E1 O ffi*(oe00) = 8oe 6= 0
and so there is a short exact sequence
0 ____Z=2 ____ß12(P 6(2)) ____Z=2 ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 79
Consider the commutative diagram of exact sequences
Z=2 _____ß12(F 6{2}) = Z=2 2______ß12(P 6(2))_________ß12(S6)__________0
  
  
~H2* H~2* 0H
  
?? ? ?
0 _______ß12(F 11{2}) = Z=2__ß12(P 11(2)) = Z=4___ß12(S11) = Z=2______0.
We have ß12(P 6(2)) = Z=2 Z=2. By [12, Lemma 2.1], the composite
~6 6 5
S9 ____ P (2) ____Q(P (2))
is null homotopic and so there is an exact sequence
E1 s 6
0 ____Z=2 ____ß12(P 6(2)) ____ß12(P (2)) ____Z=2 ____0.
5. ß13(P 6(2)): Consider the commutative diagram
[2]* 5
Z=2 = ß13(S5) ____________ß13(S )
 
 
 
 
 
 
? ffi ?
Z=2 Z=8 = ß14(S6) _______*ß13(F 6{2})_____ß13(P 6(2))__2 . ß13(S6)____ 0
   
   
   
 H2*  0
   
   
? 2 ? ? ?
Z=8 = ß14(S11)______Z=8 = ß13(F 11{2})___ß13(P 11(2))__ß13(S11)_______0,
where ß14(S6) ! ß14(S11) and ß13(F 6{2}) ! ß13(F 11{2}) are epimorphisms and
E :ß13(S5) ! ß14(S6) is a monomorphism. Observe that
ß13(F 6{2}) ~=ß12( S5xS9) = Z=2 Z=8 and [2]*(ffl5) = 2ffl5+( !5OH2)*(ffl5) *
*= 0.
Thus ffi*(~ 6) = 2~6O 9+kffl5 for some k = 0, 1 and ffi(ffl6) = 0. Consider th*
*e commutative
diagram
ffi* 6
ß14(S6)____ß13(F {2})
 
 
 
 E1
 
 
? 2 ?
ßs13(S5)____ßs13(S5)
Since E1 (~6) = 0, we have E1 (~6 O 9) = 0 and so
E1 O ffi*(~ 6) = E1 (kffl5) = kffl = 2E1 (~ 6) = 2~ = 0
Thus k = 0 and ffi*(~ 6) = 2~6 O 9 with a short exact sequence
0 ____Z=2 Z=2 ____ß13(P 6(2)) ____2 . ß13S6 ! 0.
80 J. WU
Now consider the commutative diagram
ß13(P 6(2))_2 . ß13S6
6 6
 
 ~=E
 
 
 
ß12(P 5(2))___ß12(S5)
By Step 6 of Theorem 5.10, ß12(P 5(2)) ! ß12S5 is split onto and so
ß13(P 6(2)) ~=2 . ß13(S6) ß13(S5) ß12(S9)=2 ~=Z=2 3
with an exact sequence
0 ____Z=2 ____ß13(P 6(2)) ____ßs13(P 6(2)) ____Z=2 ____0.
6. ß14(P 6(2)): Consider the exact sequence
ffi* p* 6 ffi* 6
ß15(S6) ____ß13(A6) ____ß13(Y6) ____ß14(S ) ____ß13(F {2}),
where
Ker(ffi* : ß14(S6) ! ß13(F 6{2})) = {4~ 6} {ffl5} = Z=2 Z=2.
Since ß13(S9) = 0, we have ß13(A6) ~=ß14(S5). Consider the commutative diagram
j* p*
ß14(S5) = Z=2 3____ß13(Y6)____{4~ 6} {ffl6}____0
  
  
~=E1 E1 
  
  
? j ? p ?
0 ___________ßs14(S5)_____*ßs14(P 6(2))__*ßs14(S6)______0.
Thus ß14(S5) ____ß13(Y6) is a monomorphism. Let ff1 2 ß13(Y6) so that p*(ff1) *
*= ffl6.
Then E1 (2ff1) = j O ffl6 = j* O E1 (j5 O ffl6) 6= 0 and so ff1 is of order 4. *
* Now let
ff2 2 ß13(Y6) so that p*(ff2) = 4~ 6. Then there exists fi 2 ß14(S5) so that j**
*(fi) = 2ff2.
Since p* O E1 (ff2) = E1 4~ 6= 0, we have E1 (ff2) 2 ßs14(S5) and so
2E1 (ff2) = E1 (2ff2) = E1 j*(fi) = j*E1 (fi) = 0.
Thus fi = 0 or 2ff2 = 0. Hence
ß13(Y6) ~={4~ 6} 1=2{ffl5} ß14(S5)={j5 O ffl6} ~=Z=2 3 Z=4 and
ß14(P 6(2)) ~=ß13(Y6) ß14(P 15(2)) ~=Z=2 4 Z=4
with an exact sequence 0 ____Z=2 Z=2 ____ß14(P 6(2)) ____ßs14(P 6(2)). We *
*finish
the proof.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 81
5.5. The Homotopy Groups ß*(P 7(2)).
Theorem 5.13. The homotopy groups ßr(P 7(2)) for r 15 are as follows.
1) ßr(P 7(2)) ~=ßsrP 7(2) for r 10.
2) ß11(P 7(2)) = Z=2.
3) ß12(P 7(2)) = Z=2.
4) ß13(P 7(2)) = Z=2 Z=2.
5) ß14(P 7(2)) = Z=2 Z=4 Z=2 Z=2.
6) ß15(P 7(2)) = Z=2 Z=4 Z=4 Z=8.
Proof.1. ßr(P 7(2)) for r 10: Since S6 is the 11skeleton of F 7{2}, assertio*
*n (1)
follows.
2. ß11(P 7(2)): By Proposition 4.15, the 17skeleton sk17(F 7{2}) of F 7{2} i*
*s the
homotopy cofibre of 2!6: S11 ! S6. Consider the homotopy commutative diagram
of fibre sequences
F 7{2}=== F 7{2} _______* ________F 7{2}
6 6 6 6
   
j OE  j
   
   2! 
S6 ________X7 ________S11 ________6S6.
Let _ be the composite
ffixJ(fi3)7 7 multi. 7
X7 x L3(P 6(2)) ______ F {2} x F {2} ____ F {2}.
Then
_* : Hr(X7 x L3(P 6(2)) ! Hr( F 7{2})
is an isomorphism for r 19. Since L3(P 6(2)) = P 18(2), the map
_*: ßr(X7) ____ßr( F 7{2})
is an isomorphism for r 15. By expecting the exact sequence
(2!6)* 6 7
ß11(S11) _____ß11(S ) ____ß10(X ) ____0,
we have ß10(X7) = Z=2. From the exact sequence
ß12(S7) = 0 ____ß10(X7) ____ß11(P 7(2)) ____ß11(S7) = 0,
we have ß11(P 7(2)) ~=ß10(X7) = Z=2.
3. ß12(P 7(2)): Consider the exact sequence
(2!6* 6 7
ß12(S11) ____ß12(S ) ____ß11(X ) ____0.
Since ß12(S11) = Z=2, (2!6)* : ß12(S11) ! ß12(S6) is zero and so ß12(S6) ~=ß11(*
*X7).
Now consider the commutative diagram
ffi* 7 7
ß13(S7)____ß11(X ) ___ß12(P (2))______0
6 6
~= ~=
 
 
 [2] 
ß12(S6)____*ß12(S6).
82 J. WU
Recall that [2]* = 2 + ( !6 O H2)* and (H2)*( 26) = 0. Thus [2]*( 26) = 2 26= *
*0 and
so ffi*( 27) = 0. Hence ß12(P 7(2)) ~=ß11(X7) ~=Z=2.
4. ß13(P 7(2)): Consider the exact sequence
(2!6)* 6 j* 7 11
ß13(S11) _____ß13(S ) ____ß12(X ) ____ß12(S ) ____0.
Since (2!6)* : ß13(S11) ! ß13(S6) is zero, there is a short exact sequence
0 ____ß13(S6) ____ß12(X7) ____ß12(S11) ____0.
From the commutative diagram of exact sequences
[2]* = 2 7 7 7
ß13(wS7) _________ß13( S )___ß13(S ){2}____ß13(S )
ww 6 6 ww
ww   ww
ww f1* f2* ww
ww   www
ffi*  7  7 7
ß13( S7) _________ß12(X ) ___ß13(P (2))___ß13(S ),
we have f1*O ffi*(oe0) = 2oe0. Consider the commutative diagram of exact sequen*
*ces
ffi* 7 7 7
ß14(S7)__________ß12(X ) __________ß13(P (2))____ß13(S )
   
   
   
  H~2* 0
   
   
? ffi0 ? ? ?
ß14(S13)___*ß13(F 13{2}) ~=ß13(S12)_ß13(P 13(2))__ß13(S13)
Since ffi0*:ß14(S13) ! ß13(F 13{2}) is zero, the composite
ffi* 7 11
ß14(S7) ____ß12(X ) ____ß12(S )
is zero and so ffi*(oe0) = j*(koe00) for some integer k. Now
2oe0= (f1)* O ffi*(oe0) = (f1)* O j*(koe00) = kE(oe00) = 2koe0.
Thus k 1(4) and so ffi*(oe0) = j*(oe00). Hence there is a short exact sequen*
*ce
0 ____Z=2 ____ß13(P 7(2)) ____ß13(S7) ____0.
Consider the commutative diagram
ß13(P 7(2))____ß13(S7)______0
6 6
 
E ~=E
 
 
ß12(P 6(2))____ß12(S6)______0
By Step 4 of Theorem 5.11, we have ß12(P 6(2)) ~=ß12(S6) Z=2 and so
ß13(P 7(2)) ~=ß12(X7)=(ß13(S6)) ß13(S7) ~=Z=2 Z=2.
5. ß14(P 7(2)): Consider the exact sequence
(2!6)* 6 7 11
ß14(S11) _____ß14(S ) ____ß13(X ) ____ß13(S ) ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 83
E 7 H 13
By [42, Lemma 6.2], P ( 13) = 2~ 6in the EHP sequence for S6 ____ S ____ S *
* .
Thus (2!6)*( 11) = 4~ 6and so there is a short exact sequence
p* 11
0 ____Z=2 Z=4 ____ß13(X7) ____ß13(S ) ____0.
Consider the commutative diagram
j* 7 p* 11
ß14(S6) ____ß13(X )____ß13(S )_______0
 
 
 
E f1*
 
? ?
ß15(S7)==== ß15(S7).
Let ff 2 ß13(X7) so that p*(ff) = j2. Then 2ff = j*(k~ 6+ lffl6) for some integ*
*ers k, l.
Then
E(k~ 6+ lffl6) = (f1)* O j*(k~ 6+ lffl6) = (f1)*(2ff) = 0 = k~ 7+ lffl*
*7.
It follows that l = 0 and k 0(2). Now p*(ffj*(k=2~ 6)) = j2 and 2(ff  j*(k=*
*2~ 6)) =.0
Thus
ß13(X7) ~=ß13(S11) Z=2 Z=4 ~=Z=2 Z=2 Z=4.
Now consider the exact sequence
ffi* 7 7 7 ffi* 7
ß15(S7) ____ß13(X ) ____ß14(P (2)) ____ß14(S ) ____ß12(X ).
By Step 4, ffi*(oe0) = j*oe00and so Ker(ffi* : ß14(S7) ! ß12(X7)) = 4 . ß14(S7*
*). By [42,
pp. 64 and Theorem 7.1], we have
ß15(S7) = {oe0O j14} {~ 7} {ffl7} = Z=2 Z=2 Z=2 and oe00O j13= 4~*
* 6.
Hence
ffi*(oe0O j14) = j*(oe00O j13) = j*(4~ 6) = 0
in ß13(X7). Now
ffi*(~ 7) = ffi*E(~ 6) = j* O [2]*(~ 6) = j*(2~ 6+ ( !6 O H2)*(~ 6))
= j*(2~ 6+ ( !6)*( 11)) = j*(2~ 6+ P ( 13)) = j*(4~ 6) = 0.
and
ffi*(ffl7) = j* O [2]*(ffl6) = j*(2ffl6) + ( !6 O H2)*(ffl6)) = 0
This shows that the boundary map ffi*: ß15(S7) ____ß13(X7) is zero and so ther*
*e is
a short exact sequence
0 ____ß13(X7) ____ß14(P 7(2)) ____4 . ß14(S7) ____0.
84 J. WU
Consider the commutative diagram
ß14(P 7(2))__4 . ß14(S7)_____0
6 6
 
E ~=E
 
 
ß13(P 6(2))__2 . ß13(S6)_____0
6 6
 
 ~=E
 
 
 
ß12(P 5(2))___ß12(S5)________0
By Step 6 of Theorem 5.10, ß12P 5(2) ! ß12S5 is split onto. Thus
ß14(P 7(2)) ____4 . ß14(S7)
is split onto and so
ß14(P 7(2)) ~=4 . ß14(S7) ß13(X7) = Z=2 Z=2 Z=2 Z=4.
6. ß15(P 7(2)): Consider the exact sequence
(2!6)* 6 7 11 (2!6)* 6
ß15(S11) = 0 _____ß15(S ) ____ß14(X ) ____ß14(S ) _____ß14(S ),
where Ker((2!6)* : ß14(S11) ! ß14(S6)) = 2.ß14(S11). There is a short exact seq*
*uence
0 ____ß15(S6) ____ß14(X7) ____2 . ß14(S11) ____0.
Now consider the commutative diagram
ß15(S6) ____ß14(X7)
 
 
 
E 
 
? ?
ß16(S7)==== ß16(S7).
By [42, Theorem 7.2], we have ß16(S7) ~=ß15(S6) Z=2 and so ß15(S6) is a summa*
*nd
of ß14(X7). Thus
ß14(X7) ~=ß15(S6) 2 . ß14(S11).
Consider the exact sequence
ffi* 7 7 7
ß16(S7) ____ß14(X ) ____ß15(P (2)) ____ß15(S ) ____0.
According to [42, Theorem 7.2],
ß16S7 = {oe0O j214} { 37} {~7} {j7 O ffl8}
= Z=2 Z=2 Z=2 Z=2.
Observe that
ffi*(oe0O j214) = ffi*(oe0O j14) O j15= 0
ffi*( 37) = ffi*E( 36) = j* O [2]*( 36) = j*(2 36+ ( !6 O H2)*( 36)) =*
* 0
ffi*(~7) = ffi*E~6 = j* O [2]*(~6) = j*(2~6 + ( !6 O H2)*(~6)) = 0
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 85
ffi*(j7 O ffl8) = ffi*E(j6 O ffl7) = j* O [2]*(j6 O ffl7) = j*(2j6 O ffl7 + ( !*
*6 O H2)*(j6 O ffl7)) = 0.
Thus ffi* : ß16(S7) ____ß14(X7) is zero and so there is a short exact sequence
0 ____ß14(X7) ____ß15(P 7(2)) ____ß15(S7) ____0.
Consider the commutative diagram of exact sequences
0_______ß14(X7) ___ß15(P 7(2))___ß15(S7)_______0
  
  
  
E1 E1 E1
  
? ? ?
0_______ßs15(S6)____ßs15(P 7(2))_ßs15(S7)______0,
where E1 : ß14(X7) ____ßs15(S6) and E1 : ß15(S7) ____ßs15(S7) are epimorphism*
*s.
It follows that
E1 :ß15(P 7(2)) ____ßs15(P 7(2))
is an epimorphism and so there is a commutative diagram of short exact sequences
p* 0
Z=4 __________________G _________Z=2 = {oe O j14}
\ \ \
  
  
  
  
  
? ? p* ?
Z=2 Z=2 Z=2 Z=4 _____ß15(P 7(2))_____Z=2 Z=2 Z=2
  
  
  
  
  
?? ?? ??
Z=2 Z=2 Z=2 _____Z=2 Z=2 Z=4 _____Z=2 Z=2,
where the rows and columns are short exact sequence. To solve the group extensi*
*on
problem, we need to compute ß15(P 7(2) ^ P 6(2)). According to [27, Lemma 1.6(i*
*i),
p.518], ß2n+3( P n(2) ^ P n(2)) = Z=4 Z=2 for n 4. In particular,
ß15(P 7(2) ^ P 6(2)) = Z=4 Z=2.
Now let ff 2 G so that p*(ff) = oe0O j14 in ß15(S7). From the commutative diagr*
*am
p* 7
ß15(P 7(2))_______ß15(S )
 
 
 
H2* H
 
? p ?
ß15(P 7(2) ^ P 6(2))_*ß15(S13),
we have
p* O H2*(ff) = H O p*(ff) = H(oe0O j14) = j213.
Observe that there is a short exact sequence
p* 13
0 ____Z=2 Z=2 = ß15(P 12(2)) ß15(S12)=2 ____ß15(P 7(2)^P 6(2)) ____ß15(S ) *
*____0,
86 J. WU
Thus 2H2*(ff) 6= 0 and so 2ff 6= 0. It follows that the short exact sequence
0 ____Z=4 ____G ____Z=2 ____0
does not split and so G = Z=8. By Barratt's exponent theorem [2, 7], we have
8ß15(P 7(2)) = 0 and so
ß15(P 7(2)) = Z=8 Z=2 Z=4 Z=4.
This completes the calculation.
Note. It was known in [12] that there is a family of Z=8summands in ß*(P n(2))
for n 3 by expecting certain spherical classes in H*( P n(2)). The Z=8summan*
*d of
ß15(P 7(2)) = ß14( P 7(2)) does not come from a spherical class because H14( P *
*7(2)) = 0.
It might be interesting to know whether this summand is detected by other homol*
*ogy
theory. According to [27, Proposition 4.2], a generator of Z=8summand of ß15(P*
* 7(2))
is given as ~j6O oe8, where ~j6is a generator for ß8(P 7(2)) = Z=4. According *
*to the
referee's comments, the other generators for ß15(P 7(2)) are the following: [~ *
*7], [ffl7]
and i7 6, where [ff] stands for a lifting of ff and [ffl7] = ~ffl6(a coextensio*
*n of ffl6).
5.6. The Homotopy Groups ß*(P 8(2)). Consider the homotopy commutative di
agram of fibre sequences
S7w________S15 ________S8
ww 6 6
ww  
 
www  pull p
ww  
i 8 q 8
S7 _________X _______P (2)
6 6
 
f 
 
 
F 8{2}===== F 8{2}
where S7 ! S15 ! S8 is the Hopf fibration.
(ff0,[2])7 14
Lemma 5.14. The space X8 is the homotopy cofibre of S14 ! S _ S and
therefore X8 is the 20skeleton of 2S9.
Proof.By Lemma 2.38, ~H*(X8) has a basis {x7, x14x15} with Sq1*x15= x14 Thus X8
is the homotopy cofibre of ffiS14:S14 ____ sk14(F 8{2}), where ffi : S15 ! F*
* 8{2} is
the boundary. Since sk14(F 8{2}) ' S7 _ S14, there is a cofibre sequence
_ 7 14 f 8
S14 ____S _ S ____X .
_ 7 14 proj.14
with the composite S14 ____S _ S ____S of degree 2. Let OE1 be the compo*
*site
_ 7 14 proj.7 0
S14 ____S _ S ____S . Then OE1 ' koe for some k. It suffices to show that*
* k 6 0
mod 2. Suppose that k were even. Let ~~8:S14 ! F 8{2} be the composite
S14 ____S7 _ S14 ' sk14(F 8{2}) ____F 8{2}.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 87
Then ~~08*('13) = [u, v] in H* F 8{2} and therefore in H* P 8(2), where ~~08is *
*the
adjoint map of ~~8. By [12, Theorem 2.2], the homotopy class
2[q O f O ~~8] 6= 0 and 4[q O f O ~~8] = 0
in ß14(P 8(2)). Since i: S7 ! X8 is of degree 2 on the bottom cell, we have
k 0 0
0 = q* O i*(__oe ) = q* O f*(koe ) = q*(2f*[~~8]) = 2[q O f O ~~] 6= 0
2
This is a contradiction and hence the result.
Corollary 5.15. There is an exact sequence
2.E2 2 9 8 7 2.E2 2 9
ßr(S7) ____ßr( S ) ____ßr(P (2)) ____ßr1(S ) ____ßr1( S )
for r 19.
Theorem 5.16. The homotopy groups ßr(P 8(2)) for r 16 are as follows.
1) ßr(P 8(2)) ~=ßsr(P 8(2)) for r 13.
2) ß14(P 8(2)) = Z=4 Z=2.
3) ß15(P 8(2)) = Z=2 Z=2 Z=2 Z=2.
4) ß16(P 8(2)) = Z=2 Z=2 Z=2 Z=4 Z=4.
Proof.1. ßr(P 8(2)) for r 13: Note that F 8{2} ' S7 _ S14 and ßr(S7) ~=ßsr(S7)
for r 13. The assertion (1) follows.
2. ß14(P 8(2)): By Corollary 5.15, there is a short exact sequence
0 ____Z=4 = ß14( 2S9)=4 ____ß14(P 8(2)) ____ß13(S7) = Z=2 ____0.
By [12, Theorem 2.2], there is a Z=4summand in ß14(P 8(2)) and so
ß14(P 8(2)) = Z=4 Z=2.
3. ß15(P 8(2)): Since 2ß15(S7) = 0 and the kernel of
2E2: ß14(S7) = Z=8 ____ß14( 2S9) = Z=16
is 4ß14(S7) ~=Z=2. There is a short exact sequence
0 ____ß17(S9) ____ß14(P 8(2)) ____4ß14(S7) ____0.
Consider the commutative diagram
E2 7 E 8
ß12(P 5(2))___ß14(P (2))__ß15(P (2))
  
  
  
  
  
  
? E2 ? ?
ß12(S5)____~4 . ß14(S7)==4 . ß14(S7).
=
By Step 6 of Theorem 5.10, the homomorphism ß12(P 5(2)) ! ß12(S5) is split onto
and so is ß15(P 8(2)) ! 4 . ß14(S7). Thus
ß15(P 8(2)) ~=ß17(S9) 4 . ß14(S7) ~=Z=2 Z=2 Z=2 Z=2.
88 J. WU
4. ß16(P 8(2)): Recall that ß15(S7) = Z=2 3 is generated by {oe0O j14, ~,7ffl*
*7} and
ß16(S7) = Z=2 4 is generated by {oe0 O j214, 37, ~7, j7 O ffl8}, see [42, Theo*
*rems 7.1
and 7.2]. Since 2ß16(S7) = 0, there is a short exact sequence
0 ____ß16( 2S9) ____ß16(P 8(2)) ____ß15(S7) ____0.
Consider the commutative diagram of fibre sequences
q 8 ffi 7 8
X8 ______ P (2) _______Sw_________X
6 6 www 6
  ww 
j OE ww j
  ww 
 g  p [2] 
S7 ______S7{2} _______S7 _________S7.
Since S7 is an Hspace, we have [2] ' 2 : S7 ! S7 and so
3(S7{2}) ' ( 3S7){2} = Map *(P 4(2); S7).
Thus the power map 2 : 3(S7{2}) ! 3(S7{2}) is homotopic to the composite
ids 3 7 idj 4 7 3 7
3(S7{2}) ' Map *(P 4(2); S7) ____ S ____ S ____ (S {2}),
where s: S3 ! P 4(2) is the canonical inclusion. Since 2ß15(S7) = 0, the map
p*: ß15(S7{2}) ! ß15(S7) is onto. Let ff 2 ß15(S7{2}) so that p*(ff) = oe0O j1*
*4.
Then 2ff = g*(oe0O j214) and so
2OE*(ff) = OE*(2ff) = OE* O g*(oe0O j214) = q* O j*(oe0O j214).
Consider the cofibre sequence
(ff0,[2])7 14 f 8
S14 _____S _ S ____X .
Then
j O oe0O j214= fS7 O oe0O j214' fS14O [2] O j214' *.
Thus j*(oe0O j214) = 0 and 2OE*(ff) = 0,that is, there exists ff1 2 ß16(P 8(2)*
*) so that
(1).ffi*(ff1) = oe0O j14 and
(2).2ff1 = 0.
Now consider the commutative diagram
ß16(P 8(2))== ß16(P 8(2))
6 
 
E ffi*
 
 p0 ?
ß15(P 7(2))____*ß15(S7)
By Step 6 of Theorem 5.13, the p0*:ß15(P 7(2)) ! ß15(S7) is onto. Let ff2, ff3*
* be
elements in ß15(P 7(2)) so that p0*(ff2) = ~ 7and p0*(ff3) = ffl7. Then
{ff2} {ff3} ~=E1 {ff2} E1 {ff3}
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 89
is a summand of ßs15(P 7(2)) and so E{ff2} E{ff3} is a summand of ß16(P 8(2)).
Thus there is a commutative diagram of short exact sequences
Z=2 Z=2 ================ Z=2 Z=2
6 6
 
 
 
 
 
Z=2 4 __________________ß16(P 8(2))__________________Z=2w3
6 6 ww
  ww
  w
  ww
  w
 
Z=2 Z=2 ___Z=2 Z=4 Z=4 = {ff1} {ff2} {ff3}_Z=2 3
and so ß16(P 8(2)) ~=Z=2 3 Z=4 2. We finish the calculation.
5.7. The Homotopy Groups ß*(P 9(2)).
Theorem 5.17. The homotopy groups ßr(P 9(2)) for r 17 are as follows.
1) ßr(P 9(2)) ~=ßsr(P 9(2)) for r 14.
2) ß15(P 9(2)) ~=Z=4 Z=2.
3) ß16(P 9(2)) ~=Z=2 Z=2 Z=2 Z=2.
4) ß17(P 9(2)) ~=Z=2 Z=2 Z=2 Z=4 Z=4.
Proof.1. ßr(P 9(2)) for r 14: Since S8 is the 15skeleton of F 9{2}. The a*
*sser
tion (1) follows.
2!8 8 j 9
2. ß15(P 9(2)): Consider the cofibre sequence S15 ____S ____ sk23(F {2}). Th*
*ere
is an exact sequence
(2!8)* 8 j* 9
ß15(S15) _____ß15(S ) ____ß15(F {2}) ____0.
Since (2!8)*('15) = 2 . (2oe8  Eoe0) = 4oe8  2Eoe0, we have
ß15(F 9{2}) = Z=16 Z=2
generated by j*(oe8) and j*(!8). Now consider the commutative diagram
ffi* 9 i* 9 p* 9
ß16(S9) ____ß15(F {2})___ß15(P (2))____ß15(S )______0
6 6
 
E j*
 
 [2] 
ß15(S8) _____*ß15(S8),
where E is onto. Since [2]*(oe8) = 2oe8 + !8, we obtain
ffi*(oe9) = ffi*Eoe8 = j*(2oe8 + !8) = 2j*(oe8) j*(!8),
and so there is a short exact sequence
0 ____Z=4 ____ß15(P 9(2)) ____Z=2 ____0.
90 J. WU
Notice that h(oe8) = '28in H14( S8). Thus h([1 O j O oe8]) = u2 in H14( P 9(2))*
* and
[i O j O oe8] is not divisible by 2. Hence ß15(P 9(2)) = Z=4 Z=2.
3. ß16(P 9(2)): Consider the exact sequence
(2!8)* 8 j* 9
ß16(S15) _____ß16(S ) ____ß16(F {2}) ____0.
Since 2ß16(S15) = 0, (2!8)* : ß16(S15) ____ß16(S8) is zero and ß16(S8) ~=ß16(F*
* 9{2}).
Now consider the exact sequence
ffi* 9 9 9
ß17(S9) ____ß16(F {2}) ____ß16(P (2)) ____8 . ß16(S ) ____0.
Observe that
ffi*(oe9 O j) = ffi*(oe9) O j = j*(!8 O j) = j*(Eoe0O j)
ffi*(~ 9) = ffi* O E2(~ 7) = j* O E(2~ 7) = 2j*(~ 8) = 0
ffi*(ffl9) = ffi* O E2(ffl7) = j* O E(2ffl7) = 0
Thus there is a short exact sequence
0 ____ß16(F 9{2})={Eoe0O j} ____ß16(P 9(2)) ____8 . ß16(S9) ____0 or
0 ____Z=2 Z=2 Z=2 ____ß16(P 9(2)) ____Z=2 ____0.
Consider the commutative diagram
ß16(P 9(2))____8ß16S9
6 6
 
 ~=
 
 
 
ß12(P 5(2))___ß12(S5).
Then ß16(P 9(2)) ____8 . ß16(S9) is split onto and so ß16(P 9(2)) = Z=2 4.
4. ß17(P 9(2)): Consider thee exact sequence
(2!8)* 8 j* 9 15
ß17(S15) _____ß17(S ) ____ß17(F {2}) ____ß16(S ) ____0.
Since 2 . ß17(S15) = 0, (2!8)* : ß17(S15) ! ß17(S8) is zero and so there is a s*
*hort
exact sequence
j* 9 15
0 ____ß17(S8) ____ß17(F {2}) ____ß16(S ) ____0.
By [42, Theorem 7.2], ß17(S8)={Eoe0O j215} ~= ß18(S9). Recall that the suspens*
*ion
E :S8 ! S9 factors through F 9{2}. Thus ß17(S8)={Eoe0O j215} is a summand of
ß17(F 9{2})=j*{Eoe0O j215} and so
ß17(F 9{2})=j*{Eoe0O j215} ~=ß17(S8)={Eoe0O j215} ß16(S15).
Now consider the exact sequence
ffi* 9 9 9 ffi* 9
ß18(S9) ____ß17(F {2}) ____ß17(P (2)) ____ß17(S ) ____ß16(F {2}).
By Step 3, Ker(ffi* : ß17(S9) ! ß16(F 9{2})) = {~} {ffl9} = Z=2 Z=2. Note t*
*hat
ffi*( 9 O j216) = ffi*( 9 O j16) O j17= j*(Eoe0O j215)
ffi*( 39) = ffi*E2 37= j*E(2 37) = 0
ffi*(~9) = ffi*E2~7 = j*E(2~7) = 0
ffi*(ffl9) = ffi*E2ffl7 = j*E(2ffl7) = 0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 91
Thus there is a short exact sequence
0 ____ß17(S8)={Eoe0O j215} ß16(S15) ____ß17(P 9(2)) ____{~ 9} {ffl9} ___*
*_0 or
0 ____Z=2 5 ____ß17(P 9(2)) ____Z=2 Z=2 ____0.
Consider the commutative diagram
ß17(P 9(2))__{~} {ffl9}____0
 
 
 
 ~=
 
 
? ?
ßs17(P 9(2))__ßs17(S9)_______0
 ww
 ww
 w
p w
 ww
 w
?
Z=4 Z=4 ___Z=2 Z=2 ______0,
where p is the projection. Thus ß17(P 9(2)) = Z=2 3 Z=4 2. This completes t*
*he
calculation.
5.8. The Homotopy Groups ß*(P n(2)) for n 10.
Theorem 5.18. The homotopy groups ßr(P n(2)) for n 10 and r n + 8 are as
follows.
1) ßr(P 10(2)) ~=ßsr(P 10(2)) for r 16.
2) ßr(P 11(2)) ~=ßsr(P 11(2)) for r 18.
3) ßr(P n(2)) ~=ßsr(P n(2)) for r n + 8 and n 12.
4) ß17(P 10(2)) = Z=2 4.
5) ß18(P 10(2)) = Z=8 Z=2 Z=4 Z=4.
6) ß19(P 11(2)) = Z=2 Z=2 Z=4 Z=4.
Proof.Assertions (1), (2) and (3) are obvious.
1. ß17(P 10(2)): Consider the exact sequence
ffi* 10 10 10
ß18(S10) ____ß17(F {2}) ____ß17(P (2)) ____8 . ß17(S ) ____0.
Since F 10{2} ' S9 _ S18, we have ß17(S9) ~=ß17(F 10{2}). Note that
ffi*(~ 10) = ffi*E3(~ 7) = j*E2(2~ 7) = 0
ffi*(ffl10) = ffi*E3(ffl7) = j*E2(2ffl7) = 0
Thus there is a short exact sequence
0 ____ß17(S9) ____ß17(P 10(2)) ____8 . ß17(S10) ____0.
92 J. WU
Consider the commutative diagram
ß17(P 10(2))__8 . ß17S10
6 6
 
 ~=
 
 
 
ß12(P 5(2))___ß12(S5).
By Step 6 of Theorem 5.10, the map ß12(P 5(2)) ! ß12(S5) is split onto. Thus
ß17(P 10(2)) ~=Z=2 4.
2. ß18(P 10(2)): Consider the commutative diagram of exact sequences
ffi* 10 j* 10 p* 10
ß19(S10)___ß18(F {2}) ___ß18(P (2))___ß18(S )_______0
  
  
  
H H~2* H2*
  
? ffi0 ? ?
ß19(S19)___*ß18(F 19{2})__ß18(P 19(2)).
By [42, Theorem 7.2], ß19(S10) = { ('21)} { 310} {~10} {ffl10~= Z Z=2 3*
*, where
= P : 2S21 ! S10. Note that
ffi*( 310) = ffi*(~10) = ffi*(ffl10) = 0
H~2*O ffi* ('21) = ffi0*O H ('21) = ffi0*(2'19) = 4'18.
Let ~~10: S18 ! F 10{2} be the composite
S18 ____S9 _ S18 ' F 10{2}(18)! F 10{2}.
By [12, Theorem 2.2], 4[j O ~~10] 6= 0 and 8[j O ~~10] = 0. Thus
ffi* ('21) = 4[~~10] + ff
for some ff 6= 0 in ß18(S9) and so there is a short exact sequence
0 ____ß18(S9)={ff} Z=8 ____ß18(P 10(2)) ____ß18(S10) ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 93
From the commutative diagram
Z=4 Z=4 === Z=4 Z=4 ___Z=2 Z=2 ______0
ww  
ww  
w  
ww  ~=
w  
ww  
ww ? ?
ww ß18(P 10(2))___ß18(S10)_______0
ww  
w  
ww  
w  ~=
ww  
w  
ww ? ?
www ßs18(P 10(2))__ßs18(S10)______0
ww  
ww  
ww  ~=
ww  
w ? ?
Z=4 Z=4 === Z=4 Z=4 ! Z=2 Z=2 ______0,
we obtain ß18(P 10(2)) ~=Z=8 Z=2 Z=4 Z=4.
3. ß19(P 11(2)): From the exact sequence
(2!10)* 10 11
ß19(S19) _____ß19(S ) ____ß19(F {2}) ____0,
we have ß19(F 11{2}) = ß19(S10)={2 ('21)}. Now consider the exact sequence
ffi* 11 11 11
ß20(S11) ____ß19(F {2}) ____ß19(P (2)) ____ß19(S ) ____0.
Since E4: ß16(S7) ! ß20(S11) is onto, ffi* : ß20(S11) ____ß19(F 11{2}) is zero*
* and so
there is a short exact sequence
0 ____ß19(F 11{2}) ____ß19(P 11(2)) ____ß19(S11) ____0.
By [42, Theorem 7.2], we have ß19(S11) ~= ßs19(S11). By Lemma 5.2, the group
extension for the short exact sequence above is essential and
ß19(P 11(2)) ~=Z=2 Z=2 Z=4 Z=4.
This completes the calculation.
5.9. Remarks on the generators for the homotopy groups. In this subsection,
we give some remarks along the suggestions given by the referee. Although our p*
*aper
essentially contains the classical methods completely, in the metastable range *
*the
classical method is very useful to settle generators of the group. The classica*
*l method
is to use the homotopy exact sequence of a pair (P n(2), Sn1) and to determine*
* the
relative homotopy group ß*(P n(2), Sn1) by using the James exact sequence [17,
Theorem 2.1]. By using this method, it seems not hard to determine the group
structures and the generators of ßn+r(P n(2)) for r 2n5 if ßn+r(Sm ) (m = n*
*1, n)
is given by Toda's composition methods. (Note. When n is small, for instance n *
*= 3
or 4, the range up to 2n  5 is not so good.) The following example was given b*
*y the
referee.
94 J. WU
Example 5.19. ß19(P 11(2)) ~=(Z=4) 2 (Z=2) 2, which is generated by the elements
~ffl10,~~ 10, i11~10 and i11O ['10, '10].
Proof.We consider the homotopy exact sequence of a pair (P 11(2), S10):
@ 10 i11* 11 j* 11 10
ß20(P 11(2), S10) ____ß19(S ) ____ß19(P (2)) ____ß19(P (2), S ).
By BlakersMassey's theorem,
ß19(P 11(2), S10) ~=ß19(S11) ~=Z=2 Z=2.
So j* is equivalent to the homomorphism p11*:ß19(P 11(2)) _____ß19(S11). Since
ß19(S11) = ß18(S10) and ß18(S10) = 4ß14(S6), there exist coextensions ~ffl10a*
*nd ~~ 10
of ffl10 and ~ 10satisfying p11*(~ffl10) = ffl11 and ß11*(~~ 10) = ~ 11respecti*
*vely. By [17,
Theorem 2.1], we obtain
ß20(P 11(2), S10) = {[!, '10]} ~=Z,
where ! 2 ß11(P 11(2), S10) is the characteristic map of the 11cell of P 11(2)*
* and [ , ]
stands for the relative Whitehead product. We obtain
@[!, '10] = [2'10, '10].
Thus, by the fact that
ß19(S10) ~=Z{[', ']} Z=2{~10} Z=2{ 310} Z=2{j10ffl11},
we conclude that
ß19(P 11(2)) ~=(Z=4) 2 (Z=2) 2,
which is generated by elements ~ffl10, ~~ 10, i11~10 and i11O ['10, '10]. The *
*relations are
the following:
2~ffl10= i11j10ffl11 and 2~~ 10= i11 310.
Here we note that ~ffl10and ~~ 10are chosen as suspended, so, by the fact [2] =*
* i11j10p11,
we obtain 2~ffl10= [2]*~ffl10= i11j10p11O ~ffl10= i11j10ffl11 and
2~~ 10= [2]*~~ 10= i11j10p11~~ 10= i11j10~ 11= i11 310.
Note. ~ffl10= 7~ffl3and ~~ 10= 4[~ 7].
6. The Homotopy Theory of RP2
In this chapter, we study the homotopy theory of the special space P 3(2) = *
*RP2.
This chapter will be divided into six sections. In the first section, we consid*
*er certain
canonical fibrations over RPn. In section 2, we consider the special case whe*
*re
n = 2. In this case, we obtain a product decomposition of the triple loop space
of RP2. This decomposition theorem will help us to compute the homotopy groups
of RP2. In section 3, we prove that the suspension E :ß*(RP2) ! ß*+1( RP2) is *
*the
trivial homomorphism for * 3. This information has been to help us to find ce*
*rtain
nonsuspension coHspaces X which admits the form P 3(2) [f en. An application
of our results to the homogeneous space SU(3)=SO(3) is given in section 4. We
will start to compute the homotopy groups from section 5. From our decomposition
theorem, the homotopy groups of ß*( RP42) for * 4 is a summand of ß*(P 3(2))
and so we first compute ß*( RP42) in section 5 and then do ß*(P 3(2)) in sectio*
*n 6.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 95
6.1. Fibrations over RPn. Let fln: RPn ! BSO(n + 1) be the adjoint map of
the canonical inclusion RPn ____SO(n + 1) = BSO(n + 1), where BSO(n + 1) is
the classifying space of SO(n + 1). Let the space Zn be the pullback of the di*
*agram
Zn ______ESO(n + 1)
 
 
 
 pull q
 
 
? fln ?
RPn ____BSO(n + 1),
where SO(n + 1) ! ESO(n + 1) ! BSO(n + 1) is the universal principle SO(n + 1)
bundle. We obtain a principal SO(n + 1)bundle over RPn. For instance,
Z1 ____ RP1 = S2
is the classical Hopf fibration. From now on, we assume that n 2.
Let F be any space with SO(n + 1)action. We have the induced fibre bundle
F ____Zn xSO(n+1)F ____ RPn.
Let Yn = Zn xSO(n+1)Sn. Then there is a homotopy commutative diagram of fibre
sequences
SO(n + 1) ______Snw______BSO(n) ___BSO(n + 1)
w
6 ww 6 6
 ww  
 w  pull fln
 ww  
 w  
 @  p 
RPn _______Sn _________Yn_________ RPn
6 6 6 6
   
   
   
   
   
   
Zn _________* _________Zn========= Zn.
@ n
Observe that the composite RPn ____ RPn ____S is the pinch map and so its
homotopy cofibre is RPn1. By Lemma 2.39, there are cofibre sequences
`n n
SO(n + 1)=RPn ____Zn ____ RP ^ SO(n + 1) and
`~n n n n+1 n
RPn1 ____Yn ____ RP ^ S = RP ,
where `n is a monomorphism in mod 2 homology and ~`nis an epimorphism in mod 2
homology. Furthermore the map Sn ! Yn is the composite
q n1
Sn ____ RP ____Yn,
where q :Sn1 ! RPn1 is the quotient map. Observe that the composite
p n
RPn1 ____Yn ____ RP
is the canonical inclusion. Recall that the Thom space Cn of the vector bundle
Zn xSO(n+1)Rn+1 ! RPn
96 J. WU
is the homotopy cofibre of the map p: Yn ! RPn. We have the homotopy commu
tative diagram of cofibre sequences
RPn1w ______Yn______ n+1RPn
ww  
ww  
ww  
ww pn ffin
w  
? ?
RPn1 _____ RPn ______Sn+1

  
  
  
  
  
  
? ? ?
*_________Cn ======== Cn.
We will give some information on H*(Cn) and some properties of the map
ffin: n+1RPn ! Sn+1.
We need a lemma. A space X is called stably atomic if X does not have a nontriv*
*ial
stable decomposition. The following lemma may be wellknown.
Lemma 6.1. If n 6= 3 or 7, then RPn is stably atomic.
Proof.It is easy to check that RP2n is stably atomic by induction and by consid
ering the Steenrod operations. Suppose that RP2n+1 is not stably atomic. Let
q :S2n ! RP2n be the quotient map. Since ß4n+1( 2n+1RP2n) ~=ßs4n+1( 2n+1RP2n),
the map 2n+1q :S4n+1 ! 2n+1RP2n is null homotopic. By [12], there is a homoto*
*py
commutative diagram
P 2n+1 E 2n+2 H 4n+3
2S4n+3 _____________S ________ S ________ S
6 6 6 6
   
   
 OE2n OE2n+1 
   
[ 2n+1q ' *   pinch [
S4n+1 ____________ 2n+1RP2n ___ 2n+1RP2n+1 ______S4n+2,
where the top row is the EHP sequence, the bottom row is the cofibre sequence *
*and
the map OEn: n+1RPn ! Sn+1 is the adjoint map of the canonical map
RPn ____O(n + 1) ____ n+1Sn+1.
It follows that the Whitehead square !2n+1 is null homotopic. Thus 2n + 1 = 3 o*
*r 7
and hence the result.
Proposition 6.2. Let n 2. Then
1) Sqn+1: Hn+1(Cn) ! H2n+2(Cn) is an isomorphism.
2) The composite
ffin n+1 H2 2n+1
RPn ____ n+1RPn ____ S ____ S
is nontrivial mod 2 homology.
3) If n 6= 3 or 7, then ffin*: ßn+2( n+1RPn) ! ßn+2(Sn+1) is onto.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 97
Proof.Consider the homotopy commutative diagram of fibre sequences
in pn n
RPn ______Sn _________Yn_______ RP
   
   
   
   
   
   
? ? ? ?
Cn======= Cn _________*_________Cn.
Let xn+1 be a generator for Hn+1(Cn) = Z=2. Then
ø(xn+1)2 = 0
because Hn+1( RPn) ~=Hn+1(Cn) . Thus the dual x*n+1has a nontrivial self cup
product and hence assertion (1).
(2). Consider the homotopy commutative diagram of fibre sequences
Yn _______ RPn ______Sn _________Yn_______ RPn
    
    
    
  OE   OE
    
    
? ffi ? ? ? ffi ?
n+1RPn _____n Sn+1_______Fn ______ n+1RPn ____nSn+1.
Let fli(u) be a basis for Hi(RPn). Then H*( RPn) is the tensor algebra generat*
*ed
by fli(u) for 1 i n. By the decomposition
H*( RPn) = H*( Yn) H*( Yn) . fln(u),
we have fln(u)2 = ff + fi . fln(u) for some ff 2 H2n( Yn) and fi 2 Hn( Yn). No*
*te
that OE* is a an algebraic map with OE*(fli(u)) = 0 for i < n and Q(Hn( Yn)) *
*= 0.
We have OE*(fi . fln(u)) = 0 and so
OE*(fln(u)2) = OE*(ff).
It follows that the composite
ffin* n+1
H2n( RPn) ~=H2n( n+1RPn) ____H2n( S )
is onto and hence assertion (2).
(3). Let f be the composite
n+1ffin1 n+1 n+1
1 RPn ____ 1 n+1 n+1RPn ______ 0 S
____ 1 Q0(S0) ____H2 1 Q(RP1 ) ____ 1 RP1
in the stable category. Then there is a self map g of 1 RPn such that f = j O*
* g,
where j :RPn ! RP1 is the inclusion. By assertion (2), f*: Hn(RPn) ! Hn(RP1 )
is an isomorphism and so is g*: Hn(RPn) ! Hn(RPn). By Lemma 6.1, the map g is
a homotopy equivalence and hence the result.
Question 6.3. Is the map ffin: n+1RPn ! Sn+1 homotopic to the adjoint map of
the canonical map RPn ____SO(n + 1) ____ n+1Sn+1?
98 J. WU
6.2. A Decomposition of 30P 3(2). Now we study the spaces Zn and Yn in the
special case where n = 2. By Lemma 2.38, Q(H*(Y2)) and Q(H*(Z2)) have bases
{u, [u, v], v2} and {u2, [u, v], v2, [[u, v], u], [[u, v], v]}, respectively, w*
*here Sq1*v2 = [u, v],
Sq2*v2 = u2 and Sq1*([[u, v], v]) = [[u, v], u]. Thus H~*(Y2) and H~*(Z2) have*
* bases
{x2, x4, x5} and {y3, y4, y5, z5, z6}, respectively, with the Steenrod operatio*
*ns Sq1*x5 = x4,
Sq1*y5 = y4, Sq2*y5 = y3 and Sq1*z6 = z5. The fibre sequence
j 1
Z2 ____Y2 ____BSO(2) = CP
shows that Z2 is the 2connected cover of Y2 with j*(y4) = x4 and j*(y5) = x5.
(''2,[2])2 4
Proposition 6.4. There is a cofibre sequence S4 _____S _ S ____Y2.
Proof.Consider the fibre sequence S2 ____Y2 ____P 3(2). The 4skeleton
sk4(Y2) ' S2 [k''e4
for some k 2 Z. If k 6= 0, then ß3(Y2) = Z=k and so the boundary map
ß4(P 3(2)) ! k . ß3(S2)
is onto. This is a contradiction because ß*(P 3(2)) is torsion group. Thus k *
*= 0
and so sk4(Y2) ' S2 _ S4. By the cofibre sequence S2 ____Y2 ____P 5(2), we ha*
*ve
Y2 = (S2 _ S4) [(ff,[2])e5 for some ff :S4 ! S2. Since ß4(S2) = Z=2, it suffic*
*es to
show that ff is essential. Suppose that ff is null homotopic. Then Y ' S2 _ P 5*
*(2)
and so the bottom cell S2 is retract of Y2. It follows that the bottom cell S3*
* is a
retract of Z2. This is impossible because Sq2*y5 = y3. Thus ff is essential and*
* hence
the result.
We compute the first three homotopy groups of P 3(2).
Proposition 6.5. The homotopy groups ßn(P 3(2)) for n 4 are as follows.
1) ß2(P 3(2)) ~=ßs2(P 3(2)) = Z=2.
2) ß3(P 3(2)) = Z=4 with a generator represented by the composite
'' 2 3
~j:S3 ____S ____P (2).
3) ß4(P 3(2)) = Z=4 with a generator represented by the composite
p 3
~3: S4 ____S2 _ S4 = sk4(Y2) ____Y2 ____P (2).
i p 3
Proof.Assertion (1) is obvious. Consider the fibre sequence S2 ____Y2 ____P (*
*2).
We have the exact sequence
i* p* 3 @=0 2 3 @=0 *
* 2
ß4(S2) ____ß4(Y2) ____ß4(P (2)) ____ß3(S ) ____ß3(Y2) ____ß3(P (2)) ____ß*
*2(S ).
Since i: S2 ! Y2 is of degree 2 on the bottom cell of Y2, we have
i*: ß3(S2) = Z ! ß3(Y2)
is of degree 4 and i*: ß4(S2) ! ß4(Y2) is zero. Thus
ß3(P 3(2)) = Z=4 and ß4(Y2) ~=ß4(P 3(2)).
By the cofibre sequence
(''2,[2])2 4
S4 _____S _ S ____Y2,
we have ß4(Y2) = Z=4 and hence the result.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 99
By expecting the homology of Z2, the classes z5 and z6 come from a basis for *
*L3(V )
by homology suspension. This information suggests a splitting of Z2.
Proposition 6.6. There is a splitting Z2 ' RP42_ P 6(2).
Proof.Since ß3(Z2) ~=ß3(Y2) ~=ß3(sk4(Y2)) = ß3(S2 _ S4) = Z, we have
sk4(Z2) ' S3 _ S4.
Observe that Sq1*y5 = y4 and Sq2*y5 = y3. We obtain that sk5(Z2) ' RP42_ S5 and
so there is a map OE: RP42! Z2 such that the image of OE* has a basis {y3, y4,*
* y5}.
Let g :S4 ! Z2 be a map such that g*('4) = y4 and let g0:S3 ! Z2 be the adjoint
map of g. Let _ = [g, idRP2]: S3 ^ RP2 ! Z2 be the relative Samelson product on
the fibre sequence Z2 ! P 3(2) ! SO(3). Then _*: ~H*(P 3(2)) ! Q(H*( Z2))
has the image with a {[[u, v], u], [[u, v], v]}. Let _0:P 6(2) ! Z2 be the adjo*
*int map
of _. Then the map
ffi__0 fold
RP2 _ P 6(2) ____Z2 _ Z2 ____Z2 ____Z2
is a homotopy equivalence and hence the result.
Theorem 6.7. [46, Theorem 1.1] There are homotopy decompositions localized at 2.
30(P 3(2)) ' 2(S3<3>) x 3( RP42<3> _ P 6(2)).
Proof.The assertion follows from the facts that
1) The map i: S2 ! Y2 is of degree 2 on the bottom cell.
2) 2([2]): 2(S2<3>) = 2(S3<3>) ! 2(S2<3>) = 2(S3<3>) is null homotopic.
3) The 2connected cover Z2 ' RP42_ P 6(2) by Proposition 6.6
There are some applications of this theorem.
Proposition 6.8. Let X be a simply connected CW complex and let E ! X be a
principle SO(3)bundle. Suppose that the second and the third Whitney classes *
*w2
and w3 are not zero in H*(X). Then
30X ' 30E x 2(S3<3>).
In particular, the higher homotopy groups of S3 are summands of ß*(X).
Proof.There is a map f :X ! BSO(3) such that E is homotopy equivalent to the
homotopy fibre of f. By the assumption, the inclusion P 3(2) ____BSO(3) lifts*
* to
X and hence the result.
Corollary 6.9. Let G be any 2connected topological group. Suppose that G con
tains SO(3) as a subgroup. Let G=SO(3) be the obtained homogeneous space. Then
30(G=SO(3)) is (weak) homotopy equivalent to the product 30G x 2(S3<3>).
Proof.Consider the fibration SO(3) ! G ! G=SO(3). By the assumption, we have
ß2(G=SO(3)) ~=ß2(BSO(3)) and hence the result.
As an example, consider that SO(3) is a canonical subgroup of SU(3) by comple*
*x
ification. We have the decomposition 30(SU(3)=SO(3)) ' 30SU(3) x 2(S3<3>). By
using the language in (unstable) Ktheory, we have
100 J. WU
Corollary 6.10. Let X be a pathconnected space. Then the complexification of a*
*ny
orientable 3dimensional vector bundle over 3X is a trivial bundle.
Consider the inclusion E :RP2 ! P 3(2). Let q :S2 ! RP2 be the quotient map.
Then there is a homotopy commutative diagram
q 2
S2 ________RP
 
 
 
[2] E
 
? ~j ?
S2 ______ P 3(2).
Observe that
1) [2]*: ßr(S2) ! ßr(S2) is 0 for r 4 and of degree 4 for r = 3;
2) q*: ßr(S2) ! ßr(RP2) is isomorphic and
3) ß4(P 3(2)) = Z=4.
We have the following suspension theorem.
Proposition 6.11. The suspension E*: ßr(RP2) ! ßr+1(P 3(2)) is zero for r 3.
Corollary 6.12. Let X be a complex which admits the form of P 3(2)[fen with n *
* 5.
If the attaching map f is essential, then X is not homotopy equivalent to a sus*
*pension.
This result has been used for finding a family of nonsuspension coHspaces X
which admits the form of P 3(2) [ en in [48].
The map ~3: S4 ! P 3(2), which represents a generator for ß4(P 3(2)), has some
special properties. First the adjoint map ~03:S3 ! P 3(2) has the property th*
*at
~03*('3) = [u, v], that is, ~3 is from the spherical class [u, v]. The homotop*
*y cofibre
of ~3 is actually the homogeneous space SU(3)=SO(3).
~3 3
Proposition 6.13. There is a cofibre sequence S4 ____P (2) ____SU(3)=SO(3).
Proof.By [24, Theorem 6.7], the cohomology H*(SU(3)=SO(3)) is the exterior alge
bra E(w2, w3), where wi is the ith Whitney class. Thus there is a cofibre sequ*
*ence
f 3
S4 ____P (2) ____SU(3)=SO(3).
Consider the homotopy commutative diagram
f 3
S4 _______Pw(2) ____SU(3)=SO(3)
 w 
 ww 
 w 
g ww 
 w 
 w 
? ?
Z2 _______P 3(2)_______BSO(3),
where the top row is a cofibre sequence and the bottom row is a fibre sequence.*
* The
map g :S4 ! Z2 is nontrivial in mod 2 homology. Thus g represents a generator
for ß4(Z2) and hence the result.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 101
The next proposition shows that ~3 represents a generator for the stable homo*
*topy
group ßs4(P 3(2)).
Proposition 6.14. There is an isomorphism ß4(P 3(2)) ~=ßs4(P 3(2)).
Proof.By Proposition 6.2, ffi2*:ß4(P 5(2)) ! ß4(S3) is an isomorphism and so the
~3 3 pinch 3
composite S4 ____P (2) ____S is homotopic to j. It follows that the homotopy
class [~3] is a generator for ßs4(P 3(2)) = Z=4 or ß4(P 3(2)) = Z=4 ! ßs4(P 3(2*
*)) is an
isomorphism and hence the result.
Example 6.15. Since ß4(P 3(2)) = Z=4, there are only three different complexes X
which admits the form X = P 3(2) [ e5 given by P 3(2) [~3 e5 ' SU(3)=SO(3),
P 3(2) [2~3e5and P 3(2)_S5. The space P 3(2)[2~3e5 has the cell structure S2 ['*
*'2e4 [ e5.
This shows that the space SU(3)=SO(3) is uniquely determined, up to homotopy, by
its fundamental group and its cohomology ring. More precisely, if X is a simply*
* con
nected complex such that the cohomology ring H*(X) is isomorphic to H*(SU(3)=SO*
*(3)),
then X is homotopy equivalent to SU(3)=SO(3). This information might be useful
in the theory of 5dimensional manifolds.
Unlike Theorem 2.19 on the Stiefel manifolds, there is a self homotopy equiva*
*lence
of SU(3)=SO(3) which changes the orientation.
Proposition 6.16. There is a self homotopy equivalence of SU(3)=SO(3) which
changes the orientation.
Proof.From the commutative diagram
~=
ß4(P 3(2))____1ßs4(P 3(2))
 E 
 
 
[1]* 1
 
? ~= ?
ß4(P 3(2))____1ßs4(P 3(2)),
E
we have [1]*(~3) = ~3 and so there is a homotopy commutative diagram of cofib*
*re
sequences
~3 3 pinch 5
S4 _______P (2)____ SU(3)=SO(3) _______S
   
   
   
[1] [1] OE [1]
   
? ~ ? ? pinch ?
S4 _______3P 3(2)__ SU(3)=SO(3) _______S5.
By the 5lemma, OE is a homotopy equivalence and hence the result.
6.3. The Homotopy Groups ß*( RP2). Now we start to compute ßr(P 3(2)) for
r 11. The first three homotopy groups have been computed in the previous sect*
*ion.
The higher homotopy groups will be computed using Theorem 6.7. Observe that the*
*re
are two factors in this decomposition theorem. The homotopy groups of the factor
2(S3<3>), up to this range, has been computed in [42]. Thus our work is to com*
*pute
102 J. WU
ß*( RP42_ P 6(2)). We first compute ß*( RP42), which is the most difficult par*
*t,
and then, by applying the HiltonMilnor theorem, determine the homotopy groups
of RP42_ P 6(2).
6.3.1. The Homotopy Groups ß*( RP42). We need some preliminary lemmas.
Notation 6.17. Let F and B denote the spaces in the homotopy commutative
diagram of fibre sequences
i 5
S3 _______SU(3) _______Sw
6 6 ww
  ww
  w
 q w
  ww
  w
  p
F ________ RP42_______S5
6 6 6
  
  
  
  
  
  
B========= B _________ *,
where p is the pinch map, S3 ! SU(3) ! S5 is the canonical fibration and q is t*
*he
composite q : RP42____ CP2 ____SU(3).
Lemma 6.18. There is a homotopy equivalence
F ' S3 x B
Proof.Since ß3(S3) ~=ß3(SU(3)), ß3(F ) ~=ß3( RP42). Thus ß3( RP42) ~=ß3(SU(3))
and hence the result.
Recall that H*( RP42) ~=T (H~*(RP42)) ~=T (x2, u3, v4) as algebras with Sq1*v*
*4 = u3
and Sq2*v4 = x2. Although H*( RP42) is not primitively generated, we will show
that both H*( F ) and H*( B) are primitively generated.
Lemma 6.19. Let ff 2 P H*( RP42). Then [ff, v4] + x2[ff, x2] is primitive.
Proof.Observe that _(v4) = v4 1 + 1 v4+ x2 x2 and _(ff) = ff 1 + 1 ff. *
*We
have
_([ff, v4]) = [ff, v4] 1 + 1 [ff, v4] + x2 [x2, ff] + [x2, ff] *
* x2,
_(x2[ff, x2]) = x2[ff, x2] 1 + 1 x2[ff, x2] + x2 [x2, ff] + [x2, ff*
*] x2
and hence the result.
___0
Notation 6.20. For ff 2 H*( RP42) ~= T (x2, u3, v4), we define ad (v4)(ff) = f*
*f,
___1 ___k ___1___k1
ad (v4)(ff) = [ff, v4] + x2[ff, x2] and ad (v4)(ff) = ad (ad (v4)(ff)) for k *
* 2.
___k
By Lemma 6.19, there are operations ad (v4): P H*( RP42) ! P H*( RP42).
Proposition 6.21. H*( F ) is primitively generated and H*( F ) ~=T (VF ) as sub
Hopf algebra of H*( RP42) ~=T (x2, u3, v4), where VF P H*( RP42) has a basis
___k ___k
ad (v4)(x2) and ad (v4)(u3) for k 0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 103
Proof.Observe that p* : H*( RP42) ! H*( S5) is an epimorphism. The Serre
spectral sequence for the fibre sequence F ! RP42! S5 collapses and so there
is a short exact sequence of Hopf algebras
H*( F ) ____H*( RP42) ! H*( S5).
Thus there is an exact sequence
0 ____P H*( F ) ____P H*( RP42) ____P H*( S5).
___k ___k
Since both x2 and u3 are primitive, the elements ad (v4)(x2), ad (v4)(u3) 2 P H*
**( F )
for k 0 by Lemma 6.19. Let A be the subalgebra of T (x2, u3, v4) generated by
___k ___k
ad (v4)(x2) and ad (v4)(u3) for k 0. We claim that Z=2 A T (x2, u3, v4) = T *
*(v4).
Observe that Z=2 A T (x2, u3, v4) = T (x2u3v4)=IA . T (x2, u3, v4). Since x2, *
*u3 2 IA,
it suffices to show that vk4. y 2 IA . T (x2, u3v4) for each k 0, where y = x*
*2 or u3.
This is shown by induction on k. If k = 0, v04y = y 2 IA . T (x2, u3, v4). Su*
*ppose
that vk4y 2 IA . T (x2, u3, v4). Then
X ___ti
vk4y = ad (v4)(yi) . ffi
i
for some ti 0, yi= x2 or u3 and ffi2 T (x2, u3, v4) and so
X ___ti
vk+14y = v4ad (v4)(yi) . ffi.
i
It suffice to show that
___ti
v4ad (v4)(yi) 2 IA . T (x2, u3, v4)
for each i. Now
___ti+1 ___ti ___ti
ad (v4)(yi) = [ad (v4)(yi), v4] + x2 . [ad (v4)(yi), x2]
___ti ___ti ___ti
= v4ad (v4)(yi) + ad (v4)(yi) . v4 + x2 . [ad (v4)(yi), x2]
Thus
___ti ___ti+1 ___ti ___ti
v4ad (v4)(yi) = ad (v4)(yi)+ad (v4)(yi)v4+x2[ad (v4)(yi), x2] 2 IA.T (x2, u*
*3, v4).
The induction is completed.
Similarly, T (x2, u3, v4) AZ=2 = T (v4). Thus A is a normal subalgebra of T (*
*x2, u3, v4).
Observe that
X1
Ø(A) = (1t2t3t4)1.(1t4) = (1 (t4k+2+t4k+3))1 = Ø(H* F ) = Ø(T (VF )).
k=0
Since A H*( F ) and T (VF ) ! A is onto, A = H*( F ) and so
T (VF ) ~=A = H*( F ).
Note that VF P H*( RP42). Thus H*( F ) is primitively generated and hence the
result.
Corollary 6.22. The Hopf algebra H*( B) is primitively generated and
H*( B) ~=T (VB )
as a subalgebra of H*( RP42) ~=T (x2, u3, v4), where VB P H*( RP42) has a b*
*asis
___j1 j ___j1
adj2(x2)ad (v4)(x2) for j2 0, j1 1 and ad 2(x2)ad (v4)(u3) for j2, j1 0.
104 J. WU
Proof.There is a short exact sequence of Hopf algebras
H*( B) ____H*( F ) ___H*( S3).
Note that H*( F ) ~=T (VF ) is primitively generated. The assertion follows fro*
*m [9,
Lemma 3.13].
Now we are going to decompose B up to dimension 11.
Lemma 6.23. There exists a map ` :B ! S5 so that the composite
j ` 5
S4 ____B ____ S
is of degree 1.
Proof.Consider the homotopy commutative diagram
2 5
SU(3) _____SU(3) _______Sw
w
6 6 ww
  ww
  w
  ww
  w
 [2]  p
RP42 ______ RP42_______S5w
  ww
  w
  ww
p p ww
  ww
? [2] ? pinch
P 5(2)______P 5(2)______S5.
The composite
2 5
RP42____SU(3) ____SU(3) ____S
is null homotopic and so there is a homotopy commutative diagram
SU(3) _______B ________ RP42______SU(3)
   
   
   
2 `  2
   
   
? ? ? ?
SU(3) ______ S5 ________S3 _______SU(3) _______S5.
Recall that ß4( RP42) = Z=4. From the commutative diagram of exact sequences
0 ______ß4( SU(3)) ____ß4(B) ____ß4( RP42) = Z=4 ___ß4(SU(3)) = 0
   
   
   
2 `*  
   
  ? 
? ? ? ?
0 ______ß4( SU(3)) ___ß4( S5) _____ß4(S3) = Z=2____ß4(SU(3)) = 0,
the map `* : ß4(B) ! ß4( S5) is of degree 1 and hence the result.
Corollary 6.24. The bottom cell S3 is a stable retract of B
Proof.There is a retraction: 1 2S5 ! 1 S3 and hence the result.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 105
Lemma 6.25. The 7skeleton sk7( B) of B is homotopy_equivalent_to S3_S7_Y 7,
where_H~*(Y 7) has a basis [x2, u3], u23, [x2, v4], ad(v4)(u3)_with Sq1*[x2, v4*
*] = [x2, u3],
Sq2*ad(v4)(u3) = [x2, u3] and the second Bockstein fi2(ad(v4)(u3)) = u23.
Proof.By the Cartan formula, we have
___
Sq1*[x2, v4] = [x2, u3] and Sq2*(ad(v4)(u3)) = [x2, u3].
___
We first show that fi2(ad(v4)(u3)) = u23. Consider the homotopy commutative di*
*a
gram of fibre sequences
B ______ RP42 ____ SU(3)
  
  
  
OE  
  
? ? ?
F 5{2} ____ P 5(2)______ S5
___
We have OE*(ad(v4)(u3)) = [u3, v4] and OE*(u23) = u23in H* F 5{2}. Since
___ 1 ___
fi1(ad(v4)(u3)) = Sq*(ad(v4)(u3)) = 0,
___ ___
we have OE*fi2(ad(v4)(u3)) = u23and so fi2(ad(v4)(u3)) = u23+ ffl[x2, v4] u2*
*3in
Bockstein spectral sequence for H* B. Now ~H*(sk7( B)) has a basis
___
{u3, [x2, u3], u23, [x2, v4], ad(v4)(u3), [[x2, u3], x2].
Since x2 and u3 are spherical in H*( RP42), the elements u3, [x2, u3], u23, [[*
*x2, u3], x2]
are spherical in ( RP42) by Samelson products and Hopf Invariants 1. Observe t*
*hat
SU(3) is a commutative Hspace. All of Samelson products lift to the fibre B.
It follows that u3, [x2, u3], u23and [[x2, u3], u3] are spherical in H*( B). Th*
*us the 5
skeleton sk5( B) ' S3 _ S5 and the 6skeleton sk6( B) ' S6 _ X, where ~H*(X) has
a basis {u3, [x2, u3], [x2, v4]}. The space X is the homotopy cofibre of the at*
*taching
f 3 5 1
map S5 ____S _ S . Since Sq*[x2, v4] = [x2, u3], the composite
f 3 5 proj.5
S5 ____S _ S ____S
f 3 5 proj.3 2
is of degree 2. Let OE be the composite S5 ____S _ S ____S . Then OE = kj *
*for
some k = 0, 1. If k = 1, then ß5(X) = Z=4 and ß5(S3) is not a summand of ß5(X),
which contradicts to the facts that
i)ß5(S3) ~=ßs5(S3) = Z=2;
ii)S3 is a stable retract of B by Corollary 6.24.
iii)ß5(X) ~=ß5( B).
Thus k = 0 and so X ' S3 _ P 6(2) or sk6( B) ' S3 _ S6 _ P 6(2). Now we have___
sk7( B) ' S7 _ Y , where H~*(Y ) has a basis u3, [x2, u3], u23, [x2, v4] and ad*
*(v4)(u3).
There is a cofibre sequence
ffi 3 6 6
S6 ____S _ S _ P (2) ____Y.
___ ffi proj.
Since Sq2*ad(v4)(u3) = [x2, u3], the composite S6 _____S3 _ S6___P 6(2) _____*
*P 6(2)
is ~jwhich is a generator for ß6(P 6(2)) = Z=2. Note that fi2(ad(v4)(u3)) = u23*
*. The
ffi 3 6 6 proj.6
composite S6 ____S _S _P (2) ____S is of degree 4. Observe that the composite
106 J. WU
Y ____ B ____ 2S5 maps into the 7skeleton sk7( 2S5) up to homotopy. There
is a commutative diagram of cofibre sequences
OE 3 6 6 7
S6 ________S _ S _ P (2)______Y _________S
   
   
   
t proj.  t
   
? ( 0, [2]) ? ? ?
S6 ____________S3 _ S6______sk7( 2S5) _____S7,
Thus the map t : S6 ! S6 is of degree 2 and there is a projection
p1 : S3 _ S6 _ P 6(2) ! S3
ffi 3 6 6 p1 3 0 3 3
such that the composite S6 ____S _ S _ P (2) ____S is 2 = j3 = Ej2. Hence
OE = (j33, 4, ~j) = E(j32, 4, ~j) : S6 ! S3 _ S6 _ P 6(2).
It follows that Y ' Y 0for some Y 0. Now let f :S3 _ S6 _ P 6(2) ! S3 be the
p 6 ~''233 2
composite S3 _ S6 _ P 6(2) ____P (2) ____S , where p is the projection and ~j*
*3is
the extension of j23: S5 ! S3. Then f O OE = j33. Define g :S3 _ S6 _ P 6(2) *
*! S3
by g = p1 + f. Then [g] O [OE] = [p1 O OE] + [f O OE] = 2 0+ j32= 4 0 = 0 and *
*so g
g 3
extends to ~g:Y ! S3. Note that the composite S3 ____S3 _ S6 _ P 6(2) ____S *
*is
a homotopy equivalence and Y is a suspension. The space S3 is a retract of Y and
hence the result.
Notation 6.26. Let ~L3denote hocolimfi3 (RP42)(3), where fi3 is the composite
id T12 (3)id ff (3)
X(3)______ X _____ X
with T12(x1 ^ x2 ^ x3) = (x2 ^ x1 ^ x3) and oe(x1 ^ x2 ^ x3) = x3 ^ x1 ^ x2. T*
*hen
( RP42)(3)' ~L3_ M3, see Chapter 3. Let L3 denote sk11( (~L3)).
Lemma 6.27. With the notations above, we have
1) L3 ' ~L3.
2) L3 is a retract of (RP42)(3).
Proof.The map L3 ____ ~L3induces an Hmap L3 ! ~L3. Consider the homo
topy commutative diagram of fibre sequences
L3 ________* _________ L3
  
  
  
  
  
  
? ? ?
~L3_________* _________~L3.
Since ~L3is 8connected and dim ~L3= 12, the homology suspension
oeH~*(L3):~H*(L3) ____H~*+1(~L3)
is an isomorphism and hence assertion 1.
(2). Note that J((RP42)(3))=(RP42)(3)is 11connected. The composite
L3 ____ ~L3____J((RP42)(3))
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 107
maps into the subcomplex (RP42)(3)up to homotopy. Since ~L3=L3 ' L3=L3
is 13connected, (RP42)(3)! ~L3maps into L3 up to homotopy. Thus there is a
commutative diagram
hocolimfi3
~L3 ____ (RP42)(3)____________ ~L3
6 6 6
  
  
  
  
  
L3 _______(RP42)(3)_____________L3
and hence the result.
Proposition 6.28. There is a map which is a homotopy equivalence through 11
skeleton
OE: S3 x Y 7x L3 x P 11(2) x 6RP53x S11 ! B.
Proof.The idea is to construct a map OE and show that OE induces an isomorphism*
* in
homology through dimension 11. The construction of the map OE is given by sever*
*al
steps.
Step 1. Let OE1: L3 ! B be the composite
S~3
L3 ____ (RP42)(3)____ B,
where ~S3is the Hmap induced by the Samelson product S3: (RP42)(3)! B. We
check the homology information. Observe that H~*(L3) = P H*(L3) has a basis
ff7, ff8, ff9, ff10, fi8, fi9, fi10 and fi11 with the Steenrod operations given*
* as follows:
Sq4*fi11= ff7, Sq3*fi11= fi8, Sq2*fi11= fi9, Sq1*fi11= fi10,
Sq3*ff10= ff7, Sq2*ff10= ff8, Sq1*ff10= ff9, Sq2*ff9 = ff8 and Sq1*ff8 *
*= ff7.
Note that H7( (RP42)(3)) ~=H7(RP42)(3)has a basis
{x2 x2 u3, x2 u3 x2, u3 x2 x2},
where both x2 and u3 are primitive. Thus ~S3:H7( RP42) ! H7( B) is the Lie
bracket and OE1*(ff7) = [u3, x22]. Now
___2
OE1*(fi11) = ad (v4)(u3) + l1[[x2, u3], [x2, v4]] + l2[[x2, v4], x2], *
*u3]+
___ 4 3
l3ad 2(x2)(ad(v4)(u3)) + l4ad (x2)(u3) + l5ad (u3)(x2),
OE1*(fi8) = Sq3*OE1*(fi11) = [x2, u3],
OE1*(fi9) = Sq2*OE1*(fi11= [[x2, v4], u3] + l3ad 3(x2)(u3),
___ 2
OE1*(fi10) = Sq1*OE1*(fi11) = [ad(v4), u3] + l2[[u3, x2], u3],
___2 2 ___ 2 2
OE1*(ff10) = ad (v4)(x2)+k1[x2, u3] +k2[ad(v4)(u3), u3]+k3[[u3, x2], u3]+k4ad (*
*x2)([v4, x2]),
OE1*(ff8) = [v4, x22] + [x2, u23] + k2[x2, u23],
___ 3
OE1*(ff9) = [ad(v4)(u3), x2] + k4ad (x2)(u3).
Step 2. Let OE2: sk8(L3) ^ S3 ! B be the Samelson product of
OE1sk8(L3):sk8(L3) ____ B
108 J. WU
and u3: S3 ! B. Then
OE2*(ff7 '3) = [[u3, x22], u3],
OE2*(ff8 '3) = [[v4, x22], u3] + (k2 + 1)[[x2, u23], u3],
OE2*(fi8 '3) = [[x2, u23], u3].
Step 3. We give a decomposition of sk9(L3) in this step. Observe that ~H*(sk8(L*
*3))
has a basis ff7, ff8, fi8 with Sq1*ff8 = ff7. Thus
sk8(L3) ' S8 _ P 8(2).
Note that ~H*(sk9(L3)) has a basis {ff7, ff8, ff9, fi8, fi9} with Sq2*(ff9) = f*
*f7, Sq1*ff8 = ff7
and Sq1*fi9 = fi8. There is a cofibre sequence
_ 8 8 9 9
S8 _ S8 ____S _ P (2) ____ sk9(L3) ____S _ S .
_ 8 8 proj. 8
Let _1 be the composite S8 _ S8 ____S _ P (2) ____P (2). Recall that
[S8 _ S8, P 8(2)] ~=Z=2 Z=2.
We have [_1] = ffl1~j(1)+ ffl2~j(2)for some ffli = 0, 1. Since Sq2*6= 0 in H*(s*
*k9(L3)), we
have (ffl1, ffl2) 6= (0, 0), that is, [_1] = ~j(1), ~j(2)or ~j(1)+ ~j(2). Let i*
*: S8 ! S8 _ S8 be
'(2)8, '(1)8or '(1)8+ '(2)8for the cases where [_1] = ~j(1), ~j(2)or ~j(1)+ ~j(*
*2), respectively.
Then _1 O i is null homotopic and so there is a homotopy commutative diagram of
cofibre sequences
[k] 8 9
S8 __________S ________P (k)
  
  
  
i  j
  
? _ ? ?
S8 _ S8___S8 _ P 8(2)____sk9(L3)
  
  
  
 proj. 
  
  
? ~j ? ?
S8 ________P 8(2)______ 4RP53.
Let v be a basis for H9(P 9(k)). Then Sq2*j*(v) = 0 and j*(v) = fi9. Thus Sq1*(*
*v) 6= 0
and so k = 2l with l 1 mod 2. Consider the cofibre sequence
j 4 5 ` 10
P 9(2) ____ sk9(L3) ____ RP3 ____P (2).
There is a homotopy commutative diagram
` 10
4RP53 _____P (2)
 
 6
 
pinch `0
 
? 
S9======== S9.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 109
Note that `*: ~H*( RP53) ____H~*(P 10(2)) is zero. Thus `0 is null homotopic a*
*nd so
is `. Observe that the 9skeleton sk9(L3) is the 14skeleton of the homotopy fi*
*bre of
` and sk9(L3) is a suspension. We have sk9(L3) ' P 9(2) _ 4RP53.
Step 4. Let ~OE3: 4RP53^ S2 ! RP42be Samelson product of the map x2: S2 !
RP42and the composite
ffi1 4
4RP53____ sk9(L3) ____L3 ____ B ____ RP2.
Then ~OE3lifts to B. Let OE3: 4RP53^ S2 ____ B be a lifting of ~OE3. Let x7,*
* x8, x9
be a basis for H~*( 4RP53). Let f : 4RP53! L3 be the inclusion. Then f*(x7) = f*
*f7,
f*(x9) = ff9 and f*(x8) = ff8 + fflfi8 for some ffl = 0, 1. Then
OE3*(x7 '2) = [OE1*(ff7), x2] = ad3(x2)(u3),
OE3*(x8 '2) = [OE1*(ff8), x2] + ffl[OE1*(fi8), x2] = [[v4, x22], x2] + (1 + f*
*fl + k2)[[x2, u23], x2],
___ 4
OE3*(x9 '2) = [OE1*(ff9), x2][[ad(v4)(u3), x2], x2] + k4ad (x2)(u3).
Note that ad4(x2)(u3) is spherical.
Step 5. Let OE4: S11 _____ B be a map such that OE4*('11) = ad4(x2)(u3) and l*
*et
ß1 :S3 ! B be a map such that OE1*('3) = u3. Let OE0: Y 7! B be the
Hmap induced by the inclusion Y 7! B. We obtain a map
OE: S3 x Y 7x L3 x P 11(2) x 6RP53x S11 ____ B
by taking the product of the maps OEj for 1 j 4. Now we show that OE induc*
*es an
isomorphism in homology. Let Z denote S3x Y 7x L3x P 11(2) x 6RP53x S11.
Let V be the image of OE*: P H*(Z) ! P H*( B). Then Clearly Vr = P Hr( B) for
r 8. Since V9 is generated by
___ 3 3
[[x2, v4], u3] + l3ad 3(x2)(u3), [[ad(v4)(u3), x2] + k4ad (x2)(u3), ad (x2*
*)(u3),
we have V9 = P H9( B). Observe that V10 is generated by
___ 2
[x2, u3]2, [ad(v4)(u3), u3] + l2[[u3, x2], u3],
___2 2 ___ 2 2
ad (v4)(x2) + k1[x2, u3] + k2[ad(v4)(u3), u3] + k3[[u3, x2], u3] + k4ad (x2)(*
*[v4, x2]),
[[u3, x22], u3], [[v4, x22], x2] + (1 + ffl + k2)[[x2, u23], x2].
Thus V10= P H10( B). Now V11 is generated by
[[x2, u3], [x2, v4]],
___2 2 ___
ad (v4)(u3) + l1[[x2, u3], [x2, v4] + l2[[x2, v4], x2], u3] + l3ad (x2)ad(v*
*4)(u3)+
+l4ad 4(x2)(u3) + l5ad 3(u3)(x2),
___ 4
[[x2, u23], u3], ad2(x2)ad(v4)(u3) + k4ad (x2)(u3),
[[v4, x22], u3] + (k2 + 1)[[x2, u23], u3], ad4(x2)(u3).
Thus V11 = P H11( B). Note that dim P Hr(Z) = dim P Hr( B) for r 11. Thus
OE*: P H*(Z) ! P H*( B) is an isomorphism and so OE*: Hr(Z) ! Hr( B) is a
monomorphism for r 11. Since dim Hr(Z) = dim Hr( B) for r 11, the map
OE*: Hr(Z) ! Hr( B) is an isomorphism for r 11. We finish the proof now.
110 J. WU
Corollary 6.29. There is a decomposition
ßr(B) ~=ßr1(S3) ßr( Y 7) ßr( L3) ßr(P 12(2)) ßr( 7RP53)
for r 11.
Observe that Y 7is 5connected.
Proposition 6.30. The homotopy groups ßr( Y 7) for 6 r 11 are as follows.
1) ß6( Y 7) = Z=2.
2) ß7( Y 7) = Z=8.
3) ß8( Y 7) = Z=2 Z=2.
4) ß9( Y 7) = Z=2 Z=2 Z=2.
5) ß10( Y 7) = Z=4 Z=4.
6) ß11( Y 7) = Z=2 Z=4.
Proof.1._ßr( Y 7) for r 9: Observe that ~H*(Y 7) has a basis [x2, u3], u23, [*
*x2, v4] and
ad(v4)(u3) with
___ ___ *
* 2
Sq1*[x2, v4] = [x2, u3], Sq2*ad(v4)(u3) = [x2, u3] and fi2ad(v4)(u3) = *
*u3.
Thus Y 7is the homotopy cofibre of (4, ~j): S6 ____S6_P 6(2) and so there is a*
* cofibre
sequence
(4,~'')7 7 7 8
S7 ____S _ P (2) ____ Y ____S .
Since S7_P 7(2) is the 12skeleton of the homotopy fibre of the pinch map Y 7!*
* S8,
the sequence
(4,~'')* 7 7 7 7
ßr(S7) ____ßr(S _ P (2)) ____ßr( Y ) ____ßr1(S ) ____ . . .
is exact for r 11. Thus ß6( Y 7) ~= ß6(P 7(2)) = Z=2. By expecting the exact
(4,~'')* 7 7 7 7
sequence ß7(S7) _____ß7(S ) ß7(P (2))ß7( Y ) ____0, we have ß7( Y ) ~=Z=8
and there is an exact sequence
(4,~'')* 7 7 7
ß8(S7) ____ß8(S ) ß8(P (2)) ____ß8( Y ) ____0.
Observe that (4, ~j)*(j7) = 2ff, where ff is a generator for ß8(P 7(2)) ~= Z=4.*
* We
obtain ß8( Y 7) ~=Z=2 Z=2 and there is an exact sequence
(4,~'')* 7 7 7
ß9(S7) ____ß9(S ) ß9(P (2)) ____ß9( Y ) ____0.
Since (4, ~j)*: ß9(S7) ! ß9(S7) ß9(P 7(2)) is zero, we have
ß9( Y 7) ~=ß9(S7) ß9(P 7(2)) ~=Z=2 Z=2 Z=2.
2. ß10( Y 7): Now there is an exact sequence
(4,~'')* 7 7 7 7
ß10(S7) ____ß10(S ) ß10(P (2)) ____ß10( Y ) ____ß9(S ) ____0,
where (4, ~j)*( 7) = 4 7 + ~j*( 7) = 4 7 + j*(j6 O 7) = 4 7 and j :S6 ! S7 is *
*the
inclusion of bottom cell. Thus there is a short exact sequence
0 ____ß10(S7)=4 ß10(P 7(2)) ____ß10( Y 7) ____ß9(S7) ____0
Note that ß10(P 7(2)) ~=ßs10(P 7(2)) ~=Z=2. There is a short exact sequence
0 ____Z=4 Z=2 ____ß10( Y 7) ____Z=2 ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 111
We need to solve the group extension problem. Consider the homotopy commutative
diagram of cofibre sequences
[2] 7 j1 6 2 p1 8
P 7(2)________P (2)_____P (2) ^ P (2)____P (2)
   
   
   
   p2
   
   
? (4, ~j) ? j ? ?
S7 _______S7 _ P 7(2)______2 Y 7_________S8w
   ww
   w
   ww
2   ww
   ww
? [2] ? ?
S7 ___________S7__________P 8(2)_________S8.
Let ff 2 ß10(P 6(2) ^ P 2(2)) such that p2*O p1*(ff) = j28. Then 2ff is the com*
*posite
ff 6 2 p1 8 ''^id 7 j1 6 2
S10 ____P (2) ^ P (2) ____P (2) ____P (2) ____P (2) ^ P (2).
Thus 2ff = j1*(~j3), where ~j32 ß10(P 7(2)) ~=Z=2 is the generator. Hence j2*(~*
*j3) is
also divisible by 2 in ß10( Y 7) and so ß10( Y 7) ~=Z=4 Z=4.
3. ß11( Y 7): Consider the exact sequence
ß11(S7) = 0 ____ß11(S7) ß11(P 7(2)) ____ß11( Y 7) ____2 . ß10S7 ____0.
There is a short exact sequence 0 ____ß11(P 7(2)) ____ß11( Y 7) ____2.ß10(S7*
*) ____0
or 0 ____Z=2 ____ß11( Y 7) ____Z=4 ____0. Thus ß11( Y 7) ~=Z=2 Z=4 or Z=8.
We show that ß11( Y 7) 6= Z=8. From this, we will obtain ß11( Y 7) ~=Z=2 Z=4.
Consider the homotopy commutative diagram
j ^ id 7 2 5 9
P 8(2)________Pw(2)_____CP ^ P (2) ____P (2)
6 ww 6 6
 ww  
 w  
 ww  
 w  
 ~j  
S7w__________P 7(2)_______ RP53 _________S8w
ww   w
ww 6 6 ww
ww   ww
ww proj.  ww
w   ww
(4, ~j)7  7 j2 7 8
S7 _______S _ P (2) ______ Y _________S ,
where the rows are cofibre sequences. Observe that P 7(2) is the 12skeleton o*
*f the
homotopy fibre of the pinch map CP2^P 5(2) ____P 9(2). There is an exact seque*
*nce
''^id* 7 2 5 9
ß11(P 8(2)) ____ß11(P (2)) ____ß11(CP ^ P (2)) ____ß11(P (2)).
Let ffn be a generator for ßn+1(P n(2)) ~=Z=4 for n 3. Note that ß11(P 8(2)) *
*= Z=2
and p*(ff8 O j2) = j3 = 4 in ß11(S8), where p : P 8(2) ! S8 is the pinch map. *
*Thus
ß11(P 8(2)) is generated by ff8 O j2. Also note that ß9(P 7(2)) ~=Z=2 Z=2 gen*
*erated
112 J. WU
by ff7 O j and j*( 6), where j :S6 ! P 7(2) the inclusion, and p*((j ^ id) O ff*
*8) = j27.
We have
(j ^ id) O ff8 = ff7 O j + ffl . j*( 6) and
(j ^ id) O ff8 O j = ff7 O j2 + fflj*( 6 O j) = ff7 O j2
for some ffl = 0, 1. Thus
(j ^ id) O ff8 O j2 = ff7 O j3 = ff7 O 4 O 8 = 0.
Hence (j ^ id)* : ß11(P 8(2)) ! ß11(P 7(2)) is zero and so there is an exact se*
*quence
0 ____ß11(P 7(2)) = Z=2 ____ß11(CP2 ^ P 5(2)) ____ß11(P 9(2)) = Z=2 Z=2.
Now suppose that ß11( Y 7) = Z=8. Then j2*(fi) is divisible by 4, where fi 2
ß11(P 7(2)) the generator. From the commutative diagram
ß11(Pw7(2))__ß11(CP2 ^ P 5(2))
ww 6
ww 
ww f*
ww 
j2*  7
ß11(P 7(2))_____ß11( Y ),
the element f*O (j2)*(fi) is divisible by 4 in ß11(CP2^ P 5(2)). By the exact s*
*equence
above, 4ß11(CP2^ P 5(2)) = 0. This is a contradiction. Thus ß11( Y 7) = Z=2 Z*
*=4.
This completes the calculation.
Now we compute ß*( L3). Observe that L3 is 7connected.
Proposition 6.31. The homotopy groups ßr( L3) for 8 r 11 are as follows.
1) ß8( L3) = Z=2.
2) ß9( L3) = Z=2.
3) ß10( L3) = Z=2.
4) ß11( L3) = Z=2 Z=2.
Proof.1. ß8( L3): By Step 3 of Proposition 6.28, sk10( L3) ' P 10(2) _ 5RP53.
Thus ß8( L3) ~=ß8( 5RP53) ~=Z=2.
2. ß9( L3): Consider the exact sequence
'~'* 9 5 5
ß9(S9) ____ß9(P (2)) ____ß9( RP3) ____0.
We have ß9( RP53) = 0 and so ß9( L3) = Z=2.
3. A decomposition of sk11( L3): There is a cofibre sequence
ffi 10 5 5
S10_ S10 ____P (2) _ RP3 ____ sk11( L3).
ffi 10 5 proj. 10
Let OE1 be the composite S10 _ S10 _____P (2) _ RP3 _____P (2). Note that
[S10 _ S10, P 10(2)] ~= Z=2 Z=2 and P 10(2) is not a retract of sk11( L3). T*
*here
exists an inclusion i : S10 ! S10_ S10 such that i*: H10(S10) ! H10(S10_ S10) i*
*s a
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 113
monomorphism and OE1O i is null homotopic. Thus there is a homotopy commutative
diagram
` 5 5 q _ 10
S10 _________ RP3 __________A oe___________S
   
   
   
i  g ~j
   
? OE ? ` ? `P10(2) ?
S10_ S10 ___P 10(2) _ 5RP53__sk11( L3)oe______P 10(2)
   
   
   
   
   
   
? ~j ? ? ?
S10 __________P 10(2)_______ 6RP53======== 6RP53,
where the columns are cofibre sequences and the rows except the terms in the la*
*st
column are cofibre sequences. We claim that _ is null.
First we compute ß10(A). Consider the exact sequence
'~'* 9 j* 5 5 p* 9
ß10(S9) ____ß10(P (2)) ____ß10( RP3) ____ß9(S ) ____Z=2 ____0.
Note that ~j*(j9) = 2ff9. We have
ß10( 5RP53) = {j*(ff9)} {~'9} ~=Z=2 Z
with p*(~'9) = 2'9. Observe that g*: ~H*(A) ! H~*(sk11( L3)) is a monomorphism.
Thus H~*(A) has a basis x8, x9, x10, x11 with Sq1*x11 = x10, Sq2*x11 = x9, Sq1**
*x9 = x8
and Sq2*x10= x8. The composite
` 5 5 pinch 10
S10 ____ RP3 ____S
is of degree 2. Thus
[`] = ~'9+ fflj*(ff9)
in ß10( 5RP53) for some ffl = 0, 1. Consider the exact sequence
`* 5 5
ß10(S10) ____ß10( RP3) ____ß10(A) ____0
with `*('9) = ~'9+ fflj*(ff9). Thus ß10(A) = Z=2 generated by q O j*(ff9).
Now suppose that _ :S10 ! A is essential. Then [_] = q O j*(ff9) and _ maps i*
*nto
the subcomplex P 9(2) of A up to homotopy. Thus there is a homotopy commutative
114 J. WU
diagram
ff9 9 0 11
S10______________Pw(2) ________C ________ Sw
ww   w
ww   ww
ww   ww
ww OE f ww
w   ww
_ ? ? 11
S10________________A __________Cw_________S
  ww
  w
  ww
~j  ww
  ww
? j0= `P10(2) ? p0
P 10(2)___________sk11( L3)______C,
where OE = q O j and the rows are cofibre sequences. Let u9 and v10 be the basi*
*s for
~H*(P 10(2)). Note that j0*is a monomorphism. Let j0*(v10) = l1ff10+ l2fi10. *
* Since
Sq2*v10 = 0, we have l1 = 0 and so j0*(v10) = fi10. It follows that j0*(u9) = f*
*i9. Now
~H*(C) has a basis p0*(ff8), p0*(ff9), p0*(ff10), p0*(ff11) and p0*(fi11). Obse*
*rve that H~*(C0)
has a basis {y8, y9, y11} Sq2*y11 = y9 and Sq1*y9 = y8. Since f*(y8) = p0*(ff8)*
* in H*C,
we have f*(y9) = p0*(ff9). Let
f*(y11) = k1p0*(ff11) + k2p0*(fi11).
Then
0 6= f*(y9) = f*(Sq2*y11) = Sq2*f*(y11) = k1Sq2*p0*(ff11) + k2Sq2*p0*(fi*
*11)
= k1p0*(ff9) + k2p0*(fi9) = k1p0*(ff9)
Thus k1 = 1 and so f*(y11) = p0*(ff11) + k2p0*(fi11). It follows that
0 = f*(Sq1*y11) = Sq1*f*(y11) = Sq1*p0*(ff11) + Sq1*k2p0*(fi11) = p0*(ff10*
*) 6= 0
This is a contradiction. Thus _ :S10 ! A is null homotopic and so is j0O ~j. He*
*nce
j0:P 10(2) ____ sk11( L3) extends to ~j: 6RP53____ sk11( L3). Now the composi*
*te
~j 6 5
6RP53____ sk11( L3) ____ RP3
is a homotopy equivalence and therefore sk11( L3) ' A _ 6RP53.
4. ß10( L3):
ß10(sk11( L3)) ~=ß10(A) ß10( 6RP53) = Z=2 0 = Z=2.
5. ß11( L3): By expecting the Steenrod operation on ~H*(A), we have
A ' CP2 ^ P 7(2).
From the exact sequence
''^id* 9 10 ''^id* 9
ß11(P 10(2)) ____ß11(P (2)) ____ß11(A) ____ß10(P (2)) ____ß10(P (2)),
we have ß11(A) = Z=2 and ß11(P 9(2)) ! ß11(A) is onto. Consider the cofibre se
quence
` 6 5
S11 ____A _ RP3 ____ L3.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 115
Since Sq4*6= 0 in ~H*( L3), the composite
` 6 5 proj.
S11 ____A _ RP3 ____A
is a generator for ß11(A) = Z=2. Note that Sq1*fi11= fi10. The composite
` 6 5 proj.6 5 proj. 11
S11 ____A _ RP3 ____ RP3 ____pS
is of degree 2. Observe that
ß11( 6RP53) = {j*(ff10)} {~'11} = Z=2 Z
with p*(~'11) = 2'11. There is an exact sequence
`*
Z ____Z=2 Z=2 Z ____ß11( L3) ____0
with the composite
`* proj.
Z ____Z=2 Z=2 Z ____Z
of degree 1. Thus ß11( L3) ~=Z=2 Z=2. This completes the calculation.
Now we are ready to compute ß*( RP42). Consider the homotopy commutative
diagram
ffi 4 5
S5 _________F ________ RP2 _______Sw
6 6 www www
  ww ww
E i ww ww
  ww ww
 (j, [2]) 
S4 ________S3 _ S4_____ RP42_______S5
where the top row is fibre sequence and the bottom row is cofibre sequence. By *
*Lem
mas 6.18 and 6.23 and Proposition 6.28, there is a homotopy commutative diagram
 ffi
2S5______________________________ F
6 6
 
 OE
 
 
 S3 x S3 x Y 7x L3 x P 11(2) x 6RP53x S11
 6
 
 
 
 
 
 S3 x S3 x Y 7
 6
 
 
 _
 
 
 (j, [2]) 
S4 __________________ S3 x S3 x S7 ' S3 x S4,
116 J. WU
where _ = (id S3, idS3, j), OE is a homotopy equivalence through the 11skelet*
*on and
j*('6) = u23in H*( Y 7).
Lemma 6.32. [23, pp.217] The homotopy group ß11(SU(3)) = Z=4.
Theorem 6.33. The homotopy groups ßr( RP42) for r 11 are as follows.
1) ß3( RP42) = Z.
2) ß4( RP42) = Z=4.
3) ß5( RP42) = Z=2.
4) ß6( RP42) = Z=2 Z=2 Z=2.
5) ß7( RP42) = Z=2 Z=8.
6) ß8( RP42) = Z=2 3 Z=4.
7) ß9( RP42) = Z=2 4.
8) ß10( RP42) = Z=2 Z=2 Z=4 Z=4.
9) ß11( RP42) = Z=2 4 Z=4 ß11(SU(3)) = Z=2 4 Z=4 2.
Proof.1. ßr( RP42) for r 7: Since L3 is 6connected, the sequence
ffi* 3 3 7 4 5
ßr(S5) ____ßr2( S x S x Y ) ____ßr1( RP2) ____ßr1(S ) ____ . . .
is exact for r 8. Hence ß3( RP42) = Z. Note that ffi*('5) = j3 + 2'3. We h*
*ave
ß4( RP42) = Z=4. Observe that ffi*(j5) = ffi*(Ej4) = j23. Thus
ß5( RP42) = Z=2.
Note that ffi*(j25) = ffi*(E2j23) = j33+ 0 + 0 = j33. Thus
ß6( RP42) = ß6(S3)=2 ß5(S3) ß6( Y 7) = Z=2 Z=2 Z=2.
Notice that ffi*( 5) = ffi*E 4 = (1 S3, 1S3, j)* O (j, [2])*( 4). Now
(j, [2])*( 4) = j*( 4) + [2]*( 4) = j3 O 4 + 4 4  E 0.
Thus
ffi*( 5) = j3 O 4  0+ 4j*('7),
where j*('7) is a generator for ß7( Y 7) = Z=8. It follows that
ß7( RP42) ~=Z=8 Z=2,
which are generated by OE* O j*('7) and OE*( 0). Furthermore
Ker (ffi*: ß8(S5) ! ß6( S3 x S3 x Y 7)) = 4 . ß8(S5)
because 2ffi*( 5) = 2 06= 0.
2. ß8( RP42): Consider the exact sequence
ffi* 3 3 7 4 5
ß9(S5) ____ß8(S ) ß7(S ) ß8( Y ) ß8( L3) ____ß8( RP2) ____4.ß8(S ) ____0
with ffi*( 5O j8) = ffi*( 5) O j8 = j3O 4O j7+ 0O j6 = 0O j26+ 0O j6 6= 0. *
*Recall that
ß8( Y 7) = Z=2 Z=2 and ß8( L3) = Z=2. There is a short exact sequence
0 ____(ß8(S3) ß7(S3))={ 0O j26+ 0O j6} ß8( Y 7) ß8( L3) ____ß8( RP42)
____4 . ß8(S5) ____0 or 0 ____Z=2 4 ____ß8( RP42) ____Z=2 ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 117
Consider the homotopy commutative diagram of fibre sequences
p 2 5 ffi
2F _______ 2 RP42_____ S _________ F
6 6 6 6
   
 e0 E2 e00
   
   
   (j, 2) 
2S3 x S3 ______X _________S3_______ S3 x S3.
('',2) 3 3
Then the universal cover X<1> is the homotopy fibre of S3 ____ (S <3>) x S and
so there is a commutative diagram of fibre sequences
i p0 3 (j, 2) 3 3
2(S3<3>) x S3 _____X<1>________Sw _______ (S <3>) x S
  ww 
  w 
  ww 
 f w ßproj.
  ww 
  w 
? ? j ?
2(S3<3>)________ 2S5 ________S3 _________ (S3<3>),
where the bottom fibre sequence is the EHP sequence
S3 ____ (S3<3>) ____ S5.
Consider the commutative diagram of exact sequence
0 3 3 i* p0* 3 (~j*, 2) 3 *
* 3
ß7(S3)w____ß8(S ) ß7(S )____ß6(X) _____ß6(S ) _______ß7(S ) ß*
*6(S )
ww   ww 
ww   ww 
ww ß* f* ww ß*
ww   www 
0 ?3 ?5 3 0 ?3
ß7(S3) ________ß8(S )________ß8(S ) _____ß6(S ) __________ß7(S ).
There is a short exact sequence
p0* 3
0 ____ß8(S3) ß7(S3) ____ß6(X) ____2 . ß6(S ) ____0.
Let ff 2 ß6(X) such that p0*(ff) = 2 0. Then f*(ff) = 2 5 and so ß*(2ff) = 0O*
* j26or
2ff = i*( 0O j26) + ffli*( 0O j6) for some ffl = 0, 1. Consider the homotopy co*
*mmutative
diagram of fibre sequences
(j, 2) 3 3
2S3 x S3 ______X _________S3w_______ S x S
  ww 
  w 
  ww 
 g w ß0proj.
  ww 
  w 
? ? p00 2 ?
S3 ________S3{2}_______S3 __________S3.
118 J. WU
There is a commutative diagram of short exact sequences
i* p0* 3
0 ______ß8(S3) ß7(S3)____ß6(X) _____2ß6(S ) _______0
  ww
  w
  ww
ß0* g* ww
  ww
? j ? p00
0 ___________ß7(S3)_______*ß6(S3{2})___*2ß6S3________0
with p00*(g*(ff)) = 2 0. Recall that (S3{2}) ' ( S3){2}. We have
2g*(ff) = j*((2 0) O j) = 0
or g*(2ff) = 0. Thus 0 = ß0*( 0O j26+ ffl 0O j6) = ffl 0O j6, that is, ffl = 0 *
*and so
2ff = i*( 0O j26).
Now consider the commutative diagram of exact sequences
ffi* OE* 4 p* 5
ß9(S5)________ ß8(F )_______ß8( RP2) ___4ß8(S ) _______0
6 6 6
  
e00* e0* ~=E2
  
 i  p0 
0 ______ß8(S3) ß7(S3)____*ß6(X) _____*2ß6(S3)______0
Since p0*(ff) = 2 0, we have p*(e0*(ff)) = 4 5. Since 2ff = i*( 0 O j26)), we *
*have
2e0*(ff) = OE*( 0O j26). Since ffi*( 5 O j8) = 0O j26+ 0O j6, the element 0O*
* j26does not
lie in the image of ffi* and so
2e0*(ff) = OE*( 0O j26) 6= 0.
Thus e0*(ff) is of order 4 and so ß8( RP42) ~=Z=4 Z=2 3.
3. ß9( RP42): Consider the exact sequence
ffi* 3 3 7 4
ß10(S5) ____ß9(S ) ß8(S ) ß9( Y ) ß9( L3) ____ß9( RP2) ____0
with ffi*( 5 O j28) = ffi( 5 O j8) O j9 = 0O j26O j8 + 0O j6 O j7 = 0*
*O j26.
Thus ß9( RP42) ~=ß9( Y 7) ß9( L3) ~=Z=2 4.
4. ß10( RP42): Consider the exact sequence
ffi* 3 3 7 7 5 4
ß11(S5) ____ß10(S ) ß9(S ) ß10( Y ) ß10( L3) ß10( RP3) ____ß10( RP2) ____0.
Recall that ß9(S3) = ß10(S3) = 0 and 4 . ß10( Y 7) = 0. We have
ffi*( 25) = ffi*(E 24) = j* O (j, 2)*( 24) = j*(4 24) = 0
and so ß10( RP42) ~=ß10( Y 7) ß10( L3) ß10( 7RP53) ~=Z=4 Z=4 Z=2 Z=2.
5. ß11( RP42): There is an exact sequence
ffi* 3 3 7 12 7 5
ß12(S5) ____ß11(S ) ß10(S ) ß11( Y ) ß11( L3) ß11(P (2)) ß11( RP3)
____ß11( RP42) ____ß11(S5) ____0.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 119
Consider the homotopy commutative diagram of fibre sequences
ffi 4 5
wS5 ________F ________ RP2 _______Sw
ww   w
ww   ww
ww   ww
ww   ww
w   ww
~j 3? ? 5
S5 ________S _______SU(3) _______S
By [42, pp.64], j4O oe000= 0 and so ~j*(oe000) = p O j4O oe000= 0, where p: S4*
* ____S3 is
a retraction. Thus
Im(ffi*: ß11( S5) ! ß11(F )) ß10( B).
Let X4 = sk7( 2S5). There is a commutative diagram
H 2 9
ß10( 2S5) ___ß10( S )
6 6
 
i* 
 
 p 
ß10(X4) ____*ß10(S7),
p* 7
where p is the pinch map. By Step 5 of Theorem 5.8, ß10(X4) ____4.ß10(S ) is o*
*nto.
By [42, Lemma 5.13], ß10( 2S5) ~=4 . ß10( 2S9) under the Hopf invariant H and so
i*: ß10(X4) ____ß10( 2S5) is onto. Let ffi0 be the composite
ffi proj.
2S5 ____ F ____ B.
Then ffi0maps X4 into the 7skeleton of B up to homotopy and so there is a hom*
*otopy
commutative diagram
ffi0
2S5 ________ B
6 6
 
 
 
 
 ffi00 
X4 _____ S3 x Y 7.
Let ffi000be the composite
ffi003 7 proj. 7
X4 ____S x Y ____ Y .
We show that ffi000*:ß10(X4) ____ß10( Y 7) is zero. Since X4 is the homotopy *
*cofibre
of ( 0, [2]): S6 ! S3 _ S6 and Y 7is 4connected, there is a homotopy commutati*
*ve
120 J. WU
diagram
ffi000 7
X4 _______ Y
 6
 
 
pinch E
 
? ~ffi000
P 7(2)_______Y 7.
Recall that ß10(P 7(2)) ~= ßs10(P 7(2)) = Z=2 generated by ff7 O j27, where ff7*
* is a
generator for ß8(P 7(2)) = Z=4. Since ~ffi000*:H7(P 7(2)) ! H7(Y 7) is zero, th*
*e composite
~ffi0007pinch7
P 7(2) ____Y ____S
is null homotopic and so ~ffi000maps into the subcomplex P 6(2) _ S6. Let OE1 *
*be the
composite
~ffi0006 6 proj. 6
P 7(2) ____P (2) _ S ____P (2)
~ffi000 proj.
and let OE2 be the composite P 7(2) ____P 6(2) _ S6 ____S6. Then OE2 is homot*
*opic
''6 6
to either the constant map or the composite P 7(2) ____S7 ____S . In the seco*
*nd
case, OE2O ff7 = j26and OE2O ff7O j28= 0. Thus, in both cases, OE2O ff7O j28= 0*
*. Observe
that [P 7(2), P 6(2)] = Z=2 Z=2. There are three possibilities of the map OE1:
Case 1. The map OE1 is homotopic to the composite
pinch 7 ff6 6
P 7(2) ____S ____P (2).
Then OE1 O ff7 = ff6j7 and OE1 O ff7 O j28= ff6 O j37= ff6 O 4 7 = 0.
'~' 5 j 6
Case 2. The map OE1 is homotopic to the composite P 7(2) _____S ____P (2),
where ~jis an extension of j :S6 ! S5. Then OE1 O ff7 = j O (k 5) for some k an*
*d so
OE1 O ff7 O j2 = j O (k 5 O j2). It follows that E(OE1 O ff7 O j2) = 0.
Case 3. The map OE1 = j ^ id:P 7(2) ____P 6(2). Then OE1 O ff7 = ff6 O j + ffl*
*j*( 5)
for some ffl = 0, 1 and OE1 O ff7 O j2 = ff6j3 + fflj*( 5 O j2) = fflj*( 5 O j2*
*). It follows that
E(OE1 O ff7 O j2 = 0.
Thus, in any case, we have
E O ~ffi000*:ß10(P 7(2)) ____ß10( Y 7)
is zero and so is ffi000*: ß10(X4) ____ß10( Y 7).
Now we show that the boundary ffi*: ß12(S5) ! ß11(F ) is zero. Since ß10(S3) is*
* zero,
ffi00*:ß10(X4) ____ß10(S3x Y 7) is zero. By using the fact that ß10(X4) ! ß10*
*( 2S5)
is onto, we have ffi*: ß10( 2S5) _____ß10( F ) is zero and so there is a short*
* exact
sequence
p* 5
0 ____ß11(S3) ß11( Y 7) ß11( L3) ß11(P 12(2)) ____ß11( RP42) ____ß11(S ) ___*
*_0.
We are going to solve the group extension problem now. Let ff 2 ß11( RP42) such
that p*(ff) = 25and let 2ff = fi1 + fi2 + fi3 + fi4, where fi1 2 ß11(S3), fi2 *
*2 ß11( Y 7),
fi3 2 ß11( L3) and fi4 2 ß11P 12(2).
First we show that fi4 = 0. Consider the Hurewicz homomorphism
h: ß10( RP42) ! H10( RP42).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 121
Then
0 = h(2ff) = h(fi1) + h(fi2) + h(fi3) + h(fi4).
If fi4 6= 0, then fi4 is a generator for ß11(P 12)(2) = Z=2 and h(fi4) = [[x2, *
*u3], u3]. Note
that h(fi1), h(fi2) and h(fi3) lie in H*(S3), H*( Y 7) and H*( L3), respectiv*
*ely.
There is a contradiction. Thus fi4 = 0.
Now we show that fi3 = 0. Let OE be the composite
H3 4 (3)
L3 ____ B ____ RP42____ (RP2) ____ L3.
Then OE*: H7( L3) ____H7( L3) is an isomorphism. Hence
OE*: P Hr( L3) ____P Hr( L3)
is an isomorphism for r 13 and so
OE*: ßr( L3) ____ßr( L3)
is an isomorphism for r 12. Let _ be the composite
H3 4 (3)
RP42____ (RP2) ____ L3.
From the commutative diagram
H3 4 (3)
RP42 ____ (RP2)
6 6
 
 
 
 
 H 
S3 _______3 S6,
we have _*(fi1) = 0. Recall that ß11( Y 7) ~=ß11(P 7(2)) 2 . ß10(S7) = Z=2 *
*Z=4.
By Theorem 5.13, j*(!6) is a generator for ß11(P 7(2)) = Z=2. From the homotopy
commutative diagram
j*(!6) 4 H3 4 (3)
S10 ____________ RP2 _______ (RP2)
  
 6 6
  
!6  
  
? J([x , u ])  H 
S6 __________23 (S2 _ S3)_3 (S2 _ S3)(3),
we have _*(j*(!6)) = 0. By expecting the homotopy commutative diagram
([4], ~j)6 6 7 pinch 7
S6 ________S _ P (2)______Y _________S
6 6 6 6
   
26  ffi 27
   
   
S9 ____________* __________S10======= S10,
122 J. WU
the element Effi is a generator for the summand 2ß10(S7) in ß11( Y 7). Since *
*L3
is 6connected, there is a homotopy commutative diagram
H3 4 (3)
RP42 _____ (RP2) ____ L3
6 6
 
 _0
 
 
 pinch 
Y 7_________________________S7.
Thus _*(Effi) = _0*(2 7) = 2_0*( 7) = 0 in ß10( L3) = Z=2 Z=2. And therefore
_*(fi2) = 0. It follows that
0 = _*(2ff) = _*(fi3) = OE*(fi3).
Thus fi3 = 0.
Continuation of the proof. Now consider the homotopy commutative diagram of fib*
*re
sequences
F 5{2} ____ P 5(2)______wS5
w
6 6 ww
  ww
 f~  f w
  ww
  w
 
F ______ RP42 _____ S5.
Since f S3: S3 ! P 5(2) is null homotopic, f*(fi1) = 0. By the relative Sa*
*mel
son product, the composite
J([x2,u3]) 4 5
S6 _______ RP2 ____ P (2)
is null homotopic and so f*(j*(!6)) = 0. Observe that there is a commutative
diagram
~f
B ________ F ______ F 5{2}=== F 5{2}
6 6 6
  
 g 
  
  
 f0  g 
Y 7______ F 5{2} ______X5
  
  
  
pinch  
  
?  ?
S7=====================?S7,
2!4 4 5 7
where X5 is the homotopy fibre of S7 ____S . Note that ß10(X ) ~=2 . ß10(S ). *
*The
element g O f0O ffi :S10 ! X5 is a generator for ß10(X5). Let fi2 = k1j*(!6) + *
*k2E(ffi).
Then
f*(2ff) = f*(fi1 + fi2) = f*(fi1 + k1j*(!6) + k2E(ffi)) = k2f*(E(ffi)).
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 123
Consider the commutative diagram of short exact sequences
0_______ß11(X5) ___ß11(P 5(2))___ß11(S5)_______0w
6 6 ww
  ww
 f* w
  ww
  w
 
0________ß11(F )___ß11( RP42)____ß11(S5)_______0
By Step 5 of Theorem 5.10, the short exact sequence
0 ____Z=4 ____ß11(P 5(2)) ____ß11(S5) ____0
splits. Since f*(Effi) is a generator for Z=4summand in ß11(P 5(2)), we have k*
*2 0(2).
Let ~ff= ff + k2=2E(ffi). Then p*(~ff) = 25and
2~ff= 2ff + k2E(ffi) = fi1 + k1j*(!6).
We Show that k1 = 0. Consider the commutative diagram
J([x2, u3]) 2 3 j 4
S6 ___________ (S _ S )__________ RP2
 
 
 
H2 H2
 
? j(2) ?
(S2 _ S3)(2)______ (RP42)(2).
By [47, Corollary 1.2], the composite H2 O J([x2, u3]) is a loop map and so
H2 O j O J([x2, u3]) = ~,
where ~: S6 ! (RP42)(2)is a map with the property that ~*('6) = s(x2 u3+u3 x2)
in H*( (RP42)(2)) and s is the suspension. Observe that s(x2 x2), s(x2 u3) *
*and
s(u3 u3) are spherical classes in H*( (RP42)(2)). There is a map
f :S5 _ S6 _ S7 ____ (RP42)(2)
such that the elements s(x2 x2), s(x2 u3) and s(u3 u3) lie in the image of f*. *
*Let
X1 = Cf be the homotopy cofibre of f. Then sk7(X1) = P 7(2) _ S7 and s(x2 v4)
is spherical in H*(X1). Let X2 = X1=S7 which cancels the element s(x2 v4). Th*
*en
~H*(X2) has a basis
{s[x2, u3], s[x2, v4] + s(u3 u3), s[u3, v4], s(v4 v4), s(u3 v4)}.
There is a cofibre sequence
S7 _ S7 ____P 7(2) ____ sk8(X2).
Since [S7_S7, P 7(2)] ~=Z=2 Z=2, sk8(X2) ' 3RP53_S8. Thus s(u3 v4) is spherical
in H*(X2). By expecting the Steenrod operations, we have X2=S8 ' CP2 ^ P 5(2).
Thus there is a pinch map g : (RP42)(2)! CP2 ^ P 5(2) such that g*(s[x2, u3]) 6*
*= 0.
Consider the exact sequence
i* 2 5 p* 9 @ 7
0 ____ß11(P 7(2)) ____ß11(CP ^ P (2)) ____ß11(P (2)) ____ß10P (2).
Recall that ß11(P 9(2)) ~=Z=2 Z=2 generated by ff9 O j10 and j*( 8). Observe *
*that
i)@ O j*( 8) = j*(j7 O 8) = 0.
124 J. WU
ii)@(ff9) = ff7 O j + fflj*( 6).
iii)@(ff9 O j10) = ffi*(ff9) O j10= ff7 O j2 + fflj*( 6 O j) = ff7 O j2.
The kernel Ker(@ :ß11(P 9(2)) ! ß10(P 7(2))) is generated by j*( 8). Let
fi 2 ß11(CP2 ^ P 5(2))
so that p*(fi) = j*( 8). Then 2fi is the composite
pinch 5 2 ''^id4 2 2 5
S11 ____CP2 ^ P 5(2) ____ CP ____ CP ____CP ^ P (2).
Note that j ^ idS7 is null homotopic and ß9(S6) _____ß9( 4CP2) is onto. Thus
j ^ id: 5CP2 ! 4CP2 is homotopic to the composite
pinch 9 k 6 6 4 2
5CP2 ____S ____S ____ CP .
Consider the commutative diagram of pinch maps
p0 9
5CP2 ________S
6 6
 
q q0
 
 p 
CP2 ^ P 5(2)____P 9(2)
Since p0*O q*(fi) = q0*O p*(fi) = q0*O j*( 8) = 0, we have 2fi = 0 and so
ß11(CP2 ^ P 5(2)) = Z=2 Z=2
______
generated by j*(!6) and j*( 8). Now let ~ be the composite
H2 4 (2) g 2 5
RP42____ (RP2) ____ (CP ^ P (2)).
Then ~*(j*(!6)) = j*(!6) the generator for ß11(P 7(2)). From the commutative di*
*a
gram
H2 4 (2)
RP42 ____ (RP2)
6 6
 
 
 
 
 H 
S3 ________2 S5,
we have ~*(fi1) = 0 and so
0 = ~*(2~ff) = ~*(fi1) + k1~*(j*(!6)) = k1j*(!6).
Thus k1 = 0 and so 2~ff= fi1 2 ß11(S3) = Z=2.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 125
Continuation for solving the group extension problem. Consider the commutative
diagram
ß11(B)w_____ß11(F )______ß11(S3)
ww  
ww  
ww OE1 OE2
ww  
? q ?
ß11(B) ___ ß11( RP42)___*ß11(SU(3))
 
 
 
 
 
 
? ?
ß11(S5)===== ß11(S5),
where OE1 and OE2 are monomorphisms. Let G be the subgroup of ß11( RP42) genera*
*ted
by ~ffand ffl3. Then G has four elements and q*G :G ! ß11(SU(3)) is onto. Th*
*us
G ~=ß11(SU(3)) and so
ß11( RP42) ~=ß11(SU(3)) ß11(B) = ß11(SU(3)) Z=2 4 Z=4.
We finish the proof now.
6.3.2. The Homotopy Groups ß*(P 3(2)). Now we compute the homotopy groups of
other factors in the decomposition of 30(P 3(2)). Observe that CP2 ^ P 6(2) i*
*s 6
connected.
Lemma 6.34. The homotopy groups of ßr(CP2 ^ P 6(2)) for 7 r 11 are as
follows.
1) ß7(CP2 ^ P 6(2)) = Z=2.
2) ß8(CP2 ^ P 6(2)) = 0.
3) ß9(CP2 ^ P 6(2)) = Z=2.
4) ß10(CP2 ^ P 6(2)) = Z=2.
5) ß11(CP2 ^ P 6(2)) = Z=2.
Proof.Consider the cofibre sequence
''^id 8 2 6 10 ''^id 9
P 9(2) ____P (2) ____CP ^ P (2) ____P (2) ____P (2).
The sequence
''^id* 8 2 6 9
ßr(P 9(2)) ____ßr(P (2)) ____ßr(CP ^ P (2)) ____ßr1(P (2)) ____ . . .
is exact for r 11. Thus
ß7(CP2 ^ P 6(2)) ~=ß7(P 8(2)) = Z=2 and ß8(CP2 ^ P 6(2)) = 0.
Consider the short exact sequence
''^id* 8 2 6
0 ____ß9(P 9(2)) ____ß9(P (2)) ____ß9(CP ^ P (2)) ____0.
We have ß9(CP2 ^ P 6(2)) = Z=2 and there is an exact sequence
''^id* 8 2 6
ß10(P 9(2)) ____ß10(P (2)) ____ß10(CP ^ P (2)) ____0.
126 J. WU
Thus ß10(CP2 ^ P 6(2)) = Z=2 and there is an exact sequence
''^id* 8 2 6 9
ß11(P 9(2)) ____ß11(P (2)) ____ß11(CP ^ P (2)) ____2 . ß10(P (2)) ____0.
Notice that (j^) O ff9 = ff8 O j + fflj*( 7). Then
(j ^ 1) O ff9 O j = ff8 O j2 + fflj*( 7 O j) = ff8 O j2.
Thus j ^ id*:ß11(P 9(2)) ____ß11(P 8(2)) = Z=2 if onto and so
ß11(CP2 ^ P 6(2)) ~=2ß10(P 9(2)) = Z=2.
This completes the calculation.
Lemma 6.35. The homotopy groups ßr((CP2)(2)^ P 6(2)) for 9 r 11 are as
follows.
1) ß9(CP2 ^ CP2 ^ P 6(2)) = Z=2.
2) ß10(CP2 ^ CP2 ^ P 6(2)) = 0.
3) ß11(CP2 ^ CP2 ^ P 6(2)) = Z=2 Z=2.
Proof.Clearly ß9(CP2 ^ CP2 ^ P 6(2)) ~=ß9(CP2 ^ P 8(2)) = Z=2. There is an exact
sequence
ß10(CP2 ^ P 9(2)) ____ß10(CP2 ^ P 8(2)) ____ß10(CP2 ^ CP2 ^ P 6(2)) ____0.
Thus ß10(CP2 ^ CP2 ^ P 6(2)) = 0 and there is an exact sequence
ß11(CP2 ^ P 9(2)) ____ß11(CP2 ^ P 8(2)) ____ß11(CP2 ^ CP2 ^ P 6(2))
____ß10(CP2 ^ P 9(2)) ____0.
Recall that ß11(CP2 ^ P 9(2)) = 0. There is a short exact sequence
0 ____ß11(CP2^ P 8(2)) ____ß11(CP2^ CP2^ P 6(2)) ____ß10(CP2^ P 9(2)) ____0.
From the commutative diagram
[2] 2 2 6
CP2 ^ CP2 ^ P 6(2)_______CP ^ CP ^ P (2)
 
 6
 
 
 
 
? id^j 
CP2 ^ CP2 ^ S6 __________CP2 ^ CP2 ^ S5
 6
 
 
 
 
 
? id^k 
CP2 ^ S10 _______________7CP2 ^ S7,
we have 2 . ß11(CP2 ^ CP2 ^ P 6(2)) = 0 and hence the result.
Now we list a table for ßr(P 3(2)) with r 11. In the following table, an in*
*teger
n indicates a cyclic group Z=n, the symbol " + " the direct sum of groups and nk
indicates the direct sum of kcopies of cyclic group Z=n.
HOMOTOPY THEORY OF THE SUSPENSIONS OF THE PROJECTIVE PLANE 127
Theorem 6.36. ß2(P 3(2)) = Z=2, ß3(P 3(2)) = Z=4, ß4(P 3(2)) = Z=4 and ßr(P 3(2*
*))
for 5 r 11 is as follows
________________________________________________________________________*
*
3
ßr(P_(2))__G1_____G2_____G3____G4____G5____G6____others_*
*
 r = 5  2  23  2  0  0  0  0 
 r = 6  2  2  2  0  0  0  0 
 r = 7  4  23+ 8  42  2  0  0  0 
 r = 8  2  2 + 44  22  0  2  0  0 
 r = 9  2  2 2 2  2  2  2  2  02 
 r = 10  0  24 + 42  82  2  42  02  23 
__r_=_11_____0___2__+_4____2_____2_____2____2_______2___
where G1 = ßr1(S3), G2 = ßr( RP42),G3 = ßr(P 6(2)), G4 = ßr(CP2^P 6(2)), G5 = *
*ßr(P 9(2)),
and G6 = ßr(CP2 ^ CP2 ^ P 6(2)).
Appendix A. The Table of the Homotopy Groups of nRP2
The homotopy groups of the mod 2 Moore spaces, up to 8 + dimension, are listed
in the following table. In the following table, an integer k indicates a cycli*
*c group
Z=kZ, the symbol " + " indicates the direct sum of groups, kl indicates the dir*
*ect
sum of lcopies of the cyclic group Z=kZ.
128 J. WU
______________________________________________________________________*
*
_r\n__n=3___4___5___6__7___8___9__10__11____12__*
*
r=1_____2__2___2___2___2___2__2___2___2_____2__*
*
___0_____4__2___2___2___2___2__2___2___2_____2__*
*
___1_____4__4___4___4___4___4__4___4___4_____4__*
*
3 2 2 2 2 2 2 2
 2  2  2  2  2  2  2  2  2  2  2 
   +  +        
____________4___4_______________________________*
*
5 2 2 2
___3____2___2__2___2____2___2__2___2___2_____2__*
*
2 3 2
 4  2  2  2  8  2  0  0  0  0  0 
  +2          
  4          
  +          
_________8______________________________________*
*
7 3 2
 5  2  2  2  2  2  2  2  2  2  2 
  +   +        
_________4______4_______________________________*
*
10 2 2 2 2 2 2 2
 6  2  2  2  2  2  2  2  2  2  2 
   +  +    +  +    
____________4___4___________4__4________________*
*
5 3 3 3 3 4 4 4 3 3
 7  2  2  2  2  2  2  2  2  2  2 
  +3  +    +      
  4  4    4      
  +          
_________8______________________________________*
*
14 5 2 4 3 3 2
 8  2  2  2  2  2  2  2  2  2  2 
  +2  +  +2  +  +2  +2  +2  +2  +2  +2 
  4  4  4  4  4  4  4  4  4  4 
      +    +   
________________________8__________8____________
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Department of Mathematics, National University of Singapore, Singapore 117543,
Republic of Singapore, matwuj@nus.edu.sg