A splitting result for the free loop space of
spheres and projective spaces
Marcel B"okstedt & Iver Ottosen
May 17, 2004
Abstract
Let X be a 1-connected compact space such that the algebra
H*(X; F2) is generated by one single element. We compute the co-
homology of the free loop space H*( X; F2) including the Steenrod
algebra action. When X is a projective space C Pn, H Pn, the Cay-
ley projective plane Ca P 2 or a sphere Sm we obtain a splitting re-
sult for integral and mod two cohomology of the suspension spectrum
1 ( X)+ . The splitting is in terms of 1 X+ and the Thom spaces
T h(qo), q 0 of the q-fold Whitney sums of the tangent bundle o
over X.
1 Introduction
The main part of this paper is a computation of mod two cohomology of the
free loop space on a space X, with the property that H*(X, F2) is a truncated
polynomial algebra.
This calculation has been done by various other methods before, see [Z ],
[KY ] and [CC ]. We want to make four points in order to explain what is new
in this paper.
First, we are using a new general method. The idea is to resolve the
space X as a degreewise free cosimplicial space, and then apply the functor
"free loop space" degreewise. The Bousfield spectral sequence associated to
this cosimplicial space gives us a spectral sequence converging to the (co)-
homology of the free loop space of X, at least if X is simply connected.
This spectral sequence is a generalization of the Eilenberg-Moore spectral
sequence. Eilenberg-Moore spectral sequences has been applied in this situa-
tion before. The main difference from previous work is that the cosimplicial
1
space we consider is not the same as the cosimplicial spaces obtained from
homotopy pull back squares like
X -- - ! X
? ?
? ?
y y
X -- - ! X x X.
The cosimplicial space we use is related to the cosimplicial resolution of X
(in the sense of [BK ]). In this paper we are using this method to treat some
serious cases. From this point of view, we are dealing with test cases for the
spectral sequence.
Secondly, the spectral sequence does give us information about the ac-
tion of the Steenrod algebra on the cohomology of the free loop space. We
determine this action for spaces belonging to the following list : X = Sn,
X = C Pn, X = H Pn and X = Ca P2. Except for the easy sphere case,
these results are new.
Thirdly, the circle group acts in an obvious way on a free loop space,
and one can consider the Borel construction of this action. The method we
consider extends in a natural way ([BO1 ]) to a spectral sequence converging
to the Borel construction. The structures discovered in [CS ] has recently
boosted the interest in the homology of the Borel construction. In spite
of this, this homology has not been calculated. We intend to study the
corresponding spectral sequence. From this point of view, this paper is a
preliminary result.
Fourthly, for the spaces in the short list above, we obtain a surprising
result. The spaces are all manifolds, and even symmetric spaces. In each
case we consider the tangent bundle o . Let qo be the q-fold Whitney sum
of the tangent bundle with itself. We can consider qo as a sub bundle of
(q + 1)o . This induces an inclusion of the corresponding Thom spaces, and
we define the space Cq(X) as the cofiber of the cofibration
T h(qo )! T h((q + 1)o.)
The result is that cohomology of 1 ( X+ ) with coefficients in F2 is the
same A-module as the cohomology of
i ` j
1 X+ _ (m-2)(q+1)Cq(X) ,
q 0
where m denotes the degree of the algebra generator for H*(X; F2).
A splitting discovered by Ziller [Z ] shows that the homology with Z-
coefficients of the two spectra also agree as abelian groups.
2
This strongly suggest the possibility that this wedge is actually homo-
topy equivalent to 1 ( X)+ . There is some additional evidence for such a
splitting. If X = Sn for n 2, the spectrum 1 ( X+ ) is known and simple.
It is a wedge of sphere spectra and Moore spectra for the group Z=2. In fact,
in this special case, we do obtain a stable splitting of spectra as above.
Remark 1.1. There are a few simply connected spaces besides the spheres and
the projective spaces considered above that have mod two cohomology rings
that are truncated polynomial algebras. But there are certain conditions
which the cohomology ring of such a space has to satisfy. Let X be a 1-
connected and compact space. Assume that H*(X; F2) = F2[x]=(xn+1) for
some n 1 and let m be the degree of x. If n 2 it is known known [A ],
[T ] that m = 2, 4 or m = 8 and n = 2. For n = 1 we have that m can be any
integer bigger that 2.
Notation: All cohomology groups are with F2-coefficients unless stated
otherwise. We write A for the mod two Steenrod algebra, U for the category
of unstable A-modules and K for the category of unstable A-algebras. The
degree of an element x is denoted |x|. The category of non negatively graded
F2-algebras with the property that a2 = a if |a| = 0 is denoted Alg . Note
that we have a forgetful functor K ! Alg .
We write : Alg ! Alg for the de Rham complex functor. Thus (A)
for an A 2 Alg means the free A-algebra on generators {dx|x 2 A} of degree
|dx| = |x|-1 modulo the relations d(x+y) = dx+dy, d(xy) = d(x)y+xd(y)
and (dx)2 = 0. We always write d for the de Rham differential in order to
distinguish it from the simplicial face maps di. By sC we mean the category
of simplicial objects in the category C.
2 Results via a simplicial resolution in sAlg
In this section we compute the E2-term of a spectral sequence form [BO1 ]
converging toward H*( X) in the case where X is a 1-connected, compact
space and H*X a truncated polynomial algebra F2[x]=(xn+1). The E2-term
is given_by_the (non abelian) derived functors of a twisted de Rham complex
functor as follows
__t __ * t
E-m,t2= Hm (H*X; ) := (Lm )(H X) .
See [BO1 ] Definition 7.1 and sections 5, 6 for the precise definitions.
Remark 1.1 shows that the operation ~(x) = Sq |x|-1x vanish on H*X.
By [BO1 ] Theorem 6.4, Theorem 7.6 and Proposition 7.3 this implies that
__ *
H*(H*X; ) ~=H*(H X; ),
3
so it suffices to compute the derived functors of the ordinary de Rham com-
plex functor over the simplicial category sAlg . It turns out that the two
cases n even, n odd are essentially different. The result of the computation
appears in Theorem 2.5.
Let Tn(x) denotes the graded commutative F2-algebra F2[x]=(xn+1) where
n 1 and |x| 1.
Theorem 2.1. There is an almost free simplicial resolution Ro 2 sAlg of
Tn(x) as follows: Rq = F2[x, y1, . .,.yq] for q 0 where |yi| = (n + 1)|x| and
si(x) = x, di(x) = x for all i and
(
yj , i j
si(yj) =
yj+1 , i < j
8
>>xn+1 , i = 0, j = 1
><
yj-1 , i < j, j > 1
di(yj) =
>>yj , i j, j < q
>:
0 , i = q, j = q.
__
Proof. Similar to the proof of Proposition 8.2 in [BO1 ]. |_*
*_|
For a simplicial F2-vector space V we write C*(VP) for the associated
chain complex with Cq(V ) = Vq and differential qi=0di. We write N*(V ) for
the normalized chain complex with Nq(V ) = \1 i qker(di) and differential
d0. Finally, we write D*(V ) for the sub complex of C*(V ) generated by the
degenerate elements [ML ].
Definition 2.2. The element !q 2 (Rq) is defined by !q = dy1 . .d.yq for
q > 0 and !0 = 1. We use the notation
(!q)i = dy1 . .c.dyi.d.y.q for 1 i q,
(!q)i,j= dy1 . .c.dyi.d.d.yj.d.y.q for 1 i < j q.
Define ffq, fiq 2 (Rq) by
Xq
ffq = dx!q , fiq = x!q + dx yi(!q)i ,
i=1
for q > 0 and ff0 = dx, fi0 = x.
Lemma 2.3. For q 0 one has that !q is a cycle in Nq (Ro) for n odd and
ffq , fiq are cycles in Nq (Ro) for n even. Furthermore, the elements (!q+1)r,
(!q+1)j,k, yr(!q+1)j,klie in Dq+1 (Ro) for any n where 1 j < k q + 1
and 1 r q + 1.
4
Proof. For n odd we have d0!q = 0 since d0dy1 = d(xn+1) = 0, di!q = 0 for
0 < i < q since (dyi)2 = 0 and dq!q = 0 since dqdyq = 0. For n even we have
difiq = 0 for all i since
(
xn+1dx!q-1 , i = 0
di(x!q) =
0 , i > 0,
8
>:
0 , otherwise.
Note that dxfiq = xffq such that xdiffq = 0 and hence diffq = 0.
The last statement follows directly by the formulas for the degeneracy
__
maps. |__|
By a (p, q)-shuffle (~, ) we understand two ordered subsequences ~1 <
. .<.~p and 1 < . .<. q of the set {0, 1, . .,.p + q - 1} such that
{~1, . .,.~p} t { 1, . .,. q} = {0, 1, . .,.p + q - 1}.
Furthermore, we use the notation s~ = s~p . .s.~1.
Lemma 2.4. Assume (~, ) is a (p, q)-shuffle and j 1. Then s~(yj) = y j+1
and hence
Y Y
s~(!q) = dyt+1 , s~((!q)j) = dyt+1.
t2{ 1,..., q} t2{ 1,..., bj,..., q}
Proof. We compute s~(yj) by the formulas for the degeneracy maps. If ~1 j
then j = j - 1 and s~(yj) = yj. If ~2 j + 1 and ~1 < j then j = j and
s~(yj) = yj+1. In general, if ~r+1 j +r and ~r < j +r -1 then j = j +r -1
__
and s~(yj) = yj+r. |__|
By results of Bousfield and Dwyer [D ] we have that H*(Tn(x) ; ) =
ss* (Ro) is a graded F2-algebra equipped with homotopy operations
ffii : Hq(Tn(x) ; ) ! Hq+i(Tn(x) ; ) , 2 i q.
The following result gives a complete determination of this structure.
Theorem 2.5. Let n be a positive integer. For n odd, there is an isomor-
phism of algebras
H*(Tn(x) ; ) ~=Tn(x) (dx) [!]
5
where x, dx 2 H0 and flq(!) 2 Hq The homotopy operations are given by
( 2q-1
fl2q(!) , i = q
ffii(flq(!)) = q
0 , i < q.
The algebra generators are represented by cycles in the normalized chain
complex (N* R, d0) with Nq R = \i>0ker di as follows: x = [x], dx = [dx]
and fli(!) = [!i].
For n even, we have
H*(Tn(x) ; ) ~=F2 F2[x]=(xn) < aq, bq|q 0 >
where x 2 H0 and aq, bq 2 Hq. The algebra structure is given by the relations
` ' ` '
p + q p + q
apaq = 0, bpbq = xbp+q, apbq = xap+q.
p p
The homotopy operations are given by ffii(aq) = 0 and
( 2q-1
xb2q , i = q
ffii(bq) = q
0 , i < q.
The generators are represented by cycles as follows: x = [x], aq = [ffq],
bq = [fiq].
Remark 2.6. The description in the theorem implies that H0(Tn(x) ; ) ~=
(Tn(x) ). Note also that
` ' ( r
2q - 1 1 , q = 2 , r 0
=
q 0 , otherwise.
Proof. We start by computing the derived functors additively. The algebra
Tn(x) is the pushout of the diagram F2 F2[y] ! F2[x] where y 7! xn+1.
Note that F2[x] is free as a module over F2[y]. By [BO1 ] Proposition 6.3
there is a Quillen spectral sequence as follows:
E2i,j= TorH*(F2[y];i)(F2, H*(F2[x]; ))j ) Hi+j(Tn(x) ; ).
The algebras F2[x] and F2[y] are free objects in Alg so we find that E2i,j= 0
for j > 0 and
Hi(Tn(x) ; ) ~=E2i,0~=Tori(F2[y])(F2, (F2[x])).
6
There is a resolution (K*, @) of F2 by free (F2[y])-modules as follows:
K* = (v) [w] (F2[y]), @v = y, @fli(w) = fli-1(w)dy
where v 2 K1 and fli(w) 2 Ki. We tensor K* with (F2[x]) over (F2[y])
and get a complex (C*, @) with
C* = (V ) [w] (F2[x]), @v = xn+1, @fli(w) = (n + 1)fli-1(w)xndx
By computing the homology of (C*, @) we find that for n even H0(Tn(x) ; ) =
(Tn(x) ) and
Hq(Tn(x) ; ) = F2[x]=(xn) < dxflq(w), xflq(w) + dxvflq-1(w) >, q > 0.
For n odd we find
H*(Tn(x) ; ) ~= (Tn(x) ) [w].
We now show that there is an additive basis for Hq(Tn(x) ; ) which is
represented by cycles in the normalized complex as stated. Lemma 2.3 shows
that these representatives are indeed cycles. It suffices to show that the
associated classes are linearly independent since, in each degree, the number
of classes agree with the dimension in that degree.
We introduce two gradings on Ro preserved by di. The wedge grading w
which equals the number of wedge factors and the polynomial grading p with
p(x) = 2, p(dx) = 1, p(yi) = 2n+2, p(dyi) = 2n+1 extended multiplicatively.
Let r,s(Rq) denote the subspace of elements ! 2 (Rq) with w(!) = r and
p(!) = s. We have defined a splitting of simplicial F2-vector spaces so it
follows that M
Hq(Tn(x) ; ) ~= Hq(Tn(x) ; r,s).
r,s 0
Assume that n is odd. We must show that for fixed q the elements
xt(dx)ffl!q, 0 t n, ffl 2 {0, 1}
are linearly independent. The bidegree is given by (q + ffl, 2t + ffl + q(2n + *
*1)).
The elements have different bigradings and it suffices to check that they
individually do not represent zero. Recall that the composite N*(V ) !
C*(V ) ! C*(V )=D*(V ) is an isomorphism for any simplicial F2-vector space
Vo so we have that N* (Ro) \ D* (Ro) = 0.
We first check that [xtdx!q] 6= 0. The chains which might hit the rep-
resentative lie in q+1,2t+1+q(2n+1)(Rq+1) which is spanned by the elements
xtdx(!q+1)j for 1 j q + 1 together with !q+1 if t = n. By Lemma 2.3 all
7
these elements are degenerate except for !q+1. It follows that the only non
trivial normalized chain in this bigrading is !q+1. But d0(dy1) = 0 such that
d0!q+1 = 0.
We then check that [xt!q] 6= 0. The chains which might hit the represen-
tative lie in q,2t+q(2n+1)(Rq+1) which is spanned by the elements xt(!q+1)r,
xt+ndx(!q+1)j,ktogether with xt-1dxyr(!q+1)j,kif t > 0 where 1 r q + 1,
1 j < k q + 1. By Lemma 2.3 all these elements are degenerate. So
there are no normalized chains that can have xt!q as boundary.
Assume that n is even. We must show that the elements
xtffq, xtfiq, 1 t n - 1
are linearly independent. The bidegrees are
||xtffq|| = (q + 1, 2t + 1 + q(2n + 1)) , ||xtfiq|| = (q, 2t + 2 + q(2n + 1*
*))
so the elements have different bigradings and it suffices to check that they
individually do not represent zero.
We first check that [xtffq] 6= 0. The chains which might hit the rep-
resentative lie in q+1,2t+1+q(2n+1)(Rq+1) which is spanned by the elements
xt+1dx(!q+1)r for 1 r q + 1 By Lemma 2.3 all these elements are degen-
erate.
We then check that [xtfiq] 6= 0. The chains which might hit the represen-
tative lie in q,2t+2+q(2n+1)(Rq+1) which is spanned by the elements
xt+1(!q+1)r, xt+1+ndx(!q+1)j,k, xtdxyr(!q+1)j,k,
for 1 r q + 1, 1 j < k q + 1. By Lemma 2.3 all these elements are
degenerate.
We now prove that the algebra structure and homotopy operations are as
stated. The product comes from the chain map
g C*(~)
ae : C*( Ro) C*( Ro) -- - ! C*( Ro Ro) --- ! C*( Ro),
where g denotes the Eilenberg-MacLane map [ML ] VIII.8. We use the results
from [BO2 ] to find the homotopy operations
ffii : Hq(Tn(x) ; ) ! Hq+i(Tn(x) ; ), 2 i q.
For a cycle z 2 Nq (Ro) we have ffii([z]) = [ffii(z)] where
X
ffii(z) = s (z)s~(z).
(~, )2V (i,q)
8
Assume that n is odd. By the formulas for the shuffle map and Lemma
2.4 we find that
X X ` p + q'
ae(!p !q) = s (!p)s~(!q) = !p+q = !p+q,
(~, ) (~, ) p
where the sums are over all (p, q) shuffles. Thus the algebra structure is as
stated.
The top homotopy operation is as stated since by Lemma 2.4 we have
X X ` 2p - 1'
ffiq(!q) = s (!q)s~(!q) = !2p = !2p.
(~, )2V (q,q) (~, ),~1=0 p - 1
For the other homotopy operations we note that when (~, ) 2 V (i, q)for
2 i q - 1 then both s (!q) and s~(!q) contains a factor dy1 by Lemma
2.4 so the operations vanish since (dy1)2 = 0.
Assume that n is even. We have that ae(ffp ffq) = 0. So the first product
formula holds. By the formulas defining fip and fiq we find that
ae(fip fiq)= x2ae(!p !q)
Xp Xq
+ xdx ae(yi(!p)i !q) + ae(!p yj(!q)j) .
i=1 j=1
p+q
The first term equals p x2!p+q. By Lemma 2.4 we find that
X X
ae(!p yj(!q)j) = s (!p)s~(yj)s~((!q)j) = y j+1(!p+q) j+1,
(~, ) (~, )
and by symmetry
X
ae(yi(!p)i !q) = y~i+1(!p+q)~i+1.
(~, )
p+q
We insert these above and find that ae(fip fiq) = p xfip+q. This gives the
second formula for the product. The third follows by multiplication with dx
on the chain level.
A similar argument shows that ffiq(fiq) = 2q-1q-1xfi2q. For 2 i q - 1
we get ffii(fiq) = 0 since (dy1)2 = 0. Finally, we have that ffik(ffq) = 0 for *
*all k
__
since (dx)2 = 0. |__|
9
3 Normalized chain complex manipulations
This very technical section is needed in the discussion of the Steenrod algebra
action in the next section.
Let Ao be a simplicial ring. In a DGA it follows directly from the def-
inition that the product of two cycles is a cycle, and that the product of
a cycle with a boundary is a boundary. The Eilenberg-Zilber equivalence
allows you to transfer this statement to simplicial rings. We will only use a
rather special case.
Definition 3.1. Let m(a, b) = s0(a)s1(b) - s0(b)s1(a) 2 An+1 and q(a) =
s0(a)s1(a) 2 An+1 for a, b 2 An.
When n = 1 the map m can be described as the shuffle map followed by
g ~
the multiplication map A1 A1 ! A2 A2 ! A2.
The bilinear map m is skew-symmetric and m(a, a) = 0. In case Ao is
a simplicial F2-algebra we have that q(a + b) = q(a) + q(b) + m(a, b) so the
two constructions formally behave like a symmetric bilinear form and the
quadratic form belonging to it.
We will show that m preserves the normalized chain complex, and pass
to well-defined operations on its homology. The situation for q is similar, but
slightly more complicated.
Lemma 3.2. Assume that a, b 2 Nn(A) and c 2 An-1 (in particular we may
take c = 1 when Ao is unital). Then
o s1s0(c)m(a, b) 2 Nn+1(A).
o If d0(a)c = 0 = d0(b)c, then s1s0(c)m(a, b) is a cycle.
o If d0(a)c = 0 is a cycle and b is a boundary, then s1s0(c)m(a, b) is a
boundary.
Proof. We compute the face maps on the element z = s0s0(c)m(a, b).
d0(z)= s0(c)(as0d0(b) - s0d0(a)b) = as0(cd0(b)) - s0(cd0(a))b,
d1(z)= s0(c)(ab - ab) = 0,
d2(z)= s0s0d0(c)[s0d1(a)b - as0d1(b)] = 0,
di(z)= s0s0di-2(c)[s0di-1(a)s1di-1(b) - s1di-1(a)s0di-1(b)] = 0 fori 3.
The first two statements follows from this.
Now assume that d0(a)c = 0 and that b = d0(x) for some x 2 Nn+1(A).
We define
y = s2s1s0(c)[s2s1(a)s0(x) - s2s0(a)s1(x) + s1s0(a)s2(x)] 2 An+2.
10
Note that this element is derived from the shuffle map. We compute face
maps:
d0(y) = s1s0(c)[s1s0d0(a)x - s1(a)s0d0(x) + s0(a)s1d0(x)]
= s1s0(c)s1s0d0(a)x + s1s0(c)[s0(a)s1(b) - s1(a)s0(b)]
= s1s0(cd0(a))x + s1s0(c)m(a, b) = s1s0(c)m(a, b),
d1(y) = s1s0(c)[s1(a)x - s1(a)x + s0(a)s1d1(x)] = 0,
d2(y) = s1s0(c)[s1(a)s0d1(x) - s0(a)x + s0(a)x] = 0,
di(y) = s1s0di-2s0(c)[s2s1di-2(a)s0di-1(x)
- s2s0di-2(a)s1di-1(x) + s1s0di-2(a)s2di-1(x)] = 0 fori 3.
__
|__|
Lemma 3.3. Assume that a 2 Nn(A), b 2 Nn+1(A) and c 2 An (in particu-
lar we may take c = 1 if Ao is unital).
o If ca2 = 0, then s0(c)q(a) 2 Nn+1(A).
o If in addition cas0d0(a) = 0 (in particular, if a is a cycle), then
s0(c)q(a) is a cycle.
o If s0(c)b2 = 0, then s0(c)q(d0(b)) is a boundary.
Proof. Put y = s0(c)q(a) = s0(c)s0(a)s1(a). The first two statements follows
by the computation
d0(y) = cas0d0(a),
d1(y) = ca2 = 0,
d2(y) = s0d1(c)s0d1(a)a = 0,
di(y) = s0di-1(c)s0di-1(a)s1di-1(a) = 0 ifi 3.
We claim that if s0(c)b2 = 0, then s1s0(c)s1(b)s2(b) is a chain with bound-
ary s0(c)q(d0(b)). The claim follows by the computation:
8
>>s0(c)s0d0(b)s1d0(b) = s0(c)q(d0(b)) ifi = 0,
><
s0(c)bs1d1(b) = s0(c)b . 0 = 0 ifi = 1,
di(s1s0(c)s1(b)s2(b)) =
>>s0(c)b2 = 0 ifi = 2,
>:
s1s0di-2(c)s1di-1(b)s2di-1(b) = 0 ifi 3.
__
|__|
11
4 The Steenrod algebra action
The algebra Tn(x) is sometimes an object in K. For n 2 this happens if
and only if |x| is a power of two. For n = 1 it happens if and only if |x| 1.
There is at most one A-action which makes Tn(x)_an_object in K.
For any K 2 K there is an A-action on H*(K; ). In this section we
determine this action when K = Tn(x) 2 K.
We first consider the case where |x| is a power of two. Define Tn,t(x)2 K
for n 1, t 0 by Tn,t(x)= F2[x]=(xn+1) with m = |x| = 2t. It is an object
in K by the formula ` '
mj j+j=m
Sqi(xj) = x .
i
Let : K ! K denote the doubling functor with (K)2j = Kj and
(K)2j+1 = 0 for K 2 K. The operation Sq2i act on (K) via the action of
Sq ion K. The operation Sq2i+1 acts trivially on (K). By [BO1 ] Theorem
6.4 and Theorem 7.6 we have the following result:
Proposition 4.1. Let A 2 K and assume that A = t(K) for some t 1
and K 2 K. Then there is an isomorphism of A-modules for all q:
__ t
Hq(A; ) ~=ssq (Bo)
where Bo 2 sK is an almost free resolution of K.
Note that Tn,t(x)= tTn,0(x)where the A-action on Tn,0(x)is given by
` '
i i+j
Sq jxi = x . (1)
j
Let RKo 2 sK denote a simplicial resolution of Tn,0(x) 2 K. There is a
simplicial algebra map Ro ! RKo which induces a homotopy equivalence.
We will abuse the notation, and denote the images in RKq of the classes
x, y1, . .,.yq 2 Rq by the same names.
By the formulas for the boundary maps and (1) we see that the element
Sq i(y1) + n+1ixiy1 2 RK1is a cycle for 1 i n + 1. Since RKois a resolution
of a discrete ring, ssk(Ro) = 0 for k > 0 and we can find elements z(i)2 RK2
for 1 i n + 1 such that
` '
n + 1 i (i) (i)
d0(z(i)) = Sqi(y1) + x y1, d1(z ) = d2(z ) = 0.
i
As we shall see below this information on RKoturns_out to be sufficient for
the computation of the A-action on H*(Tn,t(x); ).
12
Theorem 4.2. Let n 1, t 1 and assume that n is odd. Then the
A-action on __
H*(Tn,t(x); ) ~=Tn,t(x) (dx) [!]
is given by the following formulas where i, q 0:
` '
ti q(n + 1) i
Sq2 (flq(!)) = x flq(!) (2)
i
` '
ti j j i+j
Sq2 (x ) = x (3)
i
(
ti dx , i = 0
Sq2 (dx) = (4)
0 , i > 0.
__
If k 6= 0 mod 2t then Sqk(z) = 0 for all z 2 H*(Tn,t(x); ).
Proof. By Proposition 4.1 it suffices to prove the theorem for t = 1. We use
the simplicial object (RKo) for our computations. We already know that (3)
and (4) hold. Formula (2) holds for q = 0 and also for q = 1 since
` '
n + 1 j
d0(dz(j)) = Sq2jdy1 + x dy1, 1 j n + 1.
j
We now prove that (2) also holds for q = 2. Let vj and wj denote the
following elements for 0 j n + 1: vj = Sq2jdy1 and wj = n+1jxjd(y1).
Recall the constructions m and q from section 3.
Claim 1: m(vj, vk) is a boundary for j 6= k.
We have d1vj = d1wj = 0 since d1y1 = 0. Furthermore, d0vj = 0 and
d0wj = 0 since n is odd. Thus vj and wj are cycles and vj+ wj is a boundary
(actually 0 for j = 0). Note that
m(vj, vk) = m(vj + wj, vk) + m(wj, vk + wk) + m(wj, wk).
The first two terms of the right hand side are boundaries by Lemma 3.2. We
now show that the last term m(wj, wk) vanish and Claim 1 follows.
m(wj, wk) = s0(wj)s1(wk) + s0(wk)s1(wj)
` '` '
n + 1 n + 1 j k j k
= d(x y2)d(x y1) + d(x y1)d(x y2)
j k
so m(wj, wk) = 0 unless j and k are both even. If they are we get
` '` '
n + 1 n + 1 j+k
m(wj, wk) = x (dy2dy1 + dy1dy2) = 0.
i j
13
Claim 2: q(vj) is a cycle for j > 0 which represents the same homology
class as n+1jx2jdy1dy2.
By Lemma 3.3 we see that q(vj) is a cycle. Note that q(vj + wj) =
q(vj)+q(wj)+m(vj, wj). By Lemma 3.3 we have that q(vj+wj) is a boundary.
Furthermore, m(vj, wj) is a boundary since m(vj, wj) = m(vj + wj, wj) +
m(wj, wj) and m(vj+wj, wj) is a boundary by Lemma 3.2 and m(wj, wj) = 0.
Thus q(vj) + q(wj) is a boundary. But
q(wj)= s0(wj)s1(wj)
` '` ' ` '
n + 1 n + 1 j j n + 1 2j
= d(x y1)d(x y2) = x dy1dy2
j j j
so Claim 2 follows.
We can now prove formula (2) for q = 2. By the Cartan formula we have
Sq 2i(!2) =Sq2i(dy1dy2)
X
= Sq2j(dy1) Sq2k(dy2)
j+k=i
X
= Sq2j(dy1) Sq2k(dy2) + Sq2j(dy2) Sq2k(dy1)
j+k=i,j 0 the element dxq(Sq 2iy1) is a cycle representing the
same homology class as n+1ixiff2.
We have the following equation:
dxq(vi+ wi) = dxq(vi) + dxq(wi) + dxm(vi, wi).
Since (vi+ wi)2 = v2i+ w2i= 0 the left hand side is a boundary by Lemma
3.3. The last term on the right hand side is a boundary since m(vi, vi) = 0
and
dxm(vi, wi) = dxm(vi, vi) + dxm(vi, vi+ wi).
Thus dxq(vi) and dxq(wi) represent the same homology class. But dxq(wi) =
n+1 i
i x ff2 so Claim 2 follows.
We now use the homotopy operations and the product structure to prove
the remaining part of the theorem. Note that Sq2ibq = Cq,ixibq and Sq2iaq =
16
Dq,ixiaq for some constants Cq,iand Dq,iin F2. By the following argument it
suffices to show that the constants are as stated for 1 i n - 2. For i = 0
the constants are 1 and for i = n-1 they can can be found from the i = n-2
constants via the relation Sq 2Sq2(n-2)= Sq 2(n-1)which is the "double" of
Sq 1Sqn-2 = Sqn-1. For i n there is nothing to show since xn = 0.
By the product relation xa2 = dxb2 we see that
C2,ixi+1a2 = C2,ixidxb2 = Sq2i(dxb2) = Sq2i(xa2) = x2 Sq2i-2a2 + x Sq2ia2
` ' ` ' ` '
2(n + 1) i+1 2(n + 1) i+1 2(n + 1) + 1 i+1
= x a2 + x a2 = x a2.
i - 1 i i
Thus formula (6) holds for q = 2. So we know that (5) and (6) holds for
q = 0, 1, 2.
We now prove that (6) holds when q = 2r, r 1 by induction on r. The
initial case r = 2 holds by the above. Assume that we have the formula for
some r 2. By the general relations [G ] we find that
X
Sq 2i(ffiq(bq)) = ffiq(Sq ibq) + Sq t(bq) Sqi-t(bq).
t* 0 and that
d0~D(Aq) = ~D(d0 id + id d0)(Aq)
= ~D(dxd(Sq 2(y1) + xy1) !q-2) + ~D(dxdz(1) x2dx!q-3)
18
Here the last term vanish since (dx)2 = 0. So by the formula for the shuffle
map we find
q-2X
d0~D(Aq) = sq-2 . .b.si.s.0.(dxd(Sq 2(y1) + xy1))si(!q-2)
i=0
q-1X
= dx(Sq2(dyj) + xdyj)(!q-1)j
j=1
q-1X q-1X
= dx Sq2(dyj)(!q-1)j + xdx!q-1
j=1 j=1
= Sq2 ffq-1 + (q - 1)xffq-1.
We now check that di~D(Bq) = 0 for i > 0. Since dt(z(1)) = dt(!q-2) = 0
for t > 0 we get the following by Proposition 7.1:
q-2XXi
di~D(Bq) = ~Gt(dxdz(1) dt(yj(!q-2)j)).
j=1 t=1
When 1 t q - 3 one has that dt(yj(!q-2)j) = yj!q-3 for j = t, t + 1 and
zero otherwise. When t = q - 2 we always get zero. So di~D(Bq) = 0 for
i > 0 as stated.
Finally, we show that d0~D(Bq) = Sq 2fiq-1 + qxfiq-1. The right hand
side gives the following:
Sq2 fiq-1 + qxfiq-1 =
q-1X q-1X
x Sq2(!q-1) + x2!q-1 + dx Sq 2(yj)(!q-1)j + dx yjSq 2((!q-1)j)
j=1 j=1
q-1X
+ qx2!q-1 + qxdx yj(!q-1)j =
j=1
q-1X q-1X
x Sq2(!q-1) + (q + 1)x2!q-1 + qxdx yj(!q-1)j + dx Sq 2(yj)(!q-1)j
j=1 j=1
q-1X j-1X q-1X
+ dx yj( Sq2(dyk)(!q-1)k,j+ Sq2(dyk)(!q-1)j,k) (8)
j=1 k=1 k=j+1
For the left hand side, we use that d0~D = ~D(d0 id + id d0) by
Proposition 7.1. Since (dx)2 = 0 we find that
~D(id d0)(Bq) = ~D(d(xz(1)) x2dx!q-3 + dxdz(1) x3!q-3) = 0
19
and hence
d0~D(Bq) = ~D(d0 id)(Bq)
q-2X
= ~D(d(xd0z(1)) !q-2) + dx ~D(d(d0z(1)) yj(!q-2)j). (9)
j=1
The first term on the right hand side of (9) gives the following:
q-1X
~D(d(xd0z(1)) !q-2) = d(x Sq2(yj) + x2yj)(!q-1)j =
j=1
q-1X
dx Sq 2(yj)(!q-1)j + x Sq2(!q-1) + (q - 1)x2!q-1. (10)
j=1
By the formula for the shuffle map D the second term on the right hand side
of (9) equals
q-2Xq-1X
dx (Sq 2(dyk) + xdyk)sk-1(yj(!q-2)j) =
j=1k=1
q-2XXj
dx (Sq 2(dyk) + xdyk)yj+1(!q-1)k,j+1
j=1k=1
q-2Xq-1X
+ dx (Sq 2(dyk) + xdyk)yj(!q-1)j,k.
j=1k=j+1
After performing the substitution j0 = j + 1 in the first sum on the right
hand side we see that this equals
q-1X q-2X
xdx (j - 1)yj(!q-1)j + xdx (q - 1 - j)yj(!q-1)j
j=2 j=1
q-1Xj-1X q-2Xq-1X
+ dx yjSq 2(dyk)(!q-1)k,j+ dx yjSq 2(dyk)(!q-1)j,k.
j=2k=1 j=1k=j+1
By a final reduction we conclude that the second term on the right hand side
of (9) equals
q-1X
(q - 2)xdx yj(!q-1)j
j=1
q-1X j-1X q-1X
+dx yj( Sq2(dyk)(!q-1)k,j+ Sq2(dyk)(!q-1)j,k) (11)
j=1 k=1 k=j+1
20
__
Since the sum of (10) and (11) equals (8) we are done. |__|
When n = 1, |x| 1 we have that Tn(x) = (x) which is an object in K
with trivial A-action.
__
Proposition 4.5. The A-action on H*( (x); ) is trivial when |x| 2.
__
Proof. Put m = |x|. We have computed H*( (x); ) ~= (x) (dx) [!]
where ||x|| = (m, 0), ||dx|| = (m - 1, 0) and ||flq(!)|| = (-q, q(2m - 1)).
Because of the bigrading the only possible non zero squares on flq(!) are
Sq m-1 flq(!), Sq m flq(!), Sq 2m-1flq(!) which might equal dxflq(!), xflq(!),
xdxflq(!) respectively. Recall that Sqr is decomposable unless r = 2s, s 0.
So the result follows for all m except for m = 2k and m = 2k + 1 with k 1.
When m = 2k the result is a special case of Theorem 4.2.
Assume that m = 2k+1, k 1. Here we must show that Sqm-1 flq(!) = 0.
By the homotopy operations and algebra structure we see that it suffices to
show that Sqm-1 fl1(!) = 0 and Sqm-1 fl2(!) = 0.
Let RKobe a simplicial resolution of (x) in sK such that RK0is the free
object in K on the generator x (thus RK0~= H*(K(F2, m))). We have a weak
equivalence of simplicial algebras Ro ! RKo. Let y1 2 RKodenote the image
of y1 2 Ro. We have d0y1 = x2 and d1y1 = 0. Since ss0(RKo) = (x) there are
elements w(i)2 RK1for 1 i m-1 such that d0w(i)= Sqix and d1w(i)= 0.
But d0(Sq 2iy1 + (w(i))2) = Sq 2i(x2) + (Sq ix)2 = 0 and d0(Sq 2i+1y1) =
Sq 2i+1(x2) = 0 so there exist elements z(t)2 RK2for 1 t 2m - 1 with
djz(t)= 0 for j = 1, 2 and d0z(2i)= Sq2iy1 + (w(i))2, d0z(2i+1)= Sq2i+1y1.
In RKo we have dz(t)with djdz(t) = 0 and d0dz(t) = Sqtdy1. Thus
Sq m-1 fl1(!) = 0. For fl2(!) note that
2k-1-1
k X (t) (m-t) (2k-1)
Sq2 (dy1dy2) = m(d0dz , d0dz ) + q(d0dz )
t=0
__
so the result follows by Lemma 3.2 and Lemma 3.3. |__|
5 The results
Recall that if X is a space with H*X = F2[x]=(xn+1) where n 2 then
|x| = 2k for k = 0, 1, 2 or |x| = 8 and n = 2 [A ]. Examples are the projective
spaces R Pn, C Pn, H Pn and the Cayley projective plane Ca P2, but there
are other homotopy types of spaces with these cohomology rings. For n = 1
there is a space X with H*X = (x) for any positive value of |x|. The
spheres Sm for m 1 are examples of such spaces.
21
When n is odd one cannot determine the operation Sq 1by our spectral
sequence. For the projective spaces and spheres this information follows from
the results in [Z ].
Proposition 5.1. Consider the operation Sq1 : Ht( X) ! Ht+1( X).
1) X = C Pn, n 1: If n 6= 1 mod 4 then all Sq 1= 0. If n = 1 mod
4 then the only non vanishing Sq 1's appear when t 2 {2nk|k 1} and here
one has Sq1 = (0, id) : F2 F2 ! F2 F2.
2) X = H Pn, n 1: If n 6= 1 mod 4 then all Sq 1= 0. If n = 1 mod
4 then the only non vanishing Sq1's appear when t 2 {(4n + 2)k|k 1} and
here Sq1 = id : F2 ! F2.
3) X = Sm , m 3: If m is odd then all Sq 1= 0. If m is even then
the only non vanishing Sq 1's appear when t 2 {2k(m - 1)|k 1} and here
Sq 1= id : F2 ! F2.
Remark 5.2. In 3) we do not mention the case m = 2. This case is however
included in 1) since S2 = C P1.
Proof. The homology groups of Sm , C Pn and H Pn with Z-coefficients
are listed in [Z ] on page 21. By universal coefficients one finds the cohomolo*
*gy
groups with coefficients in Z=2 and Z=4. The result follows by the long exact
__
coefficient sequence associated with 0 ! Z=2 ! Z=4 ! Z=2 ! 0. |__|
Theorem 5.3. Assume that X is a 1-connected space with cohomology al-
gebra H*X = F2[x]=(xn+1) where n is a positive integer. Put m = |x| and
recall that when n = 1, n = 2, n 3 then m 2 Z+ , m 2 {2, 4, 8}, m 2 {2, 4}
respectively.
1) Assume that n is even. Then there is an isomorphism
H*( X) ~=F2 F2[x]=(xn) < aq, bq|q 0 >
where |x| = m, |aq| = qm(n + 1) - 2q + m - 1 and |bq| = qm(n + 1) - 2q + m.
The algebra structure is given by the relations
` ' ` '
p + q p + q
apaq = 0, bpbq = xbp+q, apbq = xap+q.
p p
The Steenrod algebra structure is given as follows where i, j, q 0:
` '
q(n + 1) + j i+j
Sq mi(xjaq) = x aq,
i
` '
q(n + 1) + 1 + j i+j
Sq mi(xjbq) = x bq
i
22
and Sqt z = 0 for all z 2 H*( X) when t 6= 0 mod m.
2) Assume instead that n is odd. Then there is an isomorphism of algebras
H*( X) ~=F2[x]=(xn+1) (dx) [!]
where |x| = m, |dx| = m - 1, |flq(!)| = qm(n + 1) - 2q. The Steenrod algebra
action satisfies the following for i, j, q 0 and ffl = 0, 1:
` '
q(n + 1) + j i+j ffl
Sqmi(xj(dx)fflflq(!)) = x (dx) flq(!)
i
If t 6= 0 mod m then Sq tz = 0 for all z 2 H*( X) unless z = flq(!) for a
q 1 and t = 1. Here Sq1 flq(!) = cqxndxflq-1(!) for a constant cq 2 F2. If
X = C Pn or X = H Pn then cq = 1 for n = 1 mod 4 and cq = 0 for n = 3
mod 4, for all q 1. If X = Sm , m 2 then cq = 0 for m odd and cq = 1
for m is even, for all q 1.
Proof. The distribution of zero's in the E2 therm shows that all differentials
are trivial. So the spectral sequence collapse and E2 = E1 . When n is even,
there is a unique representative in H*( X) for each element in E1 and the
result follows easily.
Assume that n is odd. We have a filtration of H* = H*( X) as follows:
H* . . .F -2H* F -1H* F 0H* F 1H* = 0
with F -pHq-p=F -p+1Hq-p ~=E-p,q1. Define an algebra map
OE : Tn(x) (dx) [!] ! H*
by x 7! x, dx 7! dx and fli(!) 7! fl0iwhere x and dx are the unique repre-
sentatives for x and dx in E1 and fl0iis one of the two representatives for
fli(!) in E1 . We must check that OE is well defined.
Since [!] ~= (flq(!)|q = 2j, j 0) it suffices to see that (fl0q)2 = 0 when
q is a power of two. From the E1 -term we find that
` '
mq(n + 1) q(n+1)-2q=m 0
(fl0q)2 = Sqq(m(n+1)-2)(fl0q)= x flq
mq(n + 1) - 2q
` '
m(n + 1)q q(n+1)-2q=m 0
= x flq = 0
2q
So OE is well defined. We can now define a filtration on its domain space such
that OE becomes a map of filtered rings. Since the associated map of graded
objects is an isomorphism, OE is an isomorphism.
It follows directly from the E1 term and Proposition 5.1 that the Steenrod
__
algebra action is as stated. |__|
23
6 A Thom space interpretation
Let Md be a closed, connected and oriented smooth manifold without bound-
ary. We denote the dual cohomology class of the fundamental class [M] 2
Hd(M; Z) by u0. Let o ! M be the tangent bundle with Thom space
T h(o )and Thom class uo 2 Hd(M; Z). The zero section defines a map
s0 : M ! T h(o ). Recall that s*0: Hd(T h(o ); Z) ! Hd(M; Z) is given by
multiplication with the Euler characteristics
s*0(uo) = O(M)u0.
Consider the q fold Whitney sum qo = o . . .o . We view qo as a sub
bundle of (q + 1)o by the identity on the first q summands and zero on the
last summand. Thus we have an inclusion sq : T h(qo )! T h((q + 1)o.)
Lemma 6.1. The map s*q: H(q+1)d(T h((q + 1)o ); Z) ! H(q+1)d(T h(qo ); Z)
is given by multiplication with the Euler characteristics
s*q(u(q+1)o) = O(M)uqou0.
Proof. The Whitney sum (q + 1)o is the pullback of the product bundle
o x(q+1)! Mx(q+1) along the diagonal q+1 : M ! Mx(q+1).
We write ffl0 ! M for the trivial 0-dimensional vector bundle over M.
This bundle simply means the identity map M ! M and its Thom space is
M with a disjoint base point T h(ffl0)= M+ .
Form the pullback of o xq! Mxq along a composite map as follows:
qo --- ! o xq --- ! o xqx ffl0---! o xq
? ? ? ?
? ? ? ?
y y y y
q xq idx 1 x(q+1) pr xq
M --- ! M --- - ! M --- ! M
where the projection pr is on the first q factors. The pulback of o xqalong
pr equals o xqx ffl0. The further pulback of this vector bundle is as stated in
the diagram since pr O (id x 1) = id and (id x 1) O q = q+1.
The vector bundle o xqx ffl0 is a sub bundle of o x(q+1)by the map id x s0.
Furthermore o xqx ffl0 o x(q+1)pulls back to qo (q + 1)o along q+1. So
we get a commutative diagram as follows (with Z-coefficients):
~=
H*(T h(o )^(q+1)) --- ! H*(T h(o x(q+1))) --- ! H*(T h((q + 1)o ))
? ? ?
? ?* ?
y (idxs0)y y
~=
H*(T h(o )^q^ M+ ) --- ! H*(T h(o xqx ffl0); Z)---! H*(T h(qo ))
24
The Thom class for o x(q+1)is u^(q+1)oand it maps to the Thom class u(q+1)o
by the upper horizontal map. But (id x s0)*(u^(q+1)o) = u^qo^ O(M)u0 so the
__
result follows by the Thom isomorphism for the lower horizontal map. |__|
Definition 6.2. Let Cq(M) denote the cofiber of sq:
T h(qo )! T h((q + 1)o )! Cq(M)
Proposition 6.3. Let Md be a compact, connected smooth manifold with-
out boundary and of dimension d 2. Assume that H1(M; Z) = 0 and
that Hd-1(M; Z) is torsion free. Assume also that the Euler characteristics
O(M) 6= 0. Put Tq = T h(qo ), Cq = Cq(M) and write OE for the Thom iso-
morphism and @ for the connecting homomorphism.
1) Z-coefficients: There are isomorphisms
eHk(Cq; Z)@*-!eH (T ; Z)-OE*!H (M; Z) , k < (q + 1)d
~= k-1 q ~= k-1-qd
eHk(Cq; Z)~=Z=O(M) , k = (q + 1)d
eHk(Cq; Z)- He (T ; Z)-OE*!H (M; Z) , k > (q + 1)d.
~= k q+1 ~= k-(q+1)d
2) F2-coefficients: There are isomorphisms
Hek(Cq) -@*Hek-1 (T )-OEHk-1-qd(M) , k < (q + 1)d
~= q ~=
Hek(Cq) -! eHk(T )- OEHk-(q+1)d(M) , k > (q + 1)d + 1.
~= q+1 ~=
If O(M) is odd then He(q+1)d(Cq)= eH(q+1)d+1(Cq)= 0. If O(M) is even then
eH(q+1)d(Cq)-! eH(q+1)d(T ) , eH(q+1)d+1(C )-@* eH(q+1)d(T.)
~= q+1 q ~= q
The operation Sq 1 : He(q+1)d(Cq) ! He(q+1)d+1(Cq) is the identity map id :
F2 ! F2 when O(M) = 2 mod 4 and the trivial map 0 : F2 ! F2 when
O(M) = 0 mod 4.
Proof. By universal coefficients H1(M; F2) = 0 so M is oriented. Poincar'e
duality gives that Hd-1(M; Z) = 0. So we have that Hd-2(M; Z) is tor-
sion free and Hd-1(M; Z) = 0 by universal coefficients. It follows that
Hd-1(M; F2) = 0.
The long exact (co)homology sequence and the Thom isomorphism gives
the results except for the statement regarding Sq1. By universal coefficients
one can determine the cohomology groups with Z=4-coefficients. The long
exact coefficient sequence associated with 0 ! Z=2 ! Z=4 ! Z=2 ! 0 then
__
determines Sq1 since it agrees with the Bockstein homomorphism. |__|
25
Theorem 6.4. Let M be a complex or quaternion projective space C Pn,
H Pn, n 1 the Cayley projective plane Ca P2 or a sphere Sm , m 2. Let
r(M) denote the degree of the algebra generator for H*(M; F2). Define the
pointed space CT (M) by
`
CT (M) = M+ _ (r(M)-2)(q+1)Cq(M).
q 0
Then the suspension spectra 1 ( M)+ and 1 CT (M) have isomorphic ho-
mology groups with Z-coefficients. Furthermore the cohomology groups with
F2-coefficients are isomorphic modules over the Steenrod algebra. For the
spheres we have a homotopy equivalence 1 ( Sm )+ ' 1 CT (Sm ).
Proof. We first prove the statement regarding the projective spaces and Z-
coefficients. Let K = C, H, Ca and put r = r(M) = dim RK.
We use Proposition 6.3 to compute the stable homology of CT (M). The
non vanishing groups eHk(Cq(K Pn); Z) equal Z for
k 2 {qd + 1 + jr, (q + 1)d + (j + 1)r|0 j n - 1},
and Z=(n + 1) for k = (q + 1)d. Thus Hek( (r-2)(q+1)Cq(K Pn); Z)equals Z
for k 2 A(r, q) [ B(r, q) and Z=(n + 1) for k = (n + 1)r - 2 (q + 1) and 0
otherwise where
A(r, q) = { (n + 1)r - 2 q + rj + r - 1|0 j n - 1},
B(r, q) = { (n + 1)r - 2 (q + 1) + r(j + 1)|0 j n - 1}.
Finally, Hk(K Pn; Z) = Z for k 2 C(r) where C(r) = {ri|0 i n} and
zero otherwise.
For K = C, r = 2 we have that C(2) [ A(2, 0) = {0, . .,.2n} and that
B(2, q - 1) [ A(2, q) = {2nq + 1, . .,.2n(q + 1)} for q 1. So the group
Hek(CT (C Pn); Z)equals Z Z=(n + 1) for k = 2nm, m 1 and Z otherwise.
For Hk( C Pn; Z) we have the same by the table in [Z ].
For K = H, r = 4 the group Hek(CT (H Pn); Z)equals Z when k belongs
to one of the sets
C(4) [ [q 0B(4, q) = {0} [ {2m(2n + 1) + 4l|m 1, 1 l n},
[q 0 A(4, q) = {2m(2n + 1) + 4l - 4n + 1|m 1, 0 l n - 1},
and Z=(n + 1) when k 2 {2m(2n + 1)|m 1}. Otherwise the group vanish.
For Hk( H Pn; Z) we have the same by [Z ].
For K = Ca , r = 8, n = 2 the group Hek(CT (Ca P2); Z)equals Z for
k 2 {0, 8, 16} [ {22m - 15, 22m - 7, 22m + 8, 22m + 16|m 1} and a Z=3
26
for k 2 {22m|m 1}. The list in [Z ] differs here by having zero's in degrees
0, 8, 16 and 22m - 15, m 1 instead of Z's.
This is because there is an error in Ziller's final table of H*( Ca P2; Z).
His result is the same as ours when one corrects this error as we will now see.
By [Z ] Theorem 8 one has
M
H*( Ca P2; Z) ~= H*-~(B)(B; Z),
B
where the sum is over all critical sub manifolds B Ca P2 and ~(M)
is the index of M. All the critical sub manifolds are diffeomorphic to the
sphere bundle T1M = S(o ). By the table on page 20 of [Z ] one sees that
Hk(T1Ca P2; Z) = Z for k = 0, 8, 23, 31 and Hk(T1Ca P2; Z) = Z=3 for k =
15 and zero otherwise. The index can be found in the table on page 11:
~(cm ) = 22m - 15.
So Ziller shows that
M
Hk( Ca P2; Z) ~=Hk(Ca P2; Z) Hk-22m+15(T1Ca P2; Z).
m 1
Thus one gets a Z when k 2 {0, 8, 16} or k = 22m-15+t for t 2 {0, 8, 23, 31},
m 1 and one gets a Z=3 when k = 22m, m 1. This is the same as our
result.
Now we prove the statements regarding F2-coefficients for the projective
spaces. We use Proposition 6.3 to find the A-action on He*(CT (K Pn)). We
have that H*(K Pn) = F2[x]=(xn+1) where |x| = r. By [MS ] page 133, Wu's
theorem implies that the total Stiefel-Whitney class of o is w(o ) = (1 + x)n+1
and hence 1 ` '
X q(n + 1)
w(qo ) = (1 + x)q(n+1)= xi.
i=0 i
We have a Thom isomorphism OE : H*(K Pn) ! He*+rnq(T h(qo ))and the
action on the fundamental class satisfies Sqk uqo = OE(wk(qo )) such that
` '
q(n + 1) i k
Sqriuq = x uq , Sq uq = 0 ifk 6= 0 mod r (12)
i
Assume that n is even. Then O(K Pn) = n + 1 is odd. Proposition 6.3
shows that eH*( (r-2)(q+1)Cq(K Pn))has F2-basis {ajq, bjq+1|0 j < n} with
ajq= oe(r-2)(q+1)@*(xjuqo) , bjq+1= oe(r-2)(q+1)(Q*)-1(xj+1u(q+1)o),
|ajq| = qr(n + 1) - 2q + r - 1 + rj,
|bjq+1| = (q + 1)r(n + 1) - 2(q + 1) + r + rj
27
where Q : T h((q + 1)o )! Cq(K Pn) denotes the quotient map. By (12) and
the Cartan formula we have
` '
q(n + 1) + j i+j
Sqri(ajq) = aq ,
i
` '
(q + 1)(n + 1) + 1 + j i+j
Sqri(bjq+1) = bq+1
i
where by definition atq= btq+1= 0 if t n. The operations Sq kfor k 6= 0
mod r sends the basis elements to zero. The reason why there is no therms
of the form ajqin the formula for Sqt(bjq+1) is that the degrees of such terms
are to small. The result follows by comparing with Theorem 5.3.
Assume that n is odd. Then O(K Pn) = n + 1 is even. Proposition 6.3
shows that eH*( (r-2)(q+1)Cq(K Pn))has F2-basis {cjq, djq+1|0 j n} with
cjq= oe(r-2)(q+1)@(xjuqo) , djq+1= oe(r-2)(q+1)Q-1(xju(q+1)o),
|cjq| = qr(n + 1) - 2q + 1 + rj,
|djq+1| = (q + 1)r(n + 1) - 2(q + 1) + rj
By (12) and the Cartan formula we have
` '
q(n + 1) + j i+j
Sqri(cjq) = cq ,
i
` '
(q + 1)(n + 1) + j i+j
Sqri(bjq+1) = bq+1
i
where by definition atq= btq+1= 0 if t > n. The operations Sq kfor k 6= 0
mod r vanish on the basis elements except for Sq1 d0q+1which equals cnqwhen
n = 1 mod 4 and zero when n = 3 mod 4. Note that we cannot hit any of
the classes bjq+1by a Steenrod operation on one of the classes ajqsince ajqis
in the image of @. The result follows by comparing with Theorem 5.3. The
class cjqcorresponds to xjdxflq(!) and djq+1corresponds to xjflq+1(!).
For the spheres we can prove directly that there is a homotopy equiv-
alence. We first identify the stable homotopy type of Cq(Sm ). We have
Sm Rm+1 with trivial one dimensional normal bundle ffl1 ! Sm and
o ffl1 ~=fflm+1 . Thus
q+tT h(qo )~=T h((q + t)ffl1 qo~)=T h(fflq(m+1)+t)~= q(m+1)+tSm+ ~=
Sq(m+1)+t_ Sq(m+1)+m+t .
By taking the (q + 1)-fold suspension in Definition 6.2 we find that
Sq(m+1)+1 _ S(q+1)(m+1)! S(q+1)(m+1)_ S(q+1)(m+1)+m ! q+1Cq(Sm ).
28
Any map Sr ! Ss is null homotopic when r < s and a map Sr ! Sr is
determined, up to homotopy, by its degree. By Proposition 6.1 we have that
the degree of the above self map of S(q+1)(m+1)is O(Sm ) which is 0 for m odd
and 2 for m even. So q+1Cq(Sm ) is homotopy equivalent to the following
wedge of spheres and Moore spaces:
(
S(q+1)(m+1)_ S(q+1)(m+1)+m _ Sq(m+1)+2 _ S(q+1)(m+1)+1 , m odd
M(Z=2, (q + 1)(m + 1)) _ S(q+1)(m+1)+m _ Sq(m+1)+2 , m even.
Thus (m-2)(q+1)Cq(Sm ) is stably equivalent to
(
S2(q+1)(m-1)_ S2(q+1)(m-1)+m_ S(2q+1)(m-1)_ S2(q+1)(m-1)+1 , m odd
M(Z=2, 2(q + 1)(m - 1)) _ S2(q+1)(m-1)+m_ S(2q+1)(m-1) , m even.
So we have determined the stable homotopy type of CT (Sm ).
On the other hand the results in [BM ], [CC ] give a stable decomposition
of Y where Y is a based connected space. Let Ck S1 denote the cyclic
group of order k and put Dk(Y ) = S1+^Ck Y ^k. Then there is a stable
equivalence `
Y ' Dk(Y ).
k 1
We use this result for Y = Sm-1 . Here each Dk(Sm-1 ) decomposes further.
The cofiber of the inclusion i : Ck+ S1+is a k-fold wedge of circles. We
have an associated Barrat-Puppe sequence
Ck+ ^Ck Y ^k -i^id--!Dk(Y )-- - ! (S1 ^ Ck+) ^Ck Y ^k --- !
fl fl fl
fl fl fl
fl fl fl
Y ^k --- ! Dk(Y ) -- - ! S1 ^ Y ^k --- !
So for Y = Sm-1 we find
@m-1 (m-1)k+1 m-1
S(m-1)k ! Dk(Sm-1 ) ! S(m-1)k+1 --- ! S ! Dk(S ) ! . . .
By analyzing the Barrat-Puppe sequence
Ck+ -i!S1+! S1 ^ Ck+ -@!S1 ^ Ck+ ! S1 ^ S1+! . . .
via the mapping cone of i we see that (@m-1 )* = id - T* on reduced integral
homology. Shifting to integral cohomology we see that the degree of @m-1 is
2 when m - 1 is odd and k - 1 is odd and 0 otherwise. Thus
(
M(Z=2, (m - 1)k + 1) , for m and k even
Dk(Sm-1 ) '
S(m-1)k+1 _ S(m-1)k+2 , otherwise.
29
So Dk(Sm-1 ) is stably equivalent to M(Z=2, (m - 1)k) when m and k are
both even and stably equivalent to S(m-1)k_ S(m-1)k+1 otherwise. The result
__
follows. |__|
7 Appendix: A property of the shuffle map
Let Ao and Bo be simplicial modules over a commutative ring k. Recall that
the shuffle map D : C*(A) C*(B) ! C*(A B) is defined as follows [ML ]
VIII.8: X
Dn(a b) = (-1)ffl(~)s q. .s. 1a s~p . .s.~1b
(~, )
where a 2 Ap and b 2 Bq with p + q = n. The sum is taken over all (p, q)
shuffles (~, ) cf. [ML ] and the sign is given by the formula
Xp
ffl(~) = (~i- (i - 1)). (13)
i=1
We want to investigate how the shuffle map commutes with the simplicial
boundary maps. Recall that
8
>:
sjdi-1 , i > j + 1.
Proposition 7.1. There is a commutative diagram as follows:
(C*(A) C*(B))n -Dn--! Cn(A B)
? ?
@0?y d0 d0?y
Dn-1
(C*(A) C*(B))n-1 --- ! Cn-1(A B)
where the map @0 is defined by
@0 : Ap Bq ! (Ap-1 Bq) (Ap Bq-1); @0 = d0 id + (-1)pid d0.
Here Aj = Bj = 0 when j < 0 by convention. When i 1 there are k-linear
maps Ft, Gt : (C*(A) C*(B))n-1 ! Cn-1(A B) for 1 t i such that
Xi
(di di) O Dn = FtO (dt id) + GtO (id dt) .
t=1
30
Proof. We rewrite the sum defining (d0 d0) O Dn by commuting d0 and s~j,
using formula (14). There are two possibilities:
(
s~p-1 . .s.~1-1d0 if~1 1
d0s~p . .s.~1=
s~p-1 . .s.~2-1 if~1 = 0
corresponding to the two cases
d0s q. .s. 1 d0s~p . .s.~1=
(
s q-1. .s. 2-1 s~p-1 . .s.~1-1d0 if 1 = 0
s q-1. .s. 1-1d0 s~p-1 . .s.~2-1 if~1 = 0.
To the shuffle (~, ) of the elements in [n - 1], we associate a new shuffle
(~ - 1, - 1) on [n - 2] by the rule (~ - 1)i = ~i- 1 and ( - 1)i = i- 1.
There is of course the problem that (~ - 1)1 can be negative. Actually, either
(~ - 1)1 = -1 or ( - 1)1 = -1, and we delete the corresponding element.
What we have left is a genuine shuffle on [n - 2].
This defines a bijection between shuffles on [n - 1] such that ~1 = 0 and
shuffles on [n - 2]. Similarly, it defines a bijection between shuffles of [n -*
* 1]
element such that 1 = 0 and shuffles on [n - 2] elements.
The sign of the shuffle (~ - 1, - 1) is, according to formula (13) given
by
( P
((~ - 1)i- (i - 1)) = ffl(~) if~1 = 0
ffl(~ - 1) = P 1 i p-1
1 i p((~ - 1)i- (i - 1)) = ffl(~) - pif 1 = 0.
So we can express the sum defining (d0 d0) O Dn as a sum of shuffles (with
sign) applied to either (d0 id) or (id d0). Thus the diagram commutes.
Let i 1. We use formula (14) to rewrite dis~p . .s.~1, moving the di
toward the right. There are two possibilities.
If {i, i - 1} \ {~1, . .,.~p} = ; we get that the element equals some
s~0p. .s.~01dr. Looking closer at the formula (14), we see that since i 1, we
have that i r 1. This means that there are maps ff and fi such that
(di di)(s q. .s. 1a s~p . .s.~1b) = (ff fi) O (id dr)(a b).
Similarly, if {i, i - 1} \ { 1, . .,. q} = ; there are maps ff0 and fi0 and *
*an
integer s with i s 1 such that
(di di)(s q. .s. 1a s~p . .s.~1b) = (ff0 fi0) O (ds id)(a b).
We now only have to take care of the summands of Dn corresponding to
shuffles such that either i = ~r and i - 1 = s for some r, s or i - 1 = ~r
31
and i = s for some r, s. These shuffles come in pairs: If (~, ) is a shuffle
with ~r = i and s = i - 1, then we can define a different shuffle (~0, 0) by
interchanging ~r and s. Thus
(
~t ift 6= r
~0t=
i - 1 ift = r.
Then (~, ) and (~0, 0) will according to formula (13) contribute equal terms
with opposite signs in the sum computing (di di) O Dn. These terms cancel,
__
and the second part of the proposition is proved. |__|
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33
*