The Hopf ring for P (n)
Douglas C. Ravenel * W. Stephen Wilson
University of Rochester Johns Hopkins University
Rochester, New York 14627 Baltimore, Maryland 21218
drav@troi.cc.rochester.edu wsw@math.jhu.edu
February 1, 1995
Abstract
We show that E*(P_(n)_*), the Ehomology of the spectrum for
P (n), is an E* free Hopf ring for E a complex oriented theory with
In sent to 0. This covers the cases P (q)*(P_(n)_*) and K(q)*(P_(n)_*),
q n. The generators of the Hopf ring are those necessary for the
stable result. The motivation for this paper is to show that P (n)
satisfies all of the conditions for the machinery of unstable cohomology
operations set up in [BJW95 ]. This can then be used to produce
splittings analogous to those for BP done in [Wil75 ].
1 Introduction
The spectrum P (n) for n > 0 is the BP module spectrum obtained by killing
the ideal
In = (p; v1; v2; . .v.n1) ss*(BP )
via the SullivanBaas construction, [Baa73 ], [BM71 ], and [JW75 ]. For odd
primes it is a nice multiplicative spectrum by [Mor79 ], [SY76 ], and [W"ur77 ].
______________________________
*Partially supported by the National Science Foundation
1
It comes equipped with a stable cofibration
n1)
2(p P (n) ! P (n) ! P (n + 1) (1:1)
which gives the following short exact sequence in homotopy
0 ! P (n)* vn!P (n)* ! P (n + 1)* ! 0:
The ith space in the spectrum for P (n) will be denoted by P_(n)_i.
Because P (n) is a ring spectrum there are maps
P_(n)_ix P_(n)_j! P_(n)_i+j
corresponding to cup product, in addition to the loop space product
P_(n)_ix P_(n)_i! P_(n)_i:
These induce pairings
O : E*(P_(n)_i) E*(P_(n)_j) ! E*(P_(n)_i+j):
and
* : E*(P_(n)_i) E*(P_(n)_i) ! E*(P_(n)_i):
for a generalized homology theory E*. If E* has a K"unneth isomorphism
for these spaces, e.g. if they are E* free, then these pairings satisfy certain
identities, making E*(P_(n)_*) into a Hopf ring [RW77 ], i.e., a ring object
in the category of coalgebras. The object of this paper is to describe this
structure explicitly for suitable theories E*(), namely, when E is a complex
orientable spectrum with In = 0.
In the next section we will define some special elements
e1 2 P (n)1(P_(n)_1);
a(i) 2 P (n)2pi(P_(n)_1) fori < n;
wi 2 P (n)0(P_(n)_2(pi1))fori n; and
b(i) 2 P (n)2pi(P_(n)_2) fori 0;
2
which have already been defined in previous papers.
Let
e"1aIwK bJ = e"1O aOi0(0)O . .O.aOin1(n1)O wknnwkn+1n+1.O.b.Oj0(0)O bOj*
*1(1). . .
where " = 0 or 1, iq = 0 or 1, wk = [vk], kq 0, and jq 0.
Definition 1.2 We say e"1aIwK bJ is nallowable if
J = pndn + pn+1dn+1+ . .+.pm dm + J0
where dn dn+1 . . .dm and J0 is nonnegative implies km = 0. In other
words,
n +pn+1 +...+pm
wm O bp dn dn+1 dm
does not divide e"1aIwK bJ when dn dn+1 . . .dm . We will denote the
set of such (K; J) by An.
We say e"1aIwK bJ is nplus allowable if e"1aIwK bJ+0 is nallowable.
We will denote the set of such (K; J) by A+n. Note that A+n An.
If we insist that k0 always equals 0 then note that 0plus allowable as
defined above coincides with allowable as defined in [RW77 ], and is the same
as 1allowable with i0 = 0. For n = 1, the allowability condition is vacuous.
k k
Let T Pk(x) = P (x)=(xp ); we say that such an x has height p . Let E be
a BP module spectrum. Then we have a map BP ! E. For p an odd prime,
P (n) is the universal multiplicative BP module spectrum with *(In) = 0.
This fact follows from the work of W"urgler in [W"ur77 , p. 477, 6.8]. Thus if
E has *(In) = 0, then the map factors through P (n).
Theorem 1.3 Let E be a multiplicative BP module spectrum with *(In) =
3
0, n > 0. Let p be an odd prime. As E*algebras:
8 O
>>> E(e1aIwK bJ)
>>> +
>>>(K;J)2An
>>>
>>>
>>> O
>>> I K J
>>> T Pae(I)(a w b )
>>> i0=0
>><(K;J)2An
E*(P_(n)_*) ' >
>>> O
>>> T P (aIwK bJ)
>>> ae(I)
>>>I6=M;i0=1+
>>>(K;J)2An
>>>
>>>
>>> O
>>> P (aM wK bJ)
>:
(K;J)2A+n
where ae(I) is the smallest t with int = 0, and M is the exponent sequence
consisting entirely of ones, i.e., aM = a(0)O a(1)O . .a.(n1).
This calculation includes the calculation of K(q)*(P_(n)_*) for q n. For
q < n, this calculation was carried out in [HRW , Theorem 1.5]. The results
have very little in common.
The theorem is also true as stated for n = 0, P (0) being BP by definition.
In that case there are no a(i)s, so we always have I = M = 0, which means
the truncated polynomials factors are trivial. As remarked above, 0plus
allowability is the same as allowability as defined in [RW77 ], so Theorem 1.3
coincides with the main theorem of [RW77 ]. Note that P (1) is just BP mod
p.
For n = 1 the theorem gives the usual Hopf ring description of the
homology of mod p EilenbergMac Lane spaces. As remarked above, the
allowability condition is vacuous in this case. There is no polynomial factor
since M is infinite, K = 0, and ae(I) = 1 for all I, so the truncated polynomial
algebras all have height p. See [Wil82 , x8] for more details.
The theorem will be proved by studying the bar spectral sequence (see
4
x3) for mod p homology going from H*(P_(n)_i) to H*(P_(n)_i+1). We then
show that the AtiyahHirzebruch spectral sequence collapses for appropriate
E*(). For n = 0 and n = 1 the bar spectral sequence always collapses,
but not for 0 < n < 1. There are no multiplicative extensions for n = 1.
The extensions differ significantly between the n = 0 case, studied in [RW77 ]
and the n > 0 cases studied here which more resemble those in [Wil84 ]. We
will give examples to illustrate in the next section.
The main theorem above contains a description of the generators. No de
scription of Hopf algebras is complete without understanding the primitives.
Although Theorem 1.3 contains all the information about the coproduct in
principle, we can be more explicit. Furthermore, we need a more explicit
description during our proof. Both Theorem 1.3 and our results about prim
itives will be proven inductively simultaneously by degree.
Theorem 1.4 Let E be as in Theorem 1.3. The primitives in E*(P_(n)_*)
have the following description:
(a) A basis for the primitive elements is given by all e"1aIwK bJ such that
(i) if " = 1 then (K; J) 2 A+n,
(ii)if " = 0 then i0 + j0 > 0 and (K; J) 2 An.
(b) A basis for the primitive elements of height p is given by all of the above
primitives aIwK bJ such that in1 = 0.
(c) In the mod p homology, H*(P_(n)_*), the iterated pth powers of the
generators in Theorem 1.3 are all primitives. They are those primitives
with " = 0 and i0 = 1 with (K; J) 2 An  A+nmodulo the vector space
generated by aIwK bJ with i0 = 1 and (K; J) 2 A+n(which are primitive
generators).
Part (a) of the theorem still holds for the n = 0 case. Part (c) is about
the nontrivial pth powers. Some generators are not primitive and so it is
interesting to note that all of their pth powers are primitive.
5
From now on we assume that n > 0.
Our proof of 1.3 will follow the lines of several previous papers; in partic
ular, [RW77 ], [RW80 ], and [Wil84 ]. As in [Wil84 ], it is enough to prove t*
*he
result for ordinary mod p homology and then show the AtiyahHirzebruch
spectral sequence collapses for E. Consequently, we will focus most of our
attention on ordinary homology. In fact, our theorem and proof lies some
where between the work in [Wil84 ] and [RW77 ]. In turn, the work of [Wil84 ]
lies somewhere between [RW77 ] and [RW80 ].
Our proof follows the philosophy of the second author that one can com
pute the homology of spaces in an spectrum if one knows the stable ho
motopy and the stable homology. The homotopy gives the zero dimensional
homology of all the spaces and if one computes by induction on degree using
the bar spectral sequence any false computation of a differential or extension
should lead to a contradiction with the stable homology. This works in many
cases, including this one. Getting a nice Hopf ring description is an entirely
different matter. It seems to depend on having the stable elements appear
at the earliest possible stage unstably.
There are a number of Hopf rings like this that have been computed.
Some are "good" and some are "bad". Examples of good ones are E*(BP__*),
E a complex orientable theory, [RW77 ]; E*(K(n)__*), E a complex ori
entable theory with In = 0, [Wil84 ]; H*(K(Z=(p); *)), [Wil82 , x8]; K(n)*()
for EilenbergMac Lane spaces, [RW80 ]; K(n)*(k(n)_*), [Kra90 ]; H*(KO__),
[Str92]; and more recently the breakthrough description of H*(QS0; Z=(2))
in [Tur ], and its sequel for H*(QS*; Z=(2)) in [ETW ]. All of these examples
can have their Hopf rings described with just a few generators and relations.
There are other similar calculations where the Hopf rings are not so nice,
for example for bo, bu, BP , and k(n). The standard mod p homology of
these does not work out so well as a Hopf ring. Despite that obstacle, the
results for bu and k(n) have been given very nice descriptions in [Har91 ].
By [W"ur77 ] and [Yag77 ] we know that P (n)*(P (n)) is free over P (n)*.
This result is all that is necessary to show that P (n) satisfies all of the
6
machinery for stable operations as in [Boa95 ]. Our results show:
Corollary 1.5 Both P (n)*(P_(n)_*) and the module of indecomposables,
QP (n)*(P_(n)_*), are free over P (n)*.
These two new conditions are enough to make all of the machinery of
unstable operations in [BJW95 ] work, which was the motivation for this
paper. In [BJW95 ], this machinery is used to reprove the second author's
splitting theorem for the spaces in the spectrum for BP , [Wil75 ]. The
second author had conjectured a similar splitting for P (n) and this paper
together with [BJW95 ] allows that splitting to be carried through; see [BW ].
The lowest cases involve Morava Ktheories; e.g. P_(n)_2pn' k(n)_2pnx Y .
The p = 2 case deserves some discussion. As the theorem is stated it is
true for p = 2 as well. In particular it is true for mod 2 homology. The
problem is that all the spectra E that we care about, such as K(n) or P (n),
are not commutative ring spectra. Even in those cases the result is true as
modules if not as algebras because we get the general E from the collapsing
of the AtiyahHirzebruch spectral sequence. As in [Wil84 ], no problems are
caused because of the lack of commutativity of P (n) in P_(n)_*. We make
other comments about p = 2 in the next section.
2 Basic properties
We have a long proposition analogous to [Wil84 , Proposition 1.1].
Proposition 2.1 Let E be as in Theorem 1.3. Let p be an odd prime. We
have elements
e1 2 E1(P_(n)_1);
ai 2 E2i(P_(n)_1) fori < pn;
wi 2 E0(P_(n)_2(pi1))fori n; and
bi 2 E2i(P_(n)_2)
7
such that (letting b(i)= bpi and a(i)= api):
(a) They are natural with respect to E.
(b) The homomorphism x 7! e1 O x is the homology suspension map.
P P
(c) The coproduct is given by ai ! aij aj and bi ! bij bj.
(d) They are all permanent cycles in the AtiyahHirzebruch spectral se
quence for P (n)*(P_(n)_*).
(e) e1 O e1 = b1.
(f) a(i)O a(j)= a(j)O a(i).
(g) b*p(i)= 0.
(h) a*p(i)= 0 for i < n  1.
n1
(i)a*p(n1)= *(vn)a(0) a(0)O wn O bOp(0).
n1
(j)*(vn)e1 = e1 O wn O bOp(0).
Proof. The proof of this proposition is identical to that of [Wil84 , Propositi*
*on
1.1]. In fact, that paper uses P (n) in the proof. For some of the results one
only needs to notice that P (n) and k(n) are the same up to degree 2(pn+11)
and that the proposition takes place in low degrees. A sign has crept into
(e) because of a correction from [BJW95 ].2
We will recall briefly the construction of these elements.
o Let L denote the (2pn  2)skeleton of K(Z=(p); 1). Then there is a
unique lifting
P_(n)_1
oeo 
oe 
oe 

oe 
oe 
oe u
L ______wK(Z=(p); 1)
8
and hence a map E*(L) ! E*(P_(n)_1). The elements e1 and ai are the
images of the standard elements in E*(L).
o wk is the Hurewicz image of
vk 2 ss2pk2(P (n)) = ss0(P_(0)_2pk+2):
o There is a canonical generator of P (n)2(CP 1), which corresponds to
a map CP 1 ! P_(n)_2. The elements bi are the images under this map
of the standard classes in E*(CP 1).
The case p = 2. The comments of [Wil84 ] are relevant here. e1 should be
incorporated into the a's for coproduct purposes and a(i)O a(i)is not zero.
One must fiddle with the proof a little, but not much.
The proof of our theorem will rely on being able to identify elements
in the bar spectral sequence, compute differentials and solve multiplicative
extension problems, all using Hopf ring techniques. The n = 0 case of [RW77 ]
has no differentials but does have extension problems. For the bar spectral
sequence going from H*(P_(0)_i) to H*(P_(0)_i+1) there are extensions when i
is odd. We will illustrate this phenomenon with an example for i = 1. For
k 0 let
gk = e1 O wk1O bOp1(0)O . .O.bOp1(k1)2 H*(P_(0)_1):
Then H*(P_(0)_1) has the exterior algebra
E(g0; g1; . .).
as a factor, and the corresponding Tor group contains the divided power
algebra
(oeg0; oeg1; . .).
as a factor; see 3.2. Now for k > 0 we have (in ordinary mod p homology)
oegk = e1 O e1 O wk1O bOp1(0)O . .O.bOp1(k1)by 2.1(b)
9
= b(0)O wk1O bOp1(0)O . .O.bOp1(k1)by 2.1(e)
= (w1 O bOp(0)) O wk11O bOp1(1)O . .O.bOp1(k1)
= b*p(0)O wk11O bOp1(1)O . .O.bOp1(k1)using 2.6
= (w1 O bOp(0))*pO wk21O bOp1(2)O . .O.bOp1(k1)
by the Hopf ring distributive law [RW77 , 1.12(c)(vi)]
2 k2 Op1 Op1
= b*p(0)O w1 O b(2) O . .O.b(k1) using 2.6
..
.
k
= b*p(0):
It follows that the divided power factor above in the E1 term corresponds
to a polynomial factor
P (b(0); b(1); . .).
in H*(P_(0)_2) and also in E*(P_(0)_2).
For P (n) with n > 0 this type of extension never occurs because of 2.1(g).
The type of extension we have is more interesting because they frequently
lead to truncated polynomial algebras and are implied by the relations 2.1(h)
and (i). Since we work in ordinary mod p homology (vn) = 0. For n = 1,
2.1(i) says
a*p(0)= a(0)O w1 O bOp1(0)
2 Op1 *p
so a*p(0)= (a(0)O w1 O b(0) )
= (a(0))*pO w1 O bOp1(1)
by the Hopf ring distributive law
= a(0)O w21O bOp1(0)O bOp1(1)
3 3 Op1 Op1 Op1
a*p(0)= a(0)O w1 O b(0) O b(1) O b(2)
..
.
so a(0)O x for suitable x could be a polynomial generator.
For n = 2 we have
a*p(0)= 0
10
21
a*p(1)= a(0)O w2 O bOp(0)
2 Op21 *p
so a*p(1)= (a(0)O w2 O b(0) )
21
= (a(0))*pO w2 O bOp(1)
= 0
and (a(0)O a(1))*p= a(1)O (a(1))*p
21
= a(1)O a(0)O w2 O bOp(0)
21
= a(0)O a(1)O w2 O bOp(0)
2 2 Op21 Op21
so (a(0)O a(1))*p = a(0)O a(1)O w2 O b(0) O b(1)
..
.
and for larger n the same thing happens with powers of aM = a(0)O a(1)O
. .a.(n1). This accounts for the polynomial factor in Theorem 1.3.
Also observe that for n = 3, 2.1(i) gives
(a(0)O a(1))*p= a(1)O (a*p(1))
= 0;
(a(0)O a(2))*p= a(1)O (a(2))*p
31
= a(1)O (a(0)O w3 O bOp(0))
31
= a(0)O a(1)O w3 O bOp(0)
2
so (a(0)O a(2))*p = 0;
and (a(1)O a(2))*p= a(2)O (a(2))*p
31
= a(2)O (a(0)O w3 O bOp(0))
31
= a(0)O a(2)O w3 O bOp(0)
3
so (a(1)O a(2))*p = 0:
This (and similar computations for larger n) accounts for the truncations in
Theorem 1.3.
For the spectral sequence computations of the next section, we will need
the following results about pth powers.
11
Define shift operators s on I (if in1 = 0) and J by
as(I)= aOi0(1)O aOi1(2)O . .a.Oin2(n1)and
(2:2)
bs(J)= bOj0(1)O bOj1(2)O . .:.
Lemma 2.3 In the mod p homology of P_(n)_*, let (K; J) be in An,
(a) then
(aIwK bJ)*p = 0 if in1 = 0;
(b) if in1 = 1 let I = I00+ n1. We have
00)n+K (pn1) +s(J)
(aIwK bJ)*p = a0+s(I w b 0 ;
and
(c) If I = M as in 1.3,
n1) +s(J)
(aM wK bJ)*p = aM wn+K b(p 0 :
Note that the potentially nonzero pth powers are not usually in allowable
form. The introduction of the wn together with the possibility that some
jk pn messes this up. Note also that if I is not M (as in 1.3) then the
iterated pth power must eventually be zero but in the case (c) we could have
a polynomial generator. Even though we do not know if these pthpowers are
really nonzero, we know something about their properties if they are.
Corollary 2.4 Each nonzero pthpower in Lemma 2.3 is divisible by a(0)and
is primitive.
Proof of Lemma 2.3. Assume first that I is empty and J = j0+ J0. Then
using the Hopf ring distributive law and 2.1(g) we have
0)
(wK bJ)*p = (b*p(j0)) O wK bs(J
= 0:
12
Similarly if I = i0+I0(where i0 is the first nonzero index in I) and in1 = 0,
we have, using 2.1(g)
0)K s(J)
(aIwK bJ)*p = (a*p(i0)) O as(I w b
= 0:
0)
However, when in1 = 1, as(I is not defined, so we must proceed differ
00
ently. We write aI = aI O a(n1)and use 2.1(i) and get
00)K s(J)
(aIwK bJ)*p = (a*p(n1)) O as(I w b
n1 s(I00)K s(J)
= (a(0)O wn O bOp(0)) O a w b
00)n+K (pn1) +s(J)
= a0+s(I w b 0 : 2
Proof of Corollary. Observation tells us that our elements are divisible
by a(0). Since a(0)is a primitive and we know that circle product with a
primitive is a primitive, all of our elements are primitive. 2
We now have to vary a little from [Wil84 ]. When we do so, we need only
go back to [RW77 ] to find what we need. Let us work, as always when we use
standard homology, in mod p homology. Let Q stand for the indecomposables
and Iw = ([p]; w1; w2; : :):. Then we have:
Theorem 2.5 In QH*(P_(n)_2)=I2wO QH*(P_(n)_*) we have:
Xk i
wiO bOp(ki)= 0:
i=n
Proof. The proof is the same as the proof of [RW77 , Theorem 3.14, page
259]. 2
This follows from the main relation, [RW77 , Theorem 3.8], which covers
our case and can be rewritten as:
13
P i
Theorem 2.6 Let b(t) = bit . Then, in P (n)*P_(n)_*[[s]],
0 1
X j [F] j
b @ Fvjsp A = *jn wj O b(s)Op :
jn
We need another piece of [RW77 ] which was not needed in [Wil84 ].
Namely, we need a theorem that allows us to reduce nonbasis elements to
basis elements. The relation we use to do this is Theorem 2.5, but it is not
an easy one to apply.
Theorem 2.7 In QH*(P_(n)_*), any e"1aIwK bJ can be written in terms of
nallowable elements.
Proof. The construction and proof of an algorithm for the reduction of non
allowable elements is done on pages 273275 of [RW77 ]. The proof applies
with only notational modification to the case of nallowable when I = 0. We
can then circle multiply by aI to get our result. 2
Theorems 1.3 and 1.4 are stated for rather general E as described in
Theorem 1.3. We do all of our calculations in mod p homology and so we
must lift our results to E.
Proof of Theorems 1.3 and 1.4 for general E from the theorems for mod p
homology. It is enough, for Theorem 1.3, to show that the AtiyahHirzebruch
spectral sequence collapses. The AtiyahHirzebruch spectral sequence re
spects the two products, O and *, and all elements in P (n)*(P_(n)_*) are
constructed using these two products from the basic elements of Proposition
2.1. Since the basic elements are all permanent cycles by Proposition 2.1 (d),
every element is a permanent cycle and the spectral sequence collapses. The
elements of Theorem 1.4 (a) are all primitive and no more can be created.
Part (b) also still holds. The only concern is the possibility that for P (n) t*
*he
truncated polynomial generators do not truncate at the same place because
of a shift in filtrations. However, Proposition 2.1 (h) and (i) tells us that
14
the height of an element is determined strictly by the pattern of its a's. The
results for general E follow by naturality from those for P (n). 2
3 The spectral sequence
All that remains is to prove Theorems 1.3 and 1.4 for mod p homology. We
prove our two theorems simultaneously by induction on degree in the bar
spectral sequence. Recall that for a loop space X with classifying space BX
the bar spectral sequence converges to H*(BX), and its E2term is
TorH*(X)*;*(Z=(p); Z=(p));
we will abbreviate this group by Tor H*(X)*;*or H*;*(H*(X)). When BX is
also a loop space, we have a spectral sequence of Hopf algebras. (See the
discussion in [HRW , x2].)
In this section we collect all of the facts that describe the complete behav
ior of this spectral sequence in the case X = P_(n)_i, including its differenti*
*als
and solutions to the algebra extension problems. Our proof is similar to that
of [Wil84 ] throughout except that we need a little more information from
time to time. We could have mimiced the proof completely but instead we
make some improvements.
We will prove Theorems 1.3 and 1.4 by induction on degree using the
bar spectral sequence with Hopf rings, [TW80 ]. We let P_(n)_0jbe the zero
component of P_(n)_j. Then P_(n)_j' P (n)jxP_(n)_0j, where P (n)j denotes the
group ssj(P (n)). We assume the calculations in Theorems 1.3, 1.4 and the
results stated below (specifically Lemma 3.6 and Theorem 3.7) are correct
for Hi(P_(n)_*) with i < k. (The value of H0(P_(n)_*) is obvious, given our
knowledge of ss*(P (n)).) The bar spectral sequence determines Hi(P_(n)_0*+1)
for i k and so we get Hi(P_(n)_*) for i k. This induction is always with
us, although frequently only implicitly. Let oe denote the suspension in the
spectral sequence and OE the transpotent. Most of our notation goes back to
[RW80 ].
15
For future use let
m(J) = min {k: jk 6= 0}; (3:1)
and define m(I) similarly.
When we are working in H0(P_(n)_*) we really need wi = [vi]  [02(pi1)]
to get it into the augmentation ideal. See [RW77 ] for further details. In
positive degrees this does not affect anything.
The following standard result enables us to compute all of the relevant
Tor groups.
Proposition 3.2 The group TorK (Z=(p); Z=(p)) TorK for a graded Z=(p)
algebra K has the following properties:
(i)It commutes with tensor products, i.e.,
TorK1K2 = TorK1 TorK2:
(ii)For K = E(x) (an exterior algebra on an odd dimensional generator
x),
TorE(x)= (oex);
a divided power algebra on the suspension of x, with fli(oex) 2 TorE(x)i;i*
*x
represented in the algebraic bar construction by ix. As an algebra,
(oex) = T P1(oex; flp(oex); flp2(oex); . .).
where fl0(oex) = oex.
(iii)For K = P (x) (a polynomial algebra on an even dimensional generator
x),
Tor P(x)= E(oex);
an exterior algebra on the suspension of x, oex 2 TorP(x)1;x.
16
(iv)For K = T Pk(x) (a truncated polynomial algebra of height pk on an
even dimensional generator x),
k1
TorK = E(oex) (OE(xp ));
where OE(y) is the transpotent of y, with fli(OE(y)) 2 TorK2i;piy. A rep*
*re
N i p1
sentative cycle for OE(y) is yp1 y and for fli(OE(y)), (y y).
The Tor groups corresponding to the homology given in Theorem 1.3 are
given below in Lemma 3.6.
In our proof we will need to be able to identify elements in the spectral
sequence. We usually cannot do this precisely but must introduce some
filtrations. In particular, we need to introduce an entirely new filtration. To
do that we need to review, following [TW80 ], how the Hopf ring pairing fits
into the bar spectral sequence. We recall how this pairing is constructed.
The bar spectral sequence converging to H*(P_(n)_q+1)) is based on the bar
filtration of the space
[
P_(n)_q+1= BP_(n)_q= BiP_(n)_q:
i
In [TW80 ] it was shown that the circle product respects this filtration, i.e.,
the map
P_(n)_q+1^ P_(n)_rO! P_(n)_q+1+r
induces maps
BiP_(n)_q^ P_(n)_rO! BiP_(n)_q+r
for each i. In the bar filtration we have cofibre sequences
Bi1P_(n)_q! BiP_(n)_q! iP_(n)_(i)q
where this last space is the ith suspension of the ifold smash product of
P_(n)_q. It follows that the pairing induces maps
iP_(n)_(i)q^ P_(n)_r! iP_(n)_(i)q+r: (3:3)
17
This map is the usual circle product on each of the i factors.
Now recall the Verschiebung map V , defined on any cocommutative coal
gebra as the dual of the pth power map. Since our bar spectral sequence is
one of bicommutative Hopf algebras, it has a Verschiebung map
Erps;ptV!Ers;t (3:4)
which, if one ignores the grading, is a Hopf algebra map. In addition, it
respects the circle product pairing, i.e.,
V (x O y) = (V x) O (V y):
The Verschiebung is a standard tool; we will also need the following
variant of it. There is an internal Verschiebung
Hs;pt(H*(X)) Vint!Hs;t(H*(X))
defined on the algebraic bar construction, i.e. E1*;*, by
Vint(x1 x2 . . .xs) = V x1 V x2 . . .V xs:
Because V (x * y) = V (x) * V (y) we see that Vintcommutes with d1 and is
defined on E2*;*, our Tor for the spectral sequence. Reviewing the definition of
the Hopf algebra structure on Tor we see that Vintis, ignoring the gradings, a
Hopf algebra map. From (3.3) we can deduce that Vintsatisfies the identity
Vint((x1 x2 . . .xs) O y) = (Vint(x1 x2 . . .xs)) O V y: (3:5)
Now we are ready to identify the Tor group for the algebra of Theorem
1.3.
18
Lemma 3.6 Let p be any prime. In the bar spectral sequence,
Er*;*(H*(P_(n)_*)) ) H*(P_(n)_0*+1);
The Tor group for the algebra of Theorem 1.3 is as follows.
H*(P(n)_)
E2*;*(H*(P_(n)_*)) ' Tor*;* *(Z=(p); Z=(p)) H*;*(H*(P_(n)_*)) '
O
(oee1 O aIwK bJ)
(K;J)2A+n
O
E(oeaIwK bJ)
i0=0
(K;J)2An
O
E(oeaIwK bJ)
i0=1
(K;J)2A+n
O
(OE(aIwK bJ)):
(K;J)2Ani
n1=0
Proof. The Tor group for each factor in 1.3 can be computed using Proposi
tion 3.2. The result is shown in the following table.
19
____________________________________________________
 Factor  Tor group 
___________________________________________________
  
 O I K J  O I K J 
 E(e1a w b )  (oee1a w b ) 
 +  + 
 (K;J)2An (K;J)2An 
  
__________________________________________________
  
  O 
  I K J 
  E(oea w b ) 
 O  i0=0 
 I K J (K;J)2An 
 T Pae(I)(a w b ) 
 i0=0  O 
 (K;J)2An  I K J 
  E(oea w b ) 
 O I6=M;i0=1 
 I K J (K;J)2A+ 
 T Pae(I)(a w b ) n 
 I6=M;i0=1  
 (K;J)2A+  O I K J 
 n  (OE(a w b )) 
  
 (K;J)2An 
  in1=0 
  
__________________________________________________
  
 O M K J  O M K J 
 P (a w b )  E(oea w b ) 
 +  + 
 (K;J)2An (K;J)2An 
  
__________________________________________________
The only difficulty here is identifying the transpotent elements. From
Lemma 2.3 we get the in1 condition. From Theorem 1.4 we can find all of
the elements of height p in the truncated polynomial factors of Theorem 1.3
except the nonprimitive generators which can then be read off directly from
Theorem 1.3.
We can consolidate the third and fifth factors on the right. 2
We will need to identify the generators, flpi, in our Tor using the Hopf
ring pairing. We can only do this modulo filtrations which is adequate for
our needs. The first filtration is given by the Verschiebung where we have
that V i(flpi(oex)) = fl0(oex) and we see that V ikills all other elements in t*
*his
filtration except those of this form. In this case, V ikills all decomposables *
*so
when we work modulo the kernel of V iwe are just working modulo decom
posables. However, when we work with flpi(OE(y)), V idoes not kill elements
of the form flpi(oex1) * flpi(oex2) but these terms will not show up and we can
20
still work modulo decomposables.
Our second filtration comes from Vintand it is important for us to see
how it acts on our Tor groups. V kills any primitive and anything with an e1,
a(0)or b(0)is a primitive so for starters, Vintkills the first and third factor*
*s of
Lemma 3.6. When using this filtration the exterior elements from the second
factor will never enter in. All we need to be concerned with then is what
happens with the fourth factor. If i0 or j0 is greater than zero then our factor
goes to zero. If both are zero then we can define the inverse shift operator,
s1 on I and J. Since V (a(i+1)) = a(i)and V (b(i+1)) = b(i), we have that Vint
induces an isomorphism:
1(I)K s1(J)
(OE(aIwK bJ)) ! (OE(as w b )):
We will be able to identify elements modulo the kernel of the Viintwhich we
now understand.
We can now describe the behavior of the entire bar spectral sequence.
Theorem 3.7 Let p be any prime.
(a) In the Hopf ring pairing of the bar spectral sequence we have:
For J 6= 0 and k = m(J), consider
O
O : H*;*(H*(P_(n)_*2)) H*(P_(n)_2) ! H*;*(H*(P_(n)_*)):
(i) flpi(oee1aIwK bJk ) O b(k+i)= flpi(oee1aIwK bJ) modulo decompos
ables for (K; J) 2 A+n.
(ii)flpi(OE(aIwK bJk )) O b(k+i+1) = flpi(OE(aIwK bJ)) modulo decom
posables and the kernel of Viqntwhere q = min (m(I); m(J)), for
(K; J) 2 An and in1 = 0.
(b) For J = 0, I 6= 0, and k = m(I), consider
O
O : H*;*(H*(P_(n)_*1)) H*(P_(n)_1) ! H*;*(H*(P_(n)_*)):
21
(i) flpi(oee1aIk wK )Oa(k+i)= (1)l(I)1flpi(oee1aIwK ) modulo decom
posables, k + i < n.
(ii)flpi(OE(aIk wK )) O a(k+i+1)= (1)l(I)flpi(OE(aIwK )) modulo decom
posables and the kernel of Viknt, for k + i + 1 < n and in1 = 0.
(c) Let q = min {jinj = 1} if I 6= 0, and n + 1 if I = 0. The following
differentials are nonzero:
q1 I K J +
(i) dp on flpq(oee1a w b ), I 6= 0 and (K; J) 2 An .
q11 I K J
(ii)d2p on flpq1(OE(a w b )), (K; J) 2 An and in1 = 0.
(d) The differential targets of (c) are all linearly independent and, modulo
the vector space generated by the oeaIwK bJ with i0 = 0 and (K; J) 2 A+n,
a basis for the vector space they generate is given by all oeaIwK bJ with
i0 = 0 such that (K; J) 2 An  A+n.
(e) Let q = min {jinj = 1}, up to sign in E1*;*
i(I)K si(J+ )
(i) flpi(oee1aIwK bJ), (K; J) 2 A+n, represents as w b 0 modulo
decomposables where if I 6= 0 then i < q.
(ii)flpi(OE(aIwK bJ)), for (K; J) 2 An, in1 = 0, represents the el
i+1(I)+ K si+1(J)
ement as iw b modulo decomposables and the kernel
of Vimin(m(I);m(J))nt, where i < q  1. (If I = 0 then q = n + 1.)
(iii)oeaIwK bJ represents e1aIwK bJ when (K; J) 2 A+n.
(f) As an algebra, E1*;*is
O O
E(e1aIwK bJ) T P1(aIwK bJ)
(K;J)2A+n (K;J)2An
Proof. (a). First we must note that both the flpi elements exist in the
spectral sequence and they do.
(a)(i). Next we work modulo decomposables so we can apply V ito both
sides and we need only show the i = 0 case, or,
oee1aIwK bJk O b(k)= oee1aIwK bJ
22
for (K; J) 2 A+nwhich is obvious.
(a)(ii). First we apply V ito see we need only show
OE(aIwK bJk ) O b(k+1)= OE(aIwK bJ):
Next we want to apply Viqnt. If k = m(J) = q then we have
OE(V q(aIwK bJk )) O b(1)
i j
= V q(aIwK bJk )*p1 V q(aIwK bJk ) O b(1)
= OE(V q(aIwK bJ)):
Since there are only p nontrivial terms in the iterated coproduct of b(1)this
is trivial. If k = m(J) > m(I) = q then we want to show
OE(V q(aIwK bJk )) O b(1+kq)
i j
= V q(aIwK bJk )*p1 V q(aIwK bJk ) O b(1+kq)
= OE(V q(aIwK bJ)):
The argument is now different. The p terms in OE are all primitive. All bj
except for j a power of p are decomposable so the circle product of a primitive
times any of them is trivial. Thus the only nontrivial term is the one we want.
(b). Again we must confirm that all of our elements are defined in the
spectral sequence or space. They are.
(b)(i). We apply V iand all we need to do is prove:
oee1aIk wK O a(k)= (1)l(I)1oee1aIwK
which follows immediately.
(b)(ii). Again we apply V ito get down to:
OE(aIk wK ) O a(k+1)= (1)l(I)OE(aIwK ):
We now apply Vikntwhere k = m(I) to get
OE(Viknt(aIk wK )) O a(1)= (1)l(I)OE(Viknt(aIwK )):
23
The result now follows like the b(1)case above.
(e)(i). For J 6= 0 this follows by our induction from (a)(i). If I = 0
we ground our induction by using the definition of the b(i)which clearly
corresponds (up to sign) to flpi(oee1). When J = 0 we use (b)(i) and induction
while I 6= 0. The induction starts with oee1 = e1 O e1 = b1 = b(0)(the wK
doesn't matter here).
(e)(ii). For J 6= 0 this follows by our induction from (a)(ii). When J = 0
we use (b)(ii) and induction while I 6= 0. The induction starts with the
recognition that OE([1]  [0]) is a(0)and so flpi(OE([1]  [0])) is a(i)(the wK
doesn't matter here either). These are low degree elements and our space is
just BZ=(p) in this range so these are easy to see.
(e)(iii). Follows by induction on degree and the definition of the suspen
sion.
(f)(truncated polynomial factor). We show that the even dimensional
generators in (f) are all there in E1*;*. Later, when we are done with the
differentials, we finish the proof. All of the elements in (e) must be infinite
cycles by the same induction we used to identify them. The elements in (e)(i)
correspond to to first term in Lemma 3.6. Because (K; J) 2 A+nin (e)(i),
we get (K; J) 2 An in (f). The terms in (e)(i) give us all the terms in (the
even part of) (f) with m(J) m(I). (e)(ii) corresponds to the fourth part of
Lemma 3.6 and gives us the m(I) < m(J) terms in (f). (e)(iii) corresponds
to the elements in the odd part of (f) (which are not hit by differentials).
(c) and (d). The guiding principle for the differentials is that any flpiwith
an a(n)in it has a differential on it. Of course a(n)doesn't exist and this is
why. We must note that in (c), the elements we assert have differentials are
the lowest flpi possible because for (c)(i), (e)(i) showed the lower ones were
infinite cycles and likewise for (c)(ii) and (e)(ii).
Differentials in our Hopf algebra must start with a generator and go to
a primitive, so the only generators which can have differentials are the flpi
and if they support a nontrivial differential then the target must be an odd
24
degree primitive. All of our odd degree primitives are located in filtration
one and are our exterior algebra generators. Note that if dr(flpm(x)) = y
where y is an exterior generator, then it follows formally that (up to nonzero
scalar multiplication)
dr(flspm(x)) = yfl(s1)pm(x)
for all s > 0. Thus the factor E(y) (x) in the E2term gets replaced by
T P1(x; flp(x); . .f.lpm1(x))
in Er+1. Each of our differentials takes this form.
Recall that in Lemma 3.6 we have two exterior factors. We want the odd
elements in (f) to survive. The remainder are the elements in (d) which we
want to be hit by differentials. To show that differentials are what we want
them to be we show that the proposed targets must indeed die. We then
show, strictly by counting, that the targets are in onetoone correspondence
with our proposed sources. Since our proposed sources are the lowest possible
degree elements which could support differentials and our targets are known
to be hit, we infer that our proposed sources are in fact our sources.
So, we have two parts left to finish our differentials. (1) We must show
the proposed targets are hit by differentials and (2) we must do the counting
argument. We defer the counting argument until later.
We want to show that all of the proposed targets, oeaIwK bJ with i0 = 0
such that (K; J) 2 An  A+n, must be zero modulo the same type of elements
in A+n. If they are not zero then they must be represented by e1aIwK bJ
and this, in turn, gives rise to an element oee1aIwK bJ in the next spectral
sequence for the next space. This element is even degree so must represent
aIwK bJ+0 . Because (K; J) 62 A+nwe have (K; J + 0) 62 An. This means
that it can be rewritten in terms of nallowable elements. The algorithm
does not affect the i0 = 0 condition or the fact that there must be a b(0).
Thus it can be rewritten in terms of elements that we are working modulo.
25
Thus we know that there is a relation somewhere. There are only two ways
for a relation to come up: (1) this last element could be a pth power or (2)
our differential is as claimed. We are done if we show this last element is not
a pth power. This follows immediately from Corollary 2.4 which says that a
pth power must have i0 = 1.
We must now do our counting. When that is complete, we'll see that
the elements in (f) are all that remain, both odd and even. So, showing
our sources are in onetoone correspondence with our targets will finish the
proof of Theorem 3.7. We have to do a similar counting argument in order to
solve all the extension problems to get Theorems 1.3 and 1.4 so we separate
out the common part.
Lemma 3.8 There is a onetoone correspondence between the set
n1)
{wK+n bJ+(p 0 : (K; J) 2 An; j0 = 0}
and the set
0 J0 +
{wK b : (K; J) 2 An  An }:
Proof. To see this, write
J = pndn + pn+1dn+1+ . .+.pm dm + J00
where m is maximal (this can be vacuous, i.e. J = J00, in which case we set
m = n  1) and dn dn+1 . . .dm and J00is nonnegative. Now let
J0 = J00+ (pn  1)0 + pn+1dn1 + pn+2dn+11+ . .+.pm+1 dm 1
and K0 = K + m+1 . 2
We will now do the counting argument for the differentials. We recall that
we have to show that the sources listed in Theorem 3.7 (c) are in onetoone
correspondence with the e1aIwK bJ which have i0 = 0 and have (K; J) 2
An  A+n. Strictly for the purposes of counting we introduce a nonelement,
26
a(n), and incorporate it into our notation, e"1aIwK bJ. We can, for counting
purposes only, identify the set of differential source elements in Theorem 3.7
(c) with the set,
{aIwK bJ: (K; J) 2 An; in = 1; i0 = 0; j0 = 0}:
Theorem 3.7 (c)(i) gives those with m(I) m(J) by looking at Theorem 3.7
(e)(i) and Theorem 3.7 (c)(ii) gives those with m(I) < m(J) by looking at
Theorem 3.7 (e)(ii). That is a onetoone correspondence. We now want to
pair these up with the targets of the differentials. To do that we just replace
n1
the a(n)with e1wnbp(0). This gives us a onetoone correspondence between
the sources of the differentials listed in Theorem 3.7 (c) and the set
n1)
{e1aIwK+n bJ+(p 0 : (K; J) 2 An; in = 0; i0 = 0; j0 = 0}:
Then Lemma 3.8 finishes our counting argument. 2
Proof of Theorems 1.3 and 1.4 for mod p homology. To finish Theorem
1.3 we must solve the extension problems remaining in Theorem 3.7 (f).
Theorem 1.4 (a) can just be read off the spectral sequence. The first part of
(c), that all pth powers are primitive, follows from Corollary 2.4 which says
that a pth power is divisible by a(0). In order to complete the proof we must
show that all generators and iterated pthpowers with an a(n1)in them must
have nontrivial pth powers. (b) will follow and so will the solution to the
extensions that we need.
First we'll do a counting argument to show that things can work out the
way we suggest. The elements we want to be pth powers are the aIwK bJ
with i0 = 1 and (K; J) 2 An  A+n. We need our correspondence to be
with generators and pth powers with in1 = 1 in this degree divided by p.
0 K0 J0 0 0
That combination consists of all aI w b with in1 = 1 with (K ; J ) 2 An
(by induction!). First we show that we have a onetoone correspondence
and then we show that our suggested pth powers must indeed be pth powers.
Since we show our elements to be in onetoone correspondence with the only
27
elements that can possibly have nontrivial pth powers, the pth powers must
be as claimed.
0 K0 J0
To do our counting we just take the pth power of aI w b . By Lemma
2.3 we have
0 K0 J0 *p +s(I0 ) K0+n (pn1) +s(J0)
(aI w b ) = a0 n1 w b 0 :
Lemma 3.8 now gives us the onetoone correspondence we seek.
We now have to finish the proof of Theorem 1.4 (c) that the iterated
pth powers of the generators in Theorem 1.3 are those primitives aIwK bJ
with i0 = 1 and (K; J) 2 An  A+nmodulo the vector space generated by
aIwK bJ with i0 = 1 and (K; J) 2 A+n. Having done our counting argument,
it is enough to show that these elements must indeed be pth powers. We
observe that all of our differentials start in total degrees divisible by 2p so
the targets must be in degrees equal to 1 mod 2p. If our elements are not
pth powers, then they will suspend to exterior generators in the next bar
spectral sequence. Here they cannot be targets of differentials because they
have degree +1 mod 2p. Thus, such an element would suspend once more
and be represented by aIwK bJ+0 and (K; J + 0) would cease to be in An.
It is now in degree 2 mod 2p so it cannot be a pth power here which is the
only way to create a relation. Thus it must be a pth power where we said it
would be. 2
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