Uniqueness of BP

0.16 MB·English

by  Vigleik Angeltveit

UNIQUENESS OF BPhni
VIGLEIKANGELTVEITANDJOHNA.LIND
5
1
0 Abstract. Fix a prime number p and an integer n ≥ 0. We prove that if
2 a p-complete spectrum X satisfying a mild finiteness condition has the same
modpcohomology as BPhnias amoduleover theSteenrod algebra, then X
n
isweakhomotopyequivalenttothep-completionofBPhni.
a
J

7

] 1. Introduction
T
A LetMU be the spectrumrepresentingcomplexcobordism,andrecallthatwhen
localized at a prime p there is an idempotent map ǫ:MU →MU with image
. (p) (p)
h
the Brown-Petersonspectrum BP [8]. The coefficient ring of BP is
t
a π BP =Z [v ,v ,...]
m ∗ (p) 1 2
with |v | = 2pk−2. For each n ≥ 0 we can quotient out by the generators v for
[ k k
k >n and construct a spectrum BPhni with

1
v π∗BPhni=Z(p)[v1,...,vn].
8
The construction of BPhni can be carried out using the Baas-Sullivan bordism
4
theory of manifolds with singularity or by taking the cofiber of the multiplication
4
1 by vk map. It is customary to let v0 = p and to let BPh−1i = HFp. Special
0 cases include BPh0i = HZ and BPh1i = ℓ, the Adams summand of connective
(p)
1. complexK-theorylocalizedatp. We thinkofBPhniforvaryingnasinterpolating
0 between mod p cohomology and the Brown-Peterson summand of p-local complex
5 cobordism.
1 Complex cobordism has played a central role in algebraic topology because of
v: its connection to formal groups. The ring MU∗(CP∞) carries a formal group and
i Quillen [8] showed that it is the universal formal group. Similarly, BP determines
X
theuniversalp-typicalformalgroupandBPhniiscloselyrelatedtoheightnformal
r
a groups. By further quotienting BPhni out by vk for k <n and inverting vn we get
the n-thMoravaK-theoryspectrumK(n), whichrepresentsthe universalheightn
formal group.
WhilethehomotopytypeofBP iswelldefinedbecauseitcomesfromtheQuillen
idempotent on MU, the classes v are not quite canonical. Indeed, there are two
k
popular choices of v given by Hazewinkel [5] and Araki [2], and these are not the
k
only ones. Hence it is not at all obvious that the homotopy type of BPhni is well
defined; the definition of BPhni appears to depend on the choice of v for k >n.
k
In this paper we show that the p-adic homotopy type of BPhni is well defined
by showing that it is determined by its mod p cohomology. To be more precise,
fix a model of the spectrum BPhni constructed using any choice for the genera-
tors v , and let BPhni∧ denote its p-completion. Recall that p-completion leaves
k p
mod p cohomology unchanged and that for spectra of finite type, such as BPhni,
p-completion has the effect of p-adically completing all homotopy groups [4]. Our
1
mainresultstatesthatanyp-completespectrumsatisfyingstandardfinitenesscon-
ditions which has the same mod p cohomology as BPhni as a module over the
Steenrod algebra is equivalent to BPhni∧.
p
Theorem A. Suppose X is a spectrum which is bounded below, and whose ho-
motopy groups are finitely generated over Z . Suppose also that there is an iso-
p
morphism
θ :H∗(BPhni;F )→H∗(X;F )
p p
of modules over the Steenrod algebra. Then there exists a weak homotopy equiv-
alence f : X → BPhni∧ of spectra which induces the isomorphism θ in mod p
p
cohomology.
The mod p cohomology of BPhni does not depend on the choices made in the
constructionofBPhni [11,Prop.1.7],sowededuce thatthe p-adichomotopytype
of BPhni does not either.
We believe that a strongerform of this theorem is true, where Z is replaced by
p
the p-localintegers and the conclusion is that X is equivalent to BPhni. The only
obstacle is a question of convergence of the Adams spectral sequence, see Remark
4.1.
One motivation for proving this result is our interest in twisted cohomology
theories, and in particular twisted BPhni-cohomology. The case p = 2 and n = 2
is of particular interest. In [10, 11] Wilson showed that up to localization at 2 we
have
Ω∞BPh2i≃Z×K(Z,2)×BSU ×B3SU ×BPh2i ,
12
where BPh2i is the twelfth space in the Ω-spectrum for BPh2i. Hence
12
GL (BPh2i)≃Z/2×K(Z,2)×BSU ×B3SU ×BPh2i .
1 12
By [6], BPh2i is an E ring spectrum at p= 2. Hence GL (BPh2i) is an infinite
∞ 1
loop space, and its classifying space BGL (BPh2i) and the associated spectrum
1
bgl (BPh2i) of units are both defined. We conjecture that the above splitting
1
deloops, so that
(1.1) BGL BPh2i≃K(Z/2,1)×K(Z,3)×B2SU ×B4SU ×BPh2i .
1 13
Ifthesplitting(1.1)exists,then,justasintegralcohomologyofX canbetwistedby
aclassinH1(X,Z/2)andcomplexK-theorycanbetwistedbyaclassinH3(X;Z),
the BPh2i-cohomology of X can be twisted by a class in
ku7(X)=[X,B4SU].
This has a nice interpretation in terms of the “Bockstein” spectral sequence
E∗,∗ =ku∗(X)[v ]⇒BPh2i∗(X),
2 2
with the first nontrivial differential being modified by the twisting. We hope to
return to this elsewhere.
1.1. Main proof idea. The case n = 1 of Theorem A is a result of Adams and
Priddy [1], and our proof is modeled on their proof, with a few enhancements. Let
H∗(−)denotemodpcohomology,alwaysconsideredasamoduleovertheSteenrod
algebra A. The proof uses the Adams spectral sequence
Es,t =Exts,t(H∗(Y),H∗(X))=⇒π Hom(X,Y∧).
2 A t−s p
2
The isomorphism θ: H∗(Y) → H∗(X) represents an element [θ] ∈ E0,0. If [θ] is
2
a permanent cycle then it represents a map f∧: X → Y∧ whose induced map on
p p
cohomology is θ. We will implement this plan for Y =BPhni.
The mod p cohomology of BPhni is given by
H∗(BPhni)=A//E ,
n
where A is the mod p Steenrod algebra and E = E(Q ,...,Q ) is the exterior
n 0 n
algebra on the first n+1 Milnor primitives. We use a change-of-rings theorem to
compute the E term of the Adams spectral sequence:
2
Es,t =Exts,t(H∗(BPhni),H∗(X))∼=Exts,t(F ,H∗(X)).
2 A En p
We are left with Ext over an exterior algebra on n+1 generators, which is much
more computable than Ext over the full Steenrod algebra.
WewillproveTheoremAbyshowingthatallthepossibledifferentialson[θ]land
intrivialgroups. Inotherwords,weprovethatEs,t =0forall(s,t)withs≥2and
2
t−s= −1. (In fact Es,t =0 also for (s,t)=(0,−1) and (s,t)=(1,0).) This was
2
also the strategy in Adams and Priddy’s paper [1]. In our case the required Ext
calculation is more difficult; in essence we have to understand an (n+1)-cube of
“commuting” spectral sequences. In §2, we make some general observations about
the kind of spectral sequences we need to analyze.
1.2. Acknowledgements. The authors would like to thank Craig Westerland for
many helpful conversations. Without him we would not have started thinking
about twisted BPhni-cohomology and we would not have been led to the main
result of this paper. The first author was supported by an ARC Discovery grant.
The second author was partially supported by the DFG through SFB1085, and
thanks the Australian National University for hosting him while this research was
conducted.
2. A cube of spectral sequences
Suppose that we are given an (n+1)-dimensional chain complex M = M∗,...,∗,
andthatweareinterestedinthehomologyofthetotalcomplex. Inotherwords,M
isanabeliangroupwithn+1differentgradingsandwehavedifferentialsd0,...,dn
onM, wheredi increasesthe i’thgradingofM by1. The differentialsarerequired
to commute in the graded sense, meaning that didj = −djdi, and we wish to
compute the homology of M with respect to the differential d=d0+···+dn.
Givena 2-dimensionalchaincomplex, i.e.,a double complex,there arestandard
spectral sequences
Es,t =Hs(Ht(M,d0),d1)⇒Hs+t(M,d)
2
and
Es,t =Hs(Ht(M,d1),d0)⇒Hs+t(M,d).
2
In the situationwhere we have n+1 directions,we get morespectralsequences.
We can first take the homology in the dn-direction, then run the dn−1 spectral
sequence to compute H (M,dn−1+dn). Next we have a dn−2 spectral sequence,
∗
which takes the form
Hs(Ht(M,dn−1+dn),dn−2)⇒Hs+t(M,dn−2+dn−1+dn),
3
andsoonuntilthefinald0 spectralsequence. Ifwecancomputeallthedifferentials
and solve all the extension problems this tells us H (M,d). Or we can choose to
∗
take homology with respect to the di’s in a different order.
We can organize all of this in an (n+1)-cube with M at the initial vertex and
H (M,d) at the terminal vertex. At the vertex corresponding to I ⊂ {0,1,...,n}
∗
we have H∗(M,dI), where dI = di. The i’th edge originating at the ini-
Pi∈I
tial vertex corresponds to taking homology with respect to di, all the other edges
correspond to spectral sequences. For each j 6∈I we have a spectral sequence
Hs(Ht(M,dI),dj)⇒Hs+t(M,dI∪j).
For n=2 the cube of spectral sequences looks as follows.
H∗(−,d2)
M ❘❘❘❘❘❘❘❘❘❘❘❘❘❘H❘❘∗(−,d1) +3H∗(M,d2)❙❙❙❙d❙❙1❙❙−❙❙SS
H∗(−d0) H$, ∗(M,d1) d2−SS +3H%- ∗(M,d1,2)
H∗(M(cid:11)(cid:19)d,1−d0S)❘❘S❘❘❘d❘❘2❘−❘❘HS$, ∗S(M(cid:11)(cid:19),dd0−0,S1)S +3dH0−∗Sd(M2S−(cid:11)(cid:19),SdS0,2❙❙)❙❙d❙❙1❙❙−❙❙❙❙S+3SH%- ∗(M(cid:11)(cid:19) d,0−d)SS
We can interpret taking the homology of M with respect to di as a degenerate
sort of spectral sequence, so that all the edges in the above diagram are spectral
sequences. The resulting “commutative diagram of spectral sequences” gives dif-
ferent ways of computing an associated graded of the total cohomology H∗(M,d).
Depending on which path we take we are going to get different representatives for
classes in H∗(M,d), and different extensions are going to be visible. In fact the
cube gives (n+1)! different ways of computing H∗(M,d) up to extensions, and
one could hope that in a given situation this is enough to understand H∗(M,d)
completely.
This is a rather complicated situation in general. Rather than aiming for a
complete understanding of H∗(M,d), we will construct an algorithm that takes as
input some x∈M that is a permanent cycle in the sequence of spectral sequences
dn−SS,...,d0−SS, and produces as output a representative for [x] ∈ H∗(M,d)
which is in some sense optimal.
To give the basic idea of our algorithm, consider the double complex M defined
by the following figure:
(cid:0)❥(cid:0)d❥(cid:0)1(cid:0)❥(cid:0)❥(cid:0)(cid:0)❥(cid:0)❥??x❥❥❥d❥2❥❥❥❥❥d✁❦❥1✁❦❥✁❦❥✁❦✁❥✁❦❥✁❦55@@❦y❦❦d❦2❦❦❦❦❦❦❦❦❦❦55z
a b
IfwefirstcomputeH∗(M,d2)weareleftwithx, anditisclearthatxalsosurvives
the d1 spectral sequence and represents a nonzero class [x] ∈H∗(M,d). But [x] is
also represented by −y or by z. Our algorithm will give z as a representative for
[x].
Here is the algorithm. It goes through n+1 steps, starting with step n and
ending with step 0. We write [x] for the class in H∗(M;dk,...,n) determined
dk,...,n
by x.
4
Definition 2.1. Start with an element x = x in M that is a permanent cycle
n+1
inthesequenceofspectralsequencesdn−SS,...,d0−SS. Wewill recursivelydefine
elements x ∈M for 0≤k ≤n so that in step k, we take as input a representative
k
x for theclass [x] andproduceas outputa representativex for theclass
k+1 dk+1,...,n k
[x] .
dk,...,n
For each homogeneous component x′ of x , check if x′ is a dk-boundary. If
k+1
x′ is not a dk-boundary, let x′′ = x′. If x′ is a dk-boundary, say, dk(a) = x′, let
x′′ = −dk+1,...,n(a). Replace x by x′′. Iterate this procedure until none of
k+1
P
the homogeneous components are dk-boundaries. The resulting element of M is the
output x of the k-th step of the algorithm.
k
There are two issues with this algorithm. First, there is a choice of the element
a satisfying dk(a) = x′ and different choices might give different outputs. Sec-
ond, without additional assumptions there is no guarantee that this process will
terminate in finitely many steps.
In the proof of Theorem A we deal with the first issue by choosing a particular
a (see the proof of Lemma 3.3). The second issue is taken care of by the following
result.
Lemma 2.2. Suppose M∗,...,∗ is bounded below in each grading. Then the above
process terminates after finitely many steps.
Proof. This follows because each time we replace x′ by x′′ = −dk+1,...,n(a) we
decrease the k’th grading by 1. (cid:3)
Now let us do an example from “nature”.
♦♦♦♦♦♦♦♦♦♦♦✆♦✆✆♦❦✆✆77❦✆ξ✆❦1BB1❦8OO❦τ❦3❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦❦✆❦✆❦❥✆❦✆55❥✆❦ξ✆❥❦✆19❥BB❦ξ❥❦2OO3❥❦τ❥❦✂3✂❦❥✂❦❥❥✂✂55❥❥✂ξ❥❥✂19@@❥❥ξ❥❥OO3❥❥τ❥❥3❥❥❥❥❥❥❥❥❥❥❥❥❥❥❥☞❥❥❥☞❦☞44❥☞❥❦☞ξ❥☞❥❦2FF6❥❥❦OOτ❥❥❦3❥❥✆❦✆❥❥❦✆✆❥44❦✆ξ✆❥❦✆23❥❦BBξ❥❦OO3❥❦τ❥❦✆3✆❥❦✆✆❥❦✆✆55❦✆ξ❦BB32❦OOτ❦3☞❦☞☞❦☞☞55☞τFF 4OO
ξ27 ξ18ξ3 ξ18ξ ξ9ξ6 ξ9ξ3ξ ξ9ξ2 ξ9 ξ6ξ ξ3ξ2 ξ3 ξ
1 1 2 1 3 1 2 1 2 3 1 3 2 2 3 2 3 3 4
Let M be as above, where each node indicates a free F [v ,v ,v ]-module on
3 0 1 2
that element. Each arrow of slope 1/4 represents the d2 differential which is mul-
tiplication by v , each arrowof slope 1 represents d1 which is multiplication by v ,
2 1
and each vertical arrow represents d0 which is multiplication by v . Consider the
0
class
x=v2v ξ18τ .
0 1 1 3
Notice that x is a permanent cycle after running the d2 spectral sequence, the d1
spectral sequence, then the d0 spectral sequence. To set up the algorithm, set
x = x. We see that x is not a boundary in the d2 spectral sequence, so step 2
3
amounts to setting x =x . In step 1, we get the representative
2 3
x =−v2v ξ9ξ3τ
1 0 2 1 2 3
for [x] by considering a=v2ξ18ξ3.
d1,2 0 1 2
In step 0, we first replace x by
1
v v v ξ9ξ τ +v v2ξ3ξ τ ,
0 1 2 1 3 3 0 2 2 3 3
then we replace each of the two homogeneous summands by −v v2ξ2τ to obtain
1 2 3 3
x =−2v v2ξ2τ =v v2ξ2τ .
0 1 2 3 3 1 2 3 3
5
(The last equality follows because we are in characteristic 3.) This is not sufficient
to conclude that x is a boundary. The class [x] ∈ H∗(M,d) is probably nonzero
butweprefernottosayanythingaboutit. The pointis thatwhile westartedwith
a representativeonthe “left hand side” of M,at eachstep the algorithmgaveus a
new representative “further to the right”.
If instead we start with
y =v2v x=v4v2ξ18τ ,
0 1 0 1 1 3
the algorithmdoesthe following. Step 2does nothing becausey is nota d2 bound-
ary. In step 1 we replace y first by −v4v v ξ9ξ3τ and then by
0 1 2 1 2 3
y =v4v2ξ6τ .
1 0 2 2 3
Then in step 0 we replace y first by −v3v v2ξ3ξ τ , then by v2v2v2ξ2τ , then
1 0 1 2 2 3 3 0 1 2 3 3
by −v v3v2τ , and finally by 0. This reflects the fact that there is a differential
0 1 2 4
d0(ξ )=[y] in the d0 spectral sequence.
4 4 d1,2
2.1. How this fits with the main theorem. In the proof of Theorem A we
will be interested in similar examples. We wish to show that there is no homology
in total (homological) degree −1, where the ξ ’s and τ ’s are in negative degrees
i j
and the v ’s are in positive degrees. We will do this by showing that in fact there
k
is nothing in odd total degree ≥ −1 by bounding the total degree of elements
vr0···vrn−1x that can survive the sequence of spectral sequences.
0 n−1
3. The homology and cohomology of BPhni
In this section we collect a few facts about the homology and cohomology of
BPhni. In particular, we will determine the action of the Milnor primitives Q on
i
H∗(BPhni) for 0 ≤ i ≤ n. Our method is to first determine the coaction of the
dual of Q on homology, then dualize.
i
Milnor showed [7] that the dual A of the mod p Steenrod algebra is the Hopf
∗
algebra
A =P(ξ ,ξ ,...)⊗E(τ ,τ ,...) |ξ |=2(pi−1),|τ |=2pi−1
∗ 1 2 0 1 i i
with coproduct
k k
ψ(ξ )= ξpi ⊗ξ and ψ(τ )=τ ⊗1+ ξpi ⊗τ .
k X k−i i k k X k−i i
i=0 i=0
We will use this notation for all primes, with the understanding that if p = 2
the multiplicative structure is different, with the element τ denoting the element
i
usually called ξ and ξ denoting the element usually called ξ2. The difference in
i+1 i i
multiplicative structure at p=2 will play no role in our arguments.
It will be useful to use the images ξ¯, τ¯ of Milnor’s generators under the conju-
i i
gation map of the Hopf algebra A . The coproduct is then given by:
∗
k k
(3.1) ψ(ξ¯ )= ξ¯ ⊗ξ¯pi and ψ(τ¯ )=1⊗τ¯ + τ¯ ⊗ξ¯pi .
k X i k−i k k X i k−i
i=0 i=0
The Milnor primitives Q ∈ A are inductively defined by Q = β (the mod p
i 0
Bockstein) and
Q =PpiQ −Q Ppi.
i i−1 i−1
6
Using the Milnorbasis, the element Q ∈Ais dualto τ ∈A and hashomological
i i ∗
degree −(2pi−1).
The mod p cohomology of BPhni as a left module over the Steenrod algebra is
H∗(BPhni)=A//E ,
n
whereE =E(Q ,...,Q )istheexterioralgebraonthefirstn+1Milnorprimitives
n 0 n
[11, Prop. 1.7]. Here we quotient out the exterior algebra on the right, while we
are interested in the left action of E on the result. Because the Steenrod algebra
n
is non-commutative the left action is still highly nontrivial.
Dually, the mod p homology of BPhni is the algebra
H (BPhni)=P(ξ¯,ξ¯,...)⊗E(τ¯ ,τ¯ ,...).
∗ 1 2 n+1 n+2
The left A comodule structure map ψ : H (BPhni)→A ⊗H (BPhni) is given
∗ L ∗ ∗ ∗
on the conjugate generatorsby formula (3.1) and is then extended multiplicatively
to all of H (BPhni).
∗
Dualizing, we derive the left action of α∈A on f ∈H∗(BPhni) by the formula
(α·f)(x)=hα⊗f,ψ (x)i= (−1)|f||ai|hα,a ihf,x i
L X i i
i
whereh−,−idenotesthepairingofamoduleanditsdualandψ (x)= a ⊗x is
L Pi i i
theleftcoactionofA onx∈H (BPhni). Forx∈A ,letx∗ ∈Adenotethe dual.
∗ ∗ ∗
Using this description, it is straightforward to calculate that Q ·(ξ¯pi)∗ = (τ¯ )∗
i j j+i
and that Q acts as zero on smaller powers of ξ , as well as τ . From here we can
i j j
extend multiplicatively to deduce the following lemma.
Lemma 3.2. Let x∗ = (ξ¯e1···ξ¯em ·τ¯ )∗ ∈ A be the dual of a monomial in the ξ¯
j1 jm J j
and the τ¯ . Then Q acts on x∗ in the following manner. For each k such that x
j i
contains a factor of ξ¯pi, replace that factor by τ¯ and dualize, then take the sum
jk jk+i
over all such expressions. In other words we have:
Q ·x∗ = ξ¯ek−piτ¯ · ξ¯eℓ ·τ¯ ∗
i X (cid:16) jk jk+i Y jℓ J(cid:17)
ksuchthat ℓ6=k
pi≤ek
For example, at p=3 we have
Q ·(ξ¯6ξ¯3)∗ =(ξ¯3ξ¯3τ¯ )∗+(ξ¯6τ¯ )∗.
1 2 3 2 3 3 2 4
Our convention is to use homologicalgrading, so H∗(BPhni) is concentrated in
non-positivedegree. Henceξ¯∗ isindegree−(2pj−2)andτ¯∗ isindegree−(2pj−1).
j j
Lemma 3.3. Fix 0 ≤ i ≤ n and suppose x∗ ∈ H∗(BPhni) is in odd degree with
Q ·x∗ =0. Then there is a not necessarily unique y∗ ∈H∗(BPhni) with Q ·y∗ =
i i
x∗.
Proof. Orderthe monomials in x∈H (BPhni) lexicographicallyby lookingat the
∗
τ¯ , starting with the largest j, and then by the ξ¯, starting with the largest i.
j i
Using this ordering, let x be the largest monomial in x and suppose τ¯ is the
1 a
largest τ¯ that divides x . (Because x is in odd degree some such τ¯ must exists.)
j 1 a
Notethatthesummandx cannotcontainanyξ¯pi forb+i>abecausethenQ ·x∗
1 b i 1
would contain a (x /ξ¯pi ·τ¯ )∗ and this cannot cancel with Q ·(x∗−x∗) because
1 b b+i i 1
every term of Q ·(x∗−x∗) is smaller.
i 1
7
Nowlety =x /τ¯ ·ξ¯pi . ThenQ ·y∗ =x∗+(x′)∗ wherex′ issmaller. Replace
1 1 a a−i i 1 1 1 1
x∗ by x∗−Q ·y∗ and proceed by induction. (cid:3)
i 1
4. The Adams spectral sequence
Under the hypotheses of Theorem A there is a conditionally convergent Adams
spectral sequence
Es,t =Exts,t(H∗(BPhni),H∗(X))=⇒π Hom(X,BPhni∧).
2 A t−s p
By a change-of-ringsisomorphism [9, A1.3.12], the E term of the Adams spectral
2
sequence is given by
Es,t =Exts,t(A//E ,H∗(X))∼=Exts,t(F ,H∗(X)).
2 A n En p
Using the given isomorphism θ, we make the identification H∗(X)∼=A//E , with
n
E acting onthe left. The F dual of the Koszulresolutionof F as an E -module
n p p n
is the chain complex P(v ,...,v ) ⊗ E graded by the degree of homogeneous
0 n n
polynomial generators and with differential
d(p⊗ω)= v p⊗(Q ω).
X i i
i
Thus the E term of the Adams spectral sequence is isomorphic to the homology
2
of the chain complex
M =H∗(BPhni)[v ,...,v ],
0 n
where the differential is given by
n
d(x∗)= Q (x∗)v .
X i i
i=0
Notice that the element v is in bidegree (1,2pi−1). We are now in the situation
i
discussed in Section 2.
Since H∗(BPhni) is finite in each degree, this description of the E page shows
2
us that the group Es,t is finite in each bidegree. This implies that Boardman’s
2
derived E term RE = 0 and so the spectral sequence converges strongly [3,
∞ ∞
Theorem7.1]. Inparticular,ifthe givenmapθ :H∗(BPhni)→H∗(X),considered
as a class [θ] ∈ E0,0, survives to E0,0, then [θ] represents a map f: X → BPhni∧
2 ∞ p
satisfyingH∗(f)=θ◦H∗(i), wherei: BPhni→BPhni∧ is the p-completionmap.
p
Inordertoprovethatthegivenisomorphismofcohomologyθ survivesthespectral
sequence, it suffices to show that there is nothing in total degree −1 in E∗,∗ and
2
thusnoroomfornontrivialdifferentialsonθ. Wewillprovethisinthenextsection,
completing the proof of Theorem A
Remark 4.1. We believe that a p-local version of the main theorem is true. The
only obstruction is the lack of convergence of the Adams spectral sequence
Exts,t(H∗(BPhni),H∗(X))=⇒π Hom(X,BPhni)
A t−s
before passage to the p-completion of BPhni in the abutment. Adams and Priddy
get around this using a clever argument involving Adams operations which is not
currently available in the general case of BPhni.
8
5. Proof of the main theorem
We need one more ingredient. As an A-module H∗(BPhni) is irreducible, but
whenconsideringH∗(BPhni)asanE -moduleitsplitsasadirectsumofsubmod-
n
ules, each one finite-dimensional over F . We make the following definition.
p
Definition 5.1. We define a new multiplicative grading on H (BPhni) called the
∗
weight by declaring that ξ¯ and τ¯ have weight 2pj.
j j
The degree δ of an element is slightly less than the weight w, and satisfies the
inequality
p−1
w ≤δ ≤w−1.
p
Weuseδforthedegreeratherthand,asdisalreadyoverloaded. LetH∗(BPhni)[w]
denote the dual of the weight w part of H (BPhni). By Lemma 3.2, the action of
∗
Q preserves the subspace H∗(BPhni)[w]. Hence
i
Exts,t(F ,H∗(BPhni))
En p
splits as a direct sum
Exts,t(F ,H∗(BPhni)[w]).
M En p
w
Lemma 5.2. Suppose 0 ≤ i ≤ n. There are no infinite v -towers in odd total
i
degree in Ext∗,∗(F ,H∗(BPhni).
En p
Proof. Lemma3.3showsthatifwecalculateExt∗,∗(F ,H∗(BPhni))byfirsttaking
En p
homologyinthei’thdirectiontherearenov -towersinodddegreeintheassociated
i
graded. The above splitting of Ext as a direct sum of Ext∗,∗(F ,H∗(BPhni)[w])
showsthatthereisnoroomforextensionsconspiringtomakeEinnfinpitev -towers. (cid:3)
i
Because H∗(BPhni) is concentrated in non-positive degree, with the first odd-
degreeelement τ¯ in degree−(2pn+1−1), the proofof TheoremA amountsto a
n+1
better understanding of the above lemma.
Proof of Theorem A. It suffices to prove that there is nothing in total degree −1
oneweightatatime. Fixaweightw,necessarilydivisibleby2p. Thelowestdegree
element of weight w in H (BPhni) is
∗
p−1
ξ¯w/2p in degree w
1 p
and the lowest odd degree element of weight w is, assuming w is large enough for
such an element to exist,
ξ¯w/2p−pnτ¯ in degree p−1w+2pn−1.
1 n+1 p
Now consider some nonzero class α ∈ E∗,∗[w] ⊂ H (H∗(BPhni)[v ,...,v ],d),
2 ∗ 0 n
andassumewithoutlossofgeneralitythatαisahomogeneouselement. ByLemma
3.3, we can represent α by an element of the form
α =vr0···vrn−1x∗
n 0 n−1 n
with x ∈ H (BPhni) in weight w and degree δ (if α contained a positive v -
n ∗ n n n
power then α would be a dn-boundary). In other words, α is a dn-cycle but
n n
not a dn-boundary, and its class in dn-homology survives the sequence of spectral
9
sequencesstartingwiththe dn−1 spectralsequenceandendingwiththed0 spectral
sequence to represent α in the associated graded. Notice that x∗ is in degree −δ ,
n n
so the class α=[α ] is in total degree
n
(5.3) deg(α)=r (2p−2)+r (2p2−2)+···+r (2pn−1−2)−δ .
1 2 n−1 n
Now we use the algorithm from Definition 2.1. Since α is not a dn-boundary,
n
step n returns α as a representative for the class [α] in the dn-homology of
n
H∗(BPhni)[v ,...,v ].
0 n
Let’s examine step n−1 for the example α = v2 x∗. We replace α by the
n n−1 n
element α =v2z∗ defined by the following picture.
n−1 n
♦❡d♦n❡♦−❡♦1♦❡α♦❡♦n❡77❡=❡❡v❡n2❡−❡❡1dx❡n❡∗❡❡❡❡❡❡❡d❡n❡−s❢❡1s❢❡s❢❡ss❢22sv❢sn❢99−❢1❢v❢n❢y❢d❢∗n❢❢❢❢❢❢❢❢33αn−1 =vn2z∗
v a∗ v b∗
n−1 n
Returning to the general sitution, Lemma 3.3 ensures that there is always such a
zig-zag,trading the factor of vrn−1 in α for a vrn−1 in α . Hence
n−1 n n n−1
α =vr0···vrn−2vrn−1x∗ ,
n−1 0 n−2 n n−1
for some class x in degree
n−1
δ =δ +r (2pn−2pn−1).
n−1 n n−1
Similarly,instepn−2wegetanewrepresentativeα byiterativelyreplacing
n−2
v x∗ with −v y∗−v z∗ as in the picture
n−2 n−1 n
αn−1 =OOvn❧❡−❧❡❧❡2❧❡x❧❡∗❧❡❧❡❧❡❧❧❡66❡v❡n❡−❡1❡y❡∗❡❡❡❡❡❡❡❡❡❡❡❡❡22vnz∗
a∗
Thus α is a sum of terms of the form
n−2
rn−2
vr0···vrn−3 vrn−2−jvj+rn−1x∗ .
0 n−3 X n−1 n n−2,j
j=0
For each j in the sum, deg(x )≥δ where
n−2,j n−2
δ =δ +r (2pn−1−2pn−2).
n−2 n−1 n−2
Continuing in this way, we end up with a representative
α = v x∗
0 X Ij 0,j
j
forα,wherev isamonomialinthev withi≥1. Alsoeachdeg(x )≥δ ,where
Ij i 0,j 0
δ is given by
0
(5.4) δ =δ +r (2p−2)+r (2p2−2p)+...+r (2pn−2pn−1).
0 n 0 1 n−1
Theboundonthedegreeofx isaconsequenceofthefactthateachtimewetrade
0,j
inav weincreasethedegreeoftheelementsinH (BPhni)byatleast2pi+1−2pi.
i ∗
Because x is in odd degree δ with weight w, we have:
n n
p−1
δ ≥ w+2pn−1.
n
p
10