To avoid exessive use of parentheses, write
LFP r hS 1
for(L(FP r )) hS 1
. ToproveonvergeneoftheMorsespetralsequenes,wewillneedthefollowing:
Lemma 4.3. Given
k
, there ism
suh that the inlusionsF m −→ LFP r
and
( F m ) hS 1 −→ LFP r hS 1
indue isomorphism onπ j
andH j
, for allj ≤ k
.Proof. Firstweshowthatthehomologygroupsof
LM
andLM hS 1
arenitelygeneratedineahdegreewhen
M = FP r
(wesayLM
andLM hS 1
areofnitetype): By Serre's spetral sequene for the bration
ΩM −→ P M −→ M
we see that
ΩM
is of nite type, and then the spetral sequene for thebration
ΩM −→ LM −→ M
showsthatLM
isofnitetype. ThebrationLM −→ LM hS 1 −→ BS 1
then showsLM hS 1
is of nite type. For theltrationspaes
F m
,( F m ) hS 1
,weanusethesamebrationsifwerestritthespaes
LM
,ΩM
,P M
tourvesof maximalenergym 2
. Thesame argumentworks for homotopy groups, using the long exat sequene for a bration
insteadof Serre's spetral sequene.
Werstshowthelemmaforhomologygroups. Write
X 0 ⊆ X 1 ⊆ · · · ⊆ X
tooverbothsituations,
F i ⊆ LFP r
and( F i ) hS 1 ⊆ LFP r hS 1
. Letk
begiven,and onsider numbers
m
,M
withk ≤ m ≤ M
, and with the followingproperties:
(i) H k (X m ) −→ H k (X)
issurjetive.(ii)
Ker(H k (X m ) −→ H k (X)) =
Ker(H k (X m ) −→ H k (X M ))
.A simplex
∆ k −→ X
is ompat, so it has nite energy. Takem
suh thatm 2
is bigger than the maximum energy over the nitely many generators ofH k (X)
, then the inlusionX m −→ X
indues a surjetive map onH k
. Wesee we an hose
m
as in(i)
. Given thism
, we onsider Ker(H k (X m ) −→
H k (X))
, whih is nitely generated, sineH k (X m )
is. Suh a generator isa formal sum of simplies
∆ k −→ X m
, whih, when inluded inX
, is theboundary of some formal sum of
(k + 1)
-simplies. Again by ompatness,these have nite energy,and wean hoose
M ≥ m
as desired.Consider a pair
(X i+1 , X i )
in the hainX m −→ X m+1 −→ · · · −→ X M
.By Morse theory we know the quotient
X i+1 /X i
is homotopy equivalent totheThomspaeofabundleofdimensionatleast
2ri
,andsuhaThomspaeanbegiventheellstruturewithone0-ell,andallotherellsofdimension
atleast
2ri
. So by ellularhomology,the relative homologygroups satisfy:H j (X i+1 , X i ) = 0,
forj < 2ri.
(47)Thenby thelongexat sequene forhomologygroups,themaps
H k (X i ) −→
H k (X i+1 )
are isomorphisms, sinek ≤ m ≤ 2ri − 2
fori ≥ m
. This meansH k (X m ) −→ ∼ = H k (X M )
, so by(ii)
, the mapH k (X m ) −→ H k (X)
is injetive,and thus by
(i)
anisomorphism.To showthe Lemmafor homotopy groups,dothe same for
π j
inplae ofH j
. UseHurewizon(47) togetπ j (X i+1 , X i ) = 0
forj < 2ri
,thenonludeas above.
Wenowstate the result aboutMorse spetralsequenes. Inohomology,
we need both the
S 1
-equivariant and the non-equivariant ase, but inK
-theory we need onlythe
S 1
-equivariant ase:Theorem 4.4. There are onvergent spetral sequenes in ohomology,
E s n,q ( M )(LHP r ) ⇒ H n+q (LHP r ) E s n,q ( M )(LHP r hS 1 ) ⇒ H n+q (LHP r hS 1 )
with
E 1
pages given by, forn ≥ 1
, respetively,E 1 n,q ∼ = ˜ H n+q (T h(µ − n )) ∼ = H n+q − (4r+2)n+4r − 1 (G n (HP r )), E 1 n,q ∼ = ˜ H n+q (T h(µ − n ) hS 1 ) ∼ = H n+q − (4r+2)n+4r − 1 (G n (HP r )),
and for
n = 0
,E 0,q = H q (HP r )
andE 0,q = H q (BS 1 × HP r )
, respetively.There is a strongly onvergent spetral sequene in
K
-theory,E s n,q ( M )(LCP r hS 1 ) ⇒ K n+q (LCP r hS 1 )
with
E 1
page given byE 1 0,q = K q (BS 1 ) ⊗ K q (CP r )
, andE 1 n,q ∼ = ˜ K n+q (T h(µ − n ) hS 1 ) ∼ = K n+q − 2r(n − 1) − 1 (G n (CP r ) hS 1 ),
forn ≥ 1,
where
G n (FP r )
denotes the spae of geodesis of lengthn
forn ≥ 1
.Proof. Alosed,simplegeodesihasenergy
1
,andwheniteratedn
timeshasenergy
n 2
. Sothe ritial values are0 < 1 2 < 2 2 < 3 2 < . . .
, and we denoteF (n 2 )
byF n
. Using the energy ltrations (37) and (38), respetively, we make an exat ouple via the long exat sequenes for the pair( F n , F n − 1 )
,and
(( F n ) hS 1 , ( F n − 1 ) hS 1 )
, respetively. For details about this proess, the reader an see e.g. [Hather2 ℄, 1.1. This gives rise to a spetral sequene{ E r p,q ( M ) } r
, whih we all a Morse spetral sequene. The proess whihonstrutsaspetralsequene fromtheexatpairsworksforanyohomology
theory, so we get spetral sequenes in both ohomology and
K
-theory. Byonstrution together with the homotopy equivalenes from Morse theory,
(39) and (40), the
E 1
page is given by, forn ≥ 1
,E 1 n,q ( M )(LM) = H ˜ n+q ( F n / F n − 1 ) ∼ = ˜ H n+q (T h(µ − n ));
E 1 n, ∗ ( M )(LM hS 1 ) = H ˜ ∗ (( F n ) hS 1 /( F n − 1 ) hS 1 ) ∼ = ˜ H ∗ (T h(µ − n ) hS 1 );
and similar for
K
-theory. The negative bundleµ − n
is a bundle over theritial submanifold
N (n 2 )
, whih is the spaeG n (r)
of geodesis of lengthn
. It follows that(µ − n ) hS 1
is abundle overG n (FP r ) hS 1
.For
n = 0
,F 0
is spae of loops of energy zero, i.e. the onstant loops, soF 0 = FP r
itself, and theS 1
ation is trivial,soES 1 × S 1 F 0 = BS 1 × FP r
.The result follows for
n = 0
.Now let
n ≥ 1
,and onsider rstHP r
. The negative bundleµ − n
is foundin[Bökstedt-Ottosen2 ℄, Thm. 6.2,andhereone anseeitisorientedandhas
dimension
(4r + 2)n − 4r + 1
. By[Bökstedt-Ottosen℄Lemma5.2,(µ − n ) hS 1
isalsooriented. Sowe an use the Thom isomorphism, whih gives:
E 1 n,q ( M )(LHP r ) ∼ = H ˜ n+q (T h(µ − n )) ∼ = H n+q − (4r+2)n+4r − 1 (G n (HP r ));
E 1 n,q ( M )(LHP r hS 1 ) ∼ = H ˜ n+q (T h(µ − n ) hS 1 ) ∼ = H n+q − (4r+2)n+4r − 1 (G n (HP r ) hS 1 );
Similarly for
K
-theory, but here we use Prop. 4.2 to get that the bundlesµ − n
and(µ − n ) hS 1
areboth the sumofa trivialreal linebundlewith aomplexbundle. This means we an use the Thom isomorphismfor
K
-theory. From[Bökstedt-Ottosen2 ℄ Thm. 6.1, we see that the negative bundles
µ − n
and(µ − n ) hS 1
have dimension2r(n − 1) + 1
forn ≥ 1
.Fortheonvergene, notethattheohomologyMorsespetralsequene is
arst quadrantspetral sequene. By[Hather2℄ Prop. 1.2the riterion for
onvergene is that the inlusions
F n ֒ → LHP r
, resp.( F n ) hS 1 ֒ → LHP r hS 1
,indueisomorphismon
H q ( − ; F p )
ifn
islargeenoughomparedtoq
. Bytheuniversal oeient theorem it sues to show this on
H q ( − ; F p )
, and thisisproved in Lemma 4.3.
The
K
-theory Morse spetral sequene is not rst quadrant, so theon-vergene question is more subtle. Note that, if we take a nite ltration
( F 0 ) hS 1 ⊆ · · · ⊆ ( F n ) hS 1
, the orresponding Morse spetral sequene on-verges toK ∗ (( F n ) hS 1 )
. The Morse spetral sequene then determines theinverse limitof the
K ∗ (( F n ) hS 1 )
. There is asurjetive mapK ∗ (LCP r hS 1 ) −→ lim
←− n K ∗ (( F n ) hS 1 ),
and we say say the spetral sequene onverges strongly, if this map is an
isomorphism. This requires some work, and will be shown in the lemmas
To show onvergene of the Morse spetral sequene in
K
-theory, letX 0 ⊂ X 1 ⊂ . . .
,andX = ∪ X i
. We want tond onditions that ensurei : K ∗ (X) −→ ∼ = lim
←− i K ∗ (X i )
(*)when
X = LCP r hS 1
. Asmentionedintheproofabove,themapisi
surjetive,so the question isinjetivity.
Lemma 4.5. Let
X = ES 1 × S 1 LCP r
. LetX n
denote then
-skeleton ofX
.Then
( ∗ )
holds.Proof. First note that the lemma is equivalent to saying that the
Atiyah-Hirzebruh spetral sequene for
X
onverges strongly. We haveK 0 (X) = [X, Z × BU ]
andK 1 (X) = [X, U ]
, so a lass inK
-theory an be onsidereda (homotopy lass of a) map from
X
to eitherY = Z × BU
orY = U
. Alass in the kernel of
i
is then amapX −→ Y
whoserestrition toeahX n
is null-homotopi. Suh a map is alled a phantom map, and we denote by
Ph
(X, Y )
the set of homotopy lasses of phantom mapsX −→ Y
. Theirexisteneisstudiedin[MGibbon-Roitberg℄,who givethefollowingriterion
(Thm. 1): The following are equivalent:
(i)
Ph(X, Y ) = 0
for everyY
with nitely generated homotopygroups.(ii)
There exists a map fromΣX
to a wedge of spheres that indues anisomorphismin rationalhomology.
Amapasin
(ii)
weallarationalequivalene. NotethatZ × BU
andU
havenitely generated homotopy groups. Let us apply this to
X = ES 1 × S 1 Z
,where wewill speializeto
Z = LCP r
.Firstweonsiderthebundle
ξ = p ∗ T
overX
,thepullbakofthestandardlinebundle
T −→ BS 1
under the mapp : ES 1 × S 1 Z −→ BS 1
. We use theober sequene
S(ξ) −→ D(ξ) −→ T h(ξ) −→ ΣS(ξ).
(48)We laim it sues to show the result for
T h(ξ)
instead ofX
:K ∗ (X) ∼ = K ∗ (T h(ξ))
by Thom isomorphism, and the ell struture onX
gives rise toa natural ell struture on
T h(ξ) ց X
, wheren
-ells inX
orrespond to(n + 2)
-ellsinT h(ξ)
. So wealsoget anisomorphism of the inverse systems{ K ∗ (X n ) }
and{ K ∗ (T h(ξ) n ) }
suh that the obvious diagramommutes:K ∗ (X)
∼ =
i // lim
←− K ∗ (X n )
∼ =
K ∗ (T h(ξ)) i // lim
←− K ∗ (T h(ξ) n )
So we investigate (48). We have of ourse
D(ξ) ≃ X = ES × S 1 Z
, andwewill showthat
S(ξ) ∼ = ES 1 × Z
: First noteS(ξ) =
([e, z], t) ∈ ES 1 × S 1 Z × T | k e k = 1, k t k = 1, t ∈
spanC e ,
where we onsider
e ∈ ES 1 = S ∞ ⊆ C ∞
andt ∈ T ⊆ C ∞
, by viewingBS 1 = CP ∞
as omplexlines inC ∞
. For([e, z], t) ∈ S(ξ)
, wesee that thereis
s ∈ S 1
withes = t
. We an then onstrut a homeomorphismF : S(ξ) −→ ES 1 × Z, F ([e, z], t) = (t, s − 1 z).
(49)This iswell-dened, with inverse
G(t, z) = ([t, z ], t)
.Now let
Z = LCP r
. By[Bökstedt-Ottosen2 ℄ Theorem 6.1, there is a ho-motopyequivaleneΣLCP r −→ Σ(CP r ) ∨ W
i ΣT h(µ − i )
,whihisthesplittingresult for the non-equivariant ase. So learly, the Atiyah-Hirzebruh
spe-tral sequene onverges in this ase, i.e. there are no phantom maps from
LCP r
, so by the riterion, there is rational equivalene fromΣLCP r
to awedgeofspheres. Sine
S(ξ) ∼ = ES 1 × LCP r ≃ LCP r
,wesee thatwehavearationalequivalene
f 2
fromΣS(ξ)
toawedge of spheres. By (48) thisgivesamap from
T h(ξ)
toa wedge of spheres,T h(ξ) f 1 // Σ(ξ) f 2 // W
i S n i .
(50)Let us onsider the inlusion
LCP r −→ ES 1 × S 1 LCP r
. One anin-vestigate this map on rational ohomology using Serre's spetral sequene
for the bration
LCP r −→ ES 1 × S 1 LCP r −→ BS 1
. This is done in[Bökstedt-Ottosen ℄ Prop. 15.2, and it emerges that
E ∞ = E 3
with allnon-trivial groups in either
E 3 0, ∗ ⊆ E 2 0, ∗ = H ∗ (LCP r ; Q)
orE 3 ∗ ,0 = H ∗ (BS 1 ; Q)
.This impliesthat the ombined map
H ˜ ∗ (LCP r ; Q) ⊕ H ˜ ∗ (BS 1 ; Q) −→ H ˜ ∗ (ES 1 × S 1 LCP r ; Q)
(51)issurjetive.
In (48),use the homotopy equivalenes
S(ξ) ∼ = ES 1 × LCP r
andD(ξ) ≃ ES 1 × S 1 LCP r
,and projeton the rst fator to getS(ξ) //
D(ξ) //
T h(ξ)
ES 1 // BS 1 // BS 1 /ES 1 ≃ BS 1
whih gives a map
g 1 : T h(ξ) −→ BS 1
. Note that the Atiyah-Hirzebruh spetral sequene forBS 1
onverges, so by the riterion, there is a rationalequivalene
g 2 : ΣBS 1 −→ W
j S n j
.Combining with(50), we an makea ompositemap
ϕ : ΣT h(ξ) ∆ // ΣT h(ξ) ∨ ΣT h(ξ) f 1 ∨ g 1 // Σ 2 S(ξ) ∨ ΣBS 1 f 2 ∨ g 2 // W
k S n k
Here
f 2 ∨ g 2
isarationalequivalene,andby (51),∆ ∗ ◦ (f 1 ∨ f 2 ) ∗
issurjetiveonredued ohomologywith rational oeients. Sothe omposite map
ϕ ∗
is surjetive onrational ohomology, and by ollapsingsome of the spheres,
weanensureitbeomesinjetive. Wehaveonstrutedthe desiredrational
equivalene.
Lemma 4.6. If
X i
is a sequene of subomplexes of the CW omplexX = LCP r hS 1
, and iffor everyk
there isanm
suhthat thek
-skeletonSk k (X) ⊆ X m
, then ondition( ∗ )
applies.Proof. We must show that the map
K ∗ (X) −→ lim
←− i K ∗ (X i )
is injetive. Let
a
be in the kernel of this map. Beause of our onditionon the ltration,
a
will restrit trivially to eah skeleton. Then Lemma 4.5shows that
a
vanishes.Now onsider the general ase. By Lemma 4.3, the ondition on
π j
issatisedfor
X = LCP r hS 1
.Lemma4.7. If
X i
isa sequeneof subspaes ofX
as above,andiffor everyk
there is anm
suh thatπ j (X m ) → π j (X)
is an isomorphism forj ≤ k
,then ondition
( ∗ )
applies.Proof. UsingrelativeCWapproximation(see[Hather1 ℄Prop. 4.13),wean
indutivelyonstrut asequene ofCWomplexes
Y i
suhthatthe followingladder ommutes,
Y 0
// Y 1 //
. . .
X 0 // X 1 // . . .
and suh that the vertial maps are weak homotopy equivalenes. F
ur-thermore, for a given
k
we have by assumption that there ism
suh thatπ j (X m ) → π j (X)
isanisomorphismforj ≤ k
,and thismeanswean ensurethatall
Y n
forn ≥ m
areonstrutedfromY n − 1
byaddingellsof dimensiongreater than
k
. So lettingY = ∪ i Y i
, we have that for eahk
there is anm
suh that
Sk k (Y ) ⊆ Y m
.The map
Y → X
is a weak homotopy equivalene. Noting that a weak homotopy equivalene preservesK
-theory, the lemma follows from thepre-viousone.
5
S 1
-equivariant ohomology ofLHP r
5.1 The Morse spetral sequenes
For
LHP r hS 1
, the Morse spetral sequene is asfollows:Theorem5.1. TheMorse spetralsequene
E r ∗ , ∗ ( M )(LHP r hS 1 )
isaspetralsequene of
H ∗ (BS 1 ; F p ) = F p [u]
-modules,and it has the followingE 1
page:Assume
p | r + 1
. ThenE 1 0, ∗ = F p [u, y]/ h y r+1 i ;
E 1 pm+k, ∗ = α pm+k F p [u, t]/ h Q r , Q r+1 i ,
form ≥ 0
,1 < k < p − 1;
E 1 pm, ∗ = α pm F p [u] { 1, y, . . . , y r , σ, . . . , σy r }
form ≥ 1.
Assume
p ∤ r + 1
. ThenE 1 0, ∗ = F p [u, y]/ h y r+1 i ;
E 1 pm+k, ∗ = α pm+k F p [u, t]/ h Q r , Q r+1 i ,
form ≥ 0
,1 < k < p − 1;
E 1 pm, ∗ = α pm F p [u] { 1, y, . . . , y r+1 , τ, . . . , τ y r+1 }
form ≥ 1.
Inltration
n = pm + k
, the elementα pm+k u i t j
has total degree(4r + 2)n − 4r + 2i + 4j + 1
. In ltrationn = pm
, the generators are freeF p [u]
-modulegenerators, whih have the followingdegrees:
Class Case Total degree
α pm y i p | r + 1, 0 ≤ i ≤ r (4r + 2)pm − 4r + 4i + 1 α pm y i p ∤ r + 1, 0 ≤ i ≤ r − 1 (4r + 2)pm − 4r + 4i + 1 α pm y i σ p | r + 1, 0 ≤ i ≤ r (4r + 2)pm + 4i
α pm y i τ p ∤ r + 1, 0 ≤ i ≤ r − 1 (4r + 2)pm + 4i + 4
Note that the olumns
E 1 pm, ∗
,m ≥ 0
, are innite, while the lassα pm+k u i t j
in
E 1 pm+k, ∗
iszero wheni ≥ 4r
orj ≥ 2r
.Remark5.2. Thesymbol
α n
referstotheThomisomorphism. Thenotationα n x
et. denotes the up produt with the Thom lass ofµ − n
in the ritialsubmanifold
N (n 2 )
. Theprodutisnotdened inthe spetralsequene, andso it is a bit of abuse of notation. But it is a very pratialway of keeping
trak of the dimension shiftand shouldbe read as suh.
Proof. TheMorse spetralsequene isdesribed inTheorem4.4. Weuse
o-homologywith
F p
oeients. Firsttakeltrationn = 0
. ThenG 0 (HP r ) hS 1 = HP r hS 1
itself,and theS 1
ation is trivial. ThusE 1 0, ∗ ( M )(LHP r hS 1 ) ∼ = H ∗ (HP r hS 1 ; F p ) = H ∗ (BS 1 × HP r ; F p )
∼ = H ∗ (BS 1 ; F p ) ⊗ H ∗ (HP r ; F p ) ∼ = F p [u] ⊗ F p [x]/ h x r i .
Nowtake
n ≥ 1
. From Theorem 4.4,E 1 n, ∗ ( M )(LHP r hS 1 ) ∼ = H n+ ∗− ((4r+2)n − 4r+1) (G(HP r ) (n) hS 1 ; F p ).
Nowweanusethe previousresultsabout thespaesofgeodesis,Theorems
2.12 and 2.14. For the ase
n = pm + k
we know from Theorem 2.12 thatu
maps tox
, and so theF p [u]
-module struture is that multipliation byu
equals multipliation by
x
. This is inorporated in the notation by writingu
for the lass previously namedx
. The lastpart of the theorem is Lemma2.9.
The next Lemmais based upon[Bökstedt-Ottosen℄, Lemma 9.8:
Lemma 5.3. In the Morse spetral sequene for
LHP r hS 1
, all dierentials starting in odd total degree are trivial.Proof. This is mostly seen for dimensional reasons. Usingthe table in
The-orem 5.1, we see that elements of odd total degree in the spetral sequene
have the form
α n y i u j
orα n u i t j
. Beause of the derivation property of thedierentials, it is enough to onsider the
F p [u]
generators, i.e.α pm y i
andα pm+k t j
form ≥ 0
.So let us prove that
d s (α pm y i )
is trivial(s ≥ 1)
. This has total degree(4r +2)pm − 4r +4i+2
andltrationdegreepm + s
. BythetableinTheorem5.1,observethatanon-triviallassofltration
n
andeventotaldegreeexistsif and only if
p | n
. Furthermore, inasep | n
we an determine the lass ofltration
n
with lowest total degree. Ifp | (r + 1)
, this lass isα n σ
of totaldegree
(4r + 2)n
,andifp ∤ n
thislass isα n τ
oftotaldegree(4r + 2)n + 4
. Soif
d s (α pm y i )
is non-trivial, its total degree must be at least the total degree mentionedabove. That is,(4r + 2)pm − 4r + 4i + 2 ≥
(4r + 2)(pm + s), p | (r + 1)
;(4r + 2)(pm + s) + 4, p ∤ (r + 1)
.Suppose
p | r + 1
. Then we an reduethe inequality to− 4r + 4i + 2 ≥ (4r + 2)s ⇔ ( − 4r + 2)(s + 1) + 4i ≥ 0.
Thisiseasiertosatisfyif
s
issmallandi
islarge,sowetrys = 1
(minimum)and
i = r
(maximum),obtainingtheequality2( − 4r + 2) + 4r = − 4r + 4 ≥ 0
,whih only holds for
r = 1
. In this ase we have equality. Ifs > 1
ori < r
, there are no solutions. So the question is whetherd 1 (α pm y)
an be anon-triviallass of even total degreeinltration
n = pm + 1
, and itannot,sine then, asnoted earlier,
p
shoulddividepm + 1
. Ifp ∤ r + 1
there are noNow take the ase
α pm+k t j
. Thend s (α pm+k t j )
has ltrationdegreepm + k + s
and totaldegree(4r + 2)(pm + k) − 4r + 4j + 2
, whih iseven. Bythesame observation as before, if
d s (α pm+k t j )
were to be non-trivial, its total degreemust satisfy(4r + 2)(pm + k) − 4r + 4j + 2 ≥
(4r + 2)(pm + s + k), p | (r + 1)
;(4r + 2)(pm + s + k) + 4, p ∤ (r + 1)
.Like before, weredue for
p | r + 1
:− 4r + 4j + 2 ≥ (4r + 2)s ⇔ (4r + 2)(s + 1) − 4 ≤ 4j
Reall
s ≥ 1
, so to satisfy this,j ≥ 2r
. But then the lassα pm+k t j
is zero,aordingtothelastpartof Theorem5.1. Likewisefor
p | r + 1
. Thisprovesthe Lemma.
Weare goingtoneed anoverview ofthe size of the
E 1
page ofthe Morsespetral sequene.
Lemma 5.4. The Poinaré series
P (t)
ofE 1 (L(HP r ) hS 1 )
is given by forp ∤ r + 1
:1 − t 4r+4 + 1 − t t 4r+2 3 (1 − t 4r )(1 − t 4r+4 ) + t p(4r+2) 1 − t p(4r+2) − 4r+1 (1 − t 4r )(t 4r+3 + t 4r+4 )
(1 − t 2 )(1 − t 4 ) .
and for
p | r + 1
,1 − t 4r+4 + 1 − t t 4r+2 3 (1 − t 4r )(1 − t 4r+4 ) + t p(4r+2) 1 − t p(4r+2) − 4r+1 (1 − t 4r+4 )(t 4r − 1 + t 4r )
(1 − t 2 )(1 − t 4 ) .
Proof. Ionlyprovethis for
p ∤ r + 1
. Theotherase isexatlythe same. Werst nd the Poinaréseries for
E 1 n, ∗
.• n = 0
: By Theorem 5.1, sineE 1 0, ∗
is afreeF p [u]
-module,P (E 1 0, ∗ )(t) = P (F p [u]) · P (F p [x]/ h x r i ) = 1
1 − t 2 · 1 − t 4(r+1) 1 − t 4 .
• p ∤ n
: By Theorem 5.1P (E 1 n, ∗ )(t) = t 4r(n − 1)+2n+1 · P (F p [t, u]/ h Q r , Q r+1 i )
= t 4r(n − 1)+2n+1 (1 + t 2 ) · 1 − t 4r
1 − t 4 · 1 − t 4(r+1) 1 − t 4
= t 4r(n − 1)+2n+1 (1 − t 4r )(1 − t 4r+4 ) (1 − t 2 )(1 − t 4 ) ,
using Lemma2.9to nd
P (F p [t, u]/ h Q r , Q r+1 i )
.• p | n
: Aording toTheorem 5.1, we obtainP (E 1 n, ∗ )(t) = t 4r(n − 1)+2n+1 · P (F p [u]
1, y, . . . , y r+1 , τ, . . . , τ y r+1 )
= t 4r(n − 1)+2n+1 1
1 − t 2 · (1 − t 4r )(1 + t 4r+3 ) 1 − t 4 .
sine
y
has degree 4andτ
has degree4r + 3
.Wemustsum over
n ≥ 1
toalulateP (E 1 )(t)
. Onlythe fatort 4r(n − 1)+2n+1
depends on
n
,so we sum that rst, inthe two asesp | n
andp ∤ n
:X
n ≥ 1,p | n
t 4r(n − 1)+2n+1 = X
m ≥ 1
t 4r(mp − 1)+2mp+1 = t p(4r+2) − 4r+1 1 − t p(4r+2) .
Usingthis, we an ompute
X
n ≥ 1, p∤n
t 4r(n − 1)+2n+1 = X
n ≥ 1
t 4r(n − 1)+2n+1 − t p(4r+2) − 4r+1
1 − t p(4r+2) = t 3
1 − t 4r+2 − t p(4r+2) − 4r+1 1 − t p(4r+2) .
Combiningthe results above and summingover
n ≥ 1
then yields:P (E 1 )(t) = P (E 1 0, ∗ )(t) + X
n ≥ 1, p | n
P (E 1 n, ∗ )(t) + X
n ≥ 1, p∤n
P (E 1 n, ∗ )(t)
= 1
(1 − t 2 )(1 − t 4 ) ·
1 − t 4(r+1) + t p(4r+2) − 4r+1
1 − t p(4r+2) (1 − t 4r )(1 + t 4r+3 ) +
t 3
1 − t 4r+2 − t p(4r+2) − 4r+1 1 − t p(4r+2)
(1 − t 4r )(1 − t 4r+4 )
= 1 − t 4r+4 + 1 − t t 4r+2 3 (1 − t 4r )(1 − t 4r+4 ) + t p(4r+2) 1 − 4r+1
− t p(4r+2) (1 − t 4r )(t 4r+3 + t 4r+4 )
(1 − t 2 )(1 − t 4 ) .
Remark 5.5. Later we are going toneed the odd and even parts of
E 1
,i.e.E 1
odd= L
p+q
oddE 1 p,q
, and likewiseforE 1
even. Notie thatK(t) := t p(4r+2) − 4r+1 1 − t p(4r+2)
has odd degree. Then we get from the above Lemmathat for
p ∤ r + 1
,P (E 1
even)(t) = 1 − t 4r+4 + K(t)(1 − t 4r )t 4r+3
(1 − t 2 )(1 − t 4 ) ; P (E 1
odd)(t) = 1 − t 4r
(1 − t 2 )(1 − t 4 )
(1 − t 4r+4 )t 3
1 − t 4r+2 + K(t)t 4r+4
.
Similarlyfor
p | r + 1
,P (E 1
even)(t) = 1 − t 4r+4
(1 − t 2 )(1 − t 4 ) 1 + K(t)t 4r − 1
; P (E 1
odd)(t) = 1 − t 4r+4
(1 − t 2 )(1 − t 4 )
(1 − t 4r )t 3
1 − t 4r+2 + K(t)t 4r
.
For omparison purposes we are also going to need the non-equivariant
ase,
H ∗ (LHP r )
.Theorem 5.6. Let
E s ∗ , ∗ = E s ∗ , ∗ ( M )(LHP r )
. Assumep | r + 1
. ThenE 1 0, ∗ = F p [y]/ h y r+1 i ;
E 1 n, ∗ = α n F p [y, σ]/ h y r+1 , σ 2 i
forn ≥ 1.
Assume
p ∤ r + 1
. ThenE 1 0, ∗ = F p [y]/ h y r+1 i ;
E 1 n, ∗ = α n F p [y, τ ]/ h y r , τ 2 i
forn ≥ 1.
where
| x | = 4
,| σ | = 4r − 1
,| τ | = 4r + 3
,| α n | = (4r + 2)n − 4r + 1
.Thisspetralsequeneollapsesfromthe
E 1
page. ThisdeterminesH ∗ (LHP r ; F p )
as an abelian group, and it has the followingPoinaré series: For
p ∤ r + 1
,P H ∗ (LHP r ) (t) = 1 − t 4r+4
1 − t 4 + (1 − t 4r )(1 + t 4r+3 )t 3 (1 − t 4 )(1 − t 4r+2 ) ;
and for
p | r + 1
,P H ∗ (LHP r ) (t) = 1 − t 4r+4
1 − t 4 + (1 − t 4r+4 )(1 + t 4r − 1 )t 3 (1 − t 4 )(1 − t 4r+2 ) .
The map indued by inlusion
i ∗ : E 1 n,
odd− n ( M )(LHP r hS 1 ) −→ E 1 n,
odd− n ( M )(LHP r )
issurjetive.
Proof. The omputationof
E 1
via Morse theory is just like the proof of theequivariant ase, Theorem 5.1. That the spetral sequene ollapses follows
froma splittingresult for
LHP r
. Suha result an be found in [Ziller℄.For the omputation of the Poinaré series, sine the spetral sequene
ollapses, we an ompute
P H ∗ (LHP r ) = P E ∞ = P E 1
. We reuse theomputa-tions from the proof of Lemma 5.4. Consider the ase
p ∤ r + 1
. (The asep | r + 1
is similar.) In ltrationn > 0
we have,P (E 1 n, ∗ )(t) = t 4r(n − 1)+2n+1 · 1 − t 4r
1 − t 4 (1 + t 4r+3 ).
Andso
P (E 1 )(t) = 1 − t 4r+4 1 − t 4 + X
n>0
t 4r(n − 1)+2n+1 · 1 − t 4r
1 − t 4 (1 + t 4r+3 )
= 1 − t 4r+4
1 − t 4 + (1 − t 4r )(1 + t 4r+3 )t 3 (1 − t 4 )(1 − t 4r+2 ) .
For the surjetivity, we prove for every
n ∈ N
that the mapE 1 n,
odd− n ( M )(LHP r hS 1 ) −→ E 1 n,
odd− n ( M )(LHP r )
is surjetive. For
n = 0
the target spae is zero, so the result is trivial. Forn > 0
, the degree of the Thom lassα n
isodd, soby the formula fortheE 1
page, the question is whether
i ∗ : H
even(G(HP r ) (n) hS 1 ) −→ H
even(G(HP r ) (n) )
is surjetive. This follows fromCorollary 2.15.
Remark 5.7. We also need the odd and even parts, so I will dothat
om-putationnow. For
p ∤ r + 1
,P H
odd∗ (LHP r ) (t) = (1 − t 4r )t 3 (1 − t 4 )(1 − t 4r+2 ) ;
and
P H
even∗ (LHP r ) (t) = 1 − t 4r+4
1 − t 4 + (1 − t 4r )t 4r+6
(1 − t 4 )(1 − t 4r+2 )
(52)= 1 + (1 − t 4r )t 4 (1 − t 4 )(1 − t 4r+2 ) .
Notethat
t · P (H
odd(LHP r ))(t) = P (H
even(LHP r ))(t) − 1,
(53)and that
P H
odd∗ (LHP r ) (t) = t 3 (1 + t 4 + · · · + t 4r − 4 ) X ∞ n=0
t n(4r+2)
(54)has all oeients equal to 0 or 1,and the dierene in degree between the
1-oeients is at least four. We have the same properties when
p | r + 1
,and for future referene, when
p | r + 1
,P H
odd∗ (LHP r ) (t) = (1 − t 4r+4 )t 3
(1 − t 4 )(1 − t 4r+2 ) = t 3 (1 + t 4 + · · · + t 4r ) X ∞
n=0
t n(4r+2)
(55)Corollary5.8. Fortheenergyltration
F 0 ⊆ F 1 ⊆ · · · ⊆ F n ⊆ · · · ⊆ LHP r
,the dimension of
H
odd( F m )
as anF p
vetor spae is as follows:dim H
odd( F m ) =
m(r + 1), p | r + 1
;mr, p ∤ r + 1
.Proof. TheMorsespetralsequene
{ E s ∗ , ∗ } = { E s ∗ , ∗ ( M )(LHP r ) }
induedbythe energy ltration of
LHP r
ollapses from theE 1
page by Theorem 5.6above. This means that
E ∞ = E 1
. Comparing with the spetral sequene{ E s ( F m ) }
of the nite ltrationF 0 ⊆ F 1 ⊆ · · · ⊆ F m
we see that itsE 1
pageisthesameas
E 1 ( M )(LHP r )
up toltrationm
. Sobynaturality,bothspetralsequenesollapsefromthe
E 1
page,andE ∞ ( F m )
equalsE ∞ (LHP r )
up to ltration
m
. Sowe an alulatethe dimension ofH
odd( F m )
asanF p
vetor spae:
dim H
odd( F m ) = dim E ∞ m,
odd− m ( F m ) + · · · + dim E ∞ 1,
odd− 1 ( F m )
=
m(r + 1), p | r + 1
;mr, p ∤ r + 1
.Here the last equalityis from (54) and (55).
To squeeze the last information out of the Morse spetral sequenes, we
are going to use loalization. The general setup is as follows: Given an
R
module
M
and amultipliativesetU ⊆ R
(i.e. ifu, v ∈ U
thenuv ∈ U
),wedene
M
loalizedaway fromU
asM [U − 1 ] = n m
u | m ∈ M, u ∈ U o / ∼
where
m
u ∼ m u ′ ′
if there isv ∈ U
suh thatvu ′ m = vum ′
. It is an elementaryalgebrai fat that loalization away from
U ⊆ R
is an exat funtor onR
-modules.We are going to use
U = { u n | n ∈ N } ⊆ F p [u]
, whereu
as usuallyde-notes our generator
u ∈ H 2 (BS 1 ; F p )
, suh thatH ∗ (BS 1 ; F p ) ∼ = F p [u]
. Themainloalizationresulthereis[Bökstedt-Ottosen ℄Theorem8.3,whihIstate
Theorem 5.9. There is an isomorphism of spetral sequenes
E ∗ ( M )(LHP r hS 1 ) 1
u
∼ = E ∗ ( M )(LHP r ) ⊗ F p [u, u − 1 ].
when re-indexing the olumns: ltration
pm
goes to ltrationm
form ∈ N
.Note: Thisimpliesthattheloalizedspetralsequene
E ∗ ( M )(LHP r hS 1 ) 1
u
ollapsesfromthe