• Ingen resultater fundet

The Morse theory spetral sequene

In document Afhandling (Sider 52-65)

To avoid exessive use of parentheses, write

LFP r hS 1

for

(L(FP r )) hS 1

. To

proveonvergeneoftheMorsespetralsequenes,wewillneedthefollowing:

Lemma 4.3. Given

k

, there is

m

suh that the inlusions

F m −→ LFP r

and

( F m ) hS 1 −→ LFP r hS 1

indue isomorphism on

π j

and

H j

, for all

j ≤ k

.

Proof. Firstweshowthatthehomologygroupsof

LM

and

LM hS 1

arenitely

generatedineahdegreewhen

M = FP r

(wesay

LM

and

LM hS 1

areofnite

type): 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 the

bration

ΩM −→ LM −→ M

showsthat

LM

isofnitetype. Thebration

LM −→ LM hS 1 −→ BS 1

then shows

LM hS 1

is of nite type. For the

ltrationspaes

F m

,

( F m ) hS 1

,weanusethesamebrationsifwerestritthe

spaes

LM

,

ΩM

,

P M

tourvesof maximalenergy

m 2

. Thesame argument

works 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

. Let

k

begiven,

and onsider numbers

m

,

M

with

k ≤ m ≤ M

, and with the following

properties:

(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. Take

m

suh that

m 2

is bigger than the maximum energy over the nitely many generators of

H k (X)

, then the inlusion

X m −→ X

indues a surjetive map on

H k

. We

see we an hose

m

as in

(i)

. Given this

m

, we onsider Ker

(H k (X m ) −→

H k (X))

, whih is nitely generated, sine

H k (X m )

is. Suh a generator is

a formal sum of simplies

k −→ X m

, whih, when inluded in

X

, is the

boundary 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 hain

X m −→ X m+1 −→ · · · −→ X M

.

By Morse theory we know the quotient

X i+1 /X i

is homotopy equivalent to

theThomspaeofabundleofdimensionatleast

2ri

,andsuhaThomspae

anbegiventheellstruturewithone0-ell,andallotherellsofdimension

atleast

2ri

. So by ellularhomology,the relative homologygroups satisfy:

H j (X i+1 , X i ) = 0,

for

j < 2ri.

(47)

Thenby thelongexat sequene forhomologygroups,themaps

H k (X i ) −→

H k (X i+1 )

are isomorphisms, sine

k ≤ m ≤ 2ri − 2

for

i ≥ m

. This means

H k (X m ) −→ = H k (X M )

, so by

(ii)

, the map

H k (X m ) −→ H k (X)

is injetive,

and thus by

(i)

anisomorphism.

To showthe Lemmafor homotopy groups,dothe same for

π j

inplae of

H j

. UseHurewizon(47) toget

π j (X i+1 , X i ) = 0

for

j < 2ri

,thenonlude

as above.

Wenowstate the result aboutMorse spetralsequenes. Inohomology,

we need both the

S 1

-equivariant and the non-equivariant ase, but in

K

-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, for

n ≥ 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 )

and

E 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 by

E 1 0,q = K q (BS 1 ) ⊗ K q (CP r )

, and

E 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 ),

for

n ≥ 1,

where

G n (FP r )

denotes the spae of geodesis of length

n

for

n ≥ 1

.

Proof. Alosed,simplegeodesihasenergy

1

,andwheniterated

n

timeshas

energy

n 2

. Sothe ritial values are

0 < 1 2 < 2 2 < 3 2 < . . .

, and we denote

F (n 2 )

by

F 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 ℄, Ÿ1.1. This gives rise to a spetral sequene

{ E r p,q ( M ) } r

, whih we all a Morse spetral sequene. The proess whih

onstrutsaspetralsequene fromtheexatpairsworksforanyohomology

theory, so we get spetral sequenes in both ohomology and

K

-theory. By

onstrution together with the homotopy equivalenes from Morse theory,

(39) and (40), the

E 1

page is given by, for

n ≥ 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 bundle

µ n

is a bundle over the

ritial submanifold

N (n 2 )

, whih is the spae

G n (r)

of geodesis of length

n

. It follows that

n ) hS 1

is abundle over

G n (FP r ) hS 1

.

For

n = 0

,

F 0

is spae of loops of energy zero, i.e. the onstant loops, so

F 0 = FP r

itself, and the

S 1

ation is trivial,so

ES 1 × S 1 F 0 = BS 1 × FP r

.

The result follows for

n = 0

.

Now let

n ≥ 1

,and onsider rst

HP r

. The negative bundle

µ n

is found

in[Bökstedt-Ottosen2 ℄, Thm. 6.2,andhereone anseeitisorientedandhas

dimension

(4r + 2)n − 4r + 1

. By[Bökstedt-Ottosen℄Lemma5.2,

n ) hS 1

is

alsooriented. 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 bundles

µ n

and

n ) hS 1

areboth the sumofa trivialreal linebundlewith aomplex

bundle. 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

n ) hS 1

have dimension

2r(n − 1) + 1

for

n ≥ 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 )

if

n

islargeenoughomparedto

q

. Bythe

universal oeient theorem it sues to show this on

H q ( − ; F p )

, and this

isproved in Lemma 4.3.

The

K

-theory Morse spetral sequene is not rst quadrant, so the

on-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 to

K (( F n ) hS 1 )

. The Morse spetral sequene then determines the

inverse limitof the

K (( F n ) hS 1 )

. There is asurjetive map

K (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, let

X 0 ⊂ X 1 ⊂ . . .

,and

X = ∪ X i

. We want tond onditions that ensure

i : K (X) −→ = lim

←− i K (X i )

(*)

when

X = LCP r hS 1

. Asmentionedintheproofabove,themapis

i

surjetive,

so the question isinjetivity.

Lemma 4.5. Let

X = ES 1 × S 1 LCP r

. Let

X n

denote the

n

-skeleton of

X

.

Then

( )

holds.

Proof. First note that the lemma is equivalent to saying that the

Atiyah-Hirzebruh spetral sequene for

X

onverges strongly. We have

K 0 (X) = [X, Z × BU ]

and

K 1 (X) = [X, U ]

, so a lass in

K

-theory an be onsidered

a (homotopy lass of a) map from

X

to either

Y = Z × BU

or

Y = U

. A

lass in the kernel of

i

is then amap

X −→ Y

whoserestrition toeah

X 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 maps

X −→ Y

. Their

existeneisstudiedin[MGibbon-Roitberg℄,who givethefollowingriterion

(Thm. 1): The following are equivalent:

(i)

Ph

(X, Y ) = 0

for every

Y

with nitely generated homotopygroups.

(ii)

There exists a map from

ΣX

to a wedge of spheres that indues an

isomorphismin rationalhomology.

Amapasin

(ii)

weallarationalequivalene. Notethat

Z × BU

and

U

have

nitely generated homotopy groups. Let us apply this to

X = ES 1 × S 1 Z

,

where wewill speializeto

Z = LCP r

.

Firstweonsiderthebundle

ξ = p T

over

X

,thepullbakofthestandard

linebundle

T −→ BS 1

under the map

p : ES 1 × S 1 Z −→ BS 1

. We use the

ober sequene

S(ξ) −→ D(ξ) −→ T h(ξ) −→ ΣS(ξ).

(48)

We laim it sues to show the result for

T h(ξ)

instead of

X

:

K (X) ∼ = K (T h(ξ))

by Thom isomorphism, and the ell struture on

X

gives rise to

a natural ell struture on

T h(ξ) ց X

, where

n

-ells in

X

orrespond to

(n + 2)

-ellsin

T 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

, and

wewill showthat

S(ξ) ∼ = ES 1 × Z

: First note

S(ξ) =

([e, z], t) ∈ ES 1 × S 1 Z × T | k e k = 1, k t k = 1, t ∈

span

C e ,

where we onsider

e ∈ ES 1 = S ⊆ C

and

t ∈ T ⊆ C

, by viewing

BS 1 = CP

as omplexlines in

C

. For

([e, z], t) ∈ S(ξ)

, wesee that there

is

s ∈ S 1

with

es = t

. We an then onstrut a homeomorphism

F : 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-motopyequivalene

ΣLCP r −→ Σ(CP r ) ∨ W

i ΣT h(µ i )

,whihisthesplitting

result 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 from

ΣLCP r

to a

wedgeofspheres. Sine

S(ξ) ∼ = ES 1 × LCP r ≃ LCP r

,wesee thatwehavea

rationalequivalene

f 2

from

ΣS(ξ)

toawedge of spheres. By (48) thisgives

amap 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 an

in-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 all

non-trivial groups in either

E 3 0, ⊆ E 2 0, = H (LCP r ; Q)

or

E 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

and

D(ξ) ≃ ES 1 × S 1 LCP r

,and projeton the rst fator to get

S(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 for

BS 1

onverges, so by the riterion, there is a rational

equivalene

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 )

issurjetive

onredued 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 omplex

X = LCP r hS 1

, and iffor every

k

there isan

m

suhthat the

k

-skeleton

Sk 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 ondition

on the ltration,

a

will restrit trivially to eah skeleton. Then Lemma 4.5

shows that

a

vanishes.

Now onsider the general ase. By Lemma 4.3, the ondition on

π j

is

satisedfor

X = LCP r hS 1

.

Lemma4.7. If

X i

isa sequeneof subspaes of

X

as above,andiffor every

k

there is an

m

suh that

π j (X m ) → π j (X)

is an isomorphism for

j ≤ k

,

then ondition

( )

applies.

Proof. UsingrelativeCWapproximation(see[Hather1 ℄Prop. 4.13),wean

indutivelyonstrut asequene ofCWomplexes

Y i

suhthatthe following

ladder 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 is

m

suh that

π j (X m ) → π j (X)

isanisomorphismfor

j ≤ k

,and thismeanswean ensure

thatall

Y n

for

n ≥ m

areonstrutedfrom

Y n − 1

byaddingellsof dimension

greater than

k

. So letting

Y = ∪ i Y i

, we have that for eah

k

there is an

m

suh that

Sk k (Y ) ⊆ Y m

.

The map

Y → X

is a weak homotopy equivalene. Noting that a weak homotopy equivalene preserves

K

-theory, the lemma follows from the

pre-viousone.

5

S 1

-equivariant ohomology of

LHP 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 )

isaspetral

sequene of

H (BS 1 ; F p ) = F p [u]

-modules,and it has the following

E 1

page:

Assume

p | r + 1

. Then

E 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 ,

for

m ≥ 0

,

1 < k < p − 1;

E 1 pm, = α pm F p [u] { 1, y, . . . , y r , σ, . . . , σy r }

for

m ≥ 1.

Assume

p ∤ r + 1

. Then

E 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 ,

for

m ≥ 0

,

1 < k < p − 1;

E 1 pm, = α pm F p [u] { 1, y, . . . , y r+1 , τ, . . . , τ y r+1 }

for

m ≥ 1.

Inltration

n = pm + k

, the element

α pm+k u i t j

has total degree

(4r + 2)n − 4r + 2i + 4j + 1

. In ltration

n = pm

, the generators are free

F p [u]

-module

generators, 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 lass

α pm+k u i t j

in

E 1 pm+k,

iszero when

i ≥ 4r

or

j ≥ 2r

.

Remark5.2. Thesymbol

α n

referstotheThomisomorphism. Thenotation

α n x

et. denotes the up produt with the Thom lass of

µ n

in the ritial

submanifold

N (n 2 )

. Theprodutisnotdened inthe spetralsequene, and

so 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. Firsttakeltration

n = 0

. Then

G 0 (HP r ) hS 1 = HP r hS 1

itself,and the

S 1

ation is trivial. Thus

E 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 that

u

maps to

x

, and so the

F p [u]

-module struture is that multipliation by

u

equals multipliation by

x

. This is inorporated in the notation by writing

u

for the lass previously named

x

. The lastpart of the theorem is Lemma

2.9.

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

or

α n u i t j

. Beause of the derivation property of the

dierentials, it is enough to onsider the

F p [u]

generators, i.e.

α pm y i

and

α pm+k t j

for

m ≥ 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

andltrationdegree

pm + s

. BythetableinTheorem

5.1,observethatanon-triviallassofltration

n

andeventotaldegreeexists

if and only if

p | n

. Furthermore, inase

p | n

we an determine the lass of

ltration

n

with lowest total degree. If

p | (r + 1)

, this lass is

α n σ

of total

degree

(4r + 2)n

,andif

p ∤ n

thislass is

α n τ

oftotaldegree

(4r + 2)n + 4

. So

if

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

issmalland

i

islarge,sowetry

s = 1

(minimum)

and

i = r

(maximum),obtainingtheequality

2( − 4r + 2) + 4r = − 4r + 4 ≥ 0

,

whih only holds for

r = 1

. In this ase we have equality. If

s > 1

or

i < r

, there are no solutions. So the question is whether

d 1 (α pm y)

an be a

non-triviallass of even total degreeinltration

n = pm + 1

, and itannot,

sine then, asnoted earlier,

p

shoulddivide

pm + 1

. If

p ∤ r + 1

there are no

Now take the ase

α pm+k t j

. Then

d spm+k t j )

has ltrationdegree

pm + k + s

and totaldegree

(4r + 2)(pm + k) − 4r + 4j + 2

, whih iseven. Bythe

same 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 lass

α pm+k t j

is zero,

aordingtothelastpartof Theorem5.1. Likewisefor

p | r + 1

. Thisproves

the Lemma.

Weare goingtoneed anoverview ofthe size of the

E 1

page ofthe Morse

spetral sequene.

Lemma 5.4. The Poinaré series

P (t)

of

E 1 (L(HP r ) hS 1 )

is given by 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 )(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. We

rst nd the Poinaréseries for

E 1 n,

.

• n = 0

: By Theorem 5.1, sine

E 1 0,

is afree

F 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 5.1

P (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 obtain

P (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 4and

τ

has degree

4r + 3

.

Wemustsum over

n ≥ 1

toalulate

P (E 1 )(t)

. Onlythe fator

t 4r(n 1)+2n+1

depends on

n

,so we sum that rst, inthe two ases

p | n

and

p ∤ 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

odd

E 1 p,q

, and likewisefor

E 1

even. Notie that

K(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 )

. Assume

p | r + 1

. Then

E 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

for

n ≥ 1.

Assume

p ∤ r + 1

. Then

E 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

for

n ≥ 1.

where

| x | = 4

,

| σ | = 4r − 1

,

| τ | = 4r + 3

,

| α n | = (4r + 2)n − 4r + 1

.

Thisspetralsequeneollapsesfromthe

E 1

page. Thisdetermines

H (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 the

equivariant 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 the

omputa-tions from the proof of Lemma 5.4. Consider the ase

p ∤ r + 1

. (The ase

p | r + 1

is similar.) In ltration

n > 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 map

E 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. For

n > 0

, the degree of the Thom lass

α n

isodd, soby the formula forthe

E 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 an

F 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 ) }

induedby

the energy ltration of

LHP r

ollapses from the

E 1

page by Theorem 5.6

above. This means that

E = E 1

. Comparing with the spetral sequene

{ E s ( F m ) }

of the nite ltration

F 0 ⊆ F 1 ⊆ · · · ⊆ F m

we see that its

E 1

pageisthesameas

E 1 ( M )(LHP r )

up toltration

m

. Sobynaturality,both

spetralsequenesollapsefromthe

E 1

page,and

E ( F m )

equals

E (LHP r )

up to ltration

m

. Sowe an alulatethe dimension of

H

odd

( F m )

asan

F 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 amultipliativeset

U ⊆ R

(i.e. if

u, v ∈ U

then

uv ∈ U

),we

dene

M

loalizedaway from

U

as

M [U 1 ] = n m

u | m ∈ M, u ∈ U o / ∼

where

m

u ∼ m u

if there is

v ∈ U

suh that

vu m = vum

. It is an elementary

algebrai fat that loalization away from

U ⊆ R

is an exat funtor on

R

-modules.

We are going to use

U = { u n | n ∈ N } ⊆ F p [u]

, where

u

as usually

de-notes our generator

u ∈ H 2 (BS 1 ; F p )

, suh that

H (BS 1 ; F p ) ∼ = F p [u]

. The

mainloalizationresulthereis[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 ltration

m

for

m ∈ N

.

Note: Thisimpliesthattheloalizedspetralsequene

E ( M )(LHP r hS 1 ) 1

u

ollapsesfromthe

E p

page, sine

E ( M )(LHP r )

ollapsesfromthe

E 1

page.

In document Afhandling (Sider 52-65)