Equivariant ohomology of spaes of geodesis

Usingourpreviousomputations(Theorems2.1and2.4)andSerre'sspetral

sequene,wewillbeabletoomputetheequivariantohomologyofthespae

of geodesis,

G(HP ^r ) ⁽ⁿ⁾

^.

Werst onsider the ase

p ∤ n

^{, sine this is the easiest. We show:}

Proposition 2.11. For

p ∤ n

^:

H ^m (BC n ; F p ) = 0,

^for

m > 0.

Proof. We are going to use that

EC n −→ BC n

^{is a overing, sine}

C n

^is

disrete. In general, given a

k

^{-sheet overing}

π : E −→ B

^(assume

B

on-neted), one an onstrut a so-alled transfer map. By baryentri

sub-division one knows that it is enough to onsider very small simplies in

B

^{. Therefore, given a simplex in}

B

^{we an assume it is ontained in a}

neighborhood

U

^{suh that}

π ^{− 1} (U )

^{is a disjoint union of open sets mapped}

homeomorphially to

U

^by

π

^{. Then we an pull the simplex in}

U

^bak

by

π

^{, yielding}

k

^{opies of the simplex in}

E

^{, whih we formally add,}

giv-ing a hain map

τ : C _m (B) −→ C _m (E)

^{. This indues the transfer map}

τ ^∗ : H ^m (E) −→ H ^m (B)

^{on ohomology. From the denition,}

π ♯ ◦ τ

^is

multipliation by

k

^{, and so}

τ ^∗ π ^∗

^{is also} multipliation by

k

^{. In our ase,}

EC n −→ BC n

^isan

n

^{-sheet overing, and sothe omposition}

H ^m (BC n ; F p ) ^τ

∗

−→ H ^m (EC n ; F p ) ^π

∗

−→ H ^m (BC n ; F p )

ismultipliationby

n

^{. Sine weare using}

F p

^-oeientsand

p ∤ n

^{, this isan}

isomorphism. On the other hand, for

m > 0

^{, the middle term is zero, sine}

EC n

^isontratible. Thus

H ^m (BC n ; F p ) = 0

^for

m > 0

^.

Withthis we an prove:

Theorem 2.12. For

p ∤ n

^{, the} equivariant ohomology with

F p

^oeients

of the

n

^{-twisted spae of geodesis on}

HP ^r

^is

H ^∗ ((G(HP ^r ) ⁽ⁿ⁾ ) hS ¹ ; F p ) ∼ = F p [x, t]/ h Q r , Q r+1 i ,

where

x

^{has degree}

2

^{, and}

t

^hasdegree

4

^{, and}

x

^{istheimage of the generator}

u ∈ H ² (BS ¹ )

^{under the map}

∆(HP ^r ) −→ BS ¹

^{in (9).}

Proof. We use the Serre's spetral sequene of the brationfrom Prop. 1.6:

BC n −→ ES ¹ × ^{S 1} G(r) ⁽ⁿ⁾ −→ ∆(r).

Proposition2.11 above now immediatelyimplies that

H ^∗ ((G(r) ⁽ⁿ⁾ ) _hS ¹ ; F p ) = H ^∗ (ES ¹ × S ¹ G(r) ⁽ⁿ⁾ ; F p ) ∼ = H ^∗ (∆(r); F p )

The ase

p | n

^{requires more work, and one needs to take into aount}

whether or not

p | r + 1

^{. But rst we need a omputationof}

H ^∗ (BC n ; F p )

^:

Proposition 2.13. For

p | n

^,

H ^∗ (BC n ; F p ) ∼ = F p [u, e]/

e ² .

Proof. Use Theorem 1.4

(i)

^{onthe bration}

S ¹ −→ ES ¹ −→ BS ¹

^{to divide}

out the ation of

C n ⊆ S ¹

^{, and obtaina bration}

S ¹ −→ BC n −→ BS ¹ .

⁽²¹⁾

Here wehave identied the quotientgroup

S ¹ /C n

^with

S ¹

^{itself viathe}

n

^th

powermap

z 7→ z ⁿ

^{. Wewillapply Serre'sspetral sequene.}

First, though, we will nd

H ¹ (BC n ; F p )

^{. Sine}

C n

^{is disrete,}

EC n −→

BC n

^{is the universal overing. From overing spae theory,}

π 1 (BC n ) ∼ = C n

^,

and sine this is abelian, it follows that

H ₁ (BC _n ; Z) ∼ = Z/nZ

^{. Using the}

Universal Coeient theorem, we an ompute

H ¹ (BC n ; F p )

^{. Note that}

H 0 (BC n ) = Z

^{, soExt}

(H 0 (BC n ), F p ) = 0

^{, and therefore,sine}

p | n

^:

H ¹ (BC n ; F p ) ∼ =

^Hom

(H 1 (BC n ), F p ) ∼ =

^Hom

(Z/nZ, Z/pZ) ∼ = F p ,

Now we turn to Serre's spetral sequene for the bration (21), with

E ₂ ^p,q = H ^p (BS ¹ ; H ^q (S ¹ ; F p )) = H ^p (BS ¹ , F p ) ⊗ H ^q (S ¹ ; F p )

^{. Note that the}

only possible non-trivial dierential is

d ²

^{, sine the}

E ²

^{page has only two}

non-zero rows. Knowing that

H ¹ (BC n ; F p ) ∼ = F p

^{, we onlude that the rst}

dierential

d ^0,1 ₂

^{must beamap}

F p −→ F p

^{withkernel isomorphito}

F p

^{. This}

fores

d ₂ (e) = 0

^,where

e

^generates

H(S ¹ ; F p )

^{. Usingthederivationproperty:}

d(eu ^j ) = d(e)u ^j ± ed(u ^j ) = 0.

So all dierentials are zero, the spetral sequene ollapses, and

E _∞ = E 2

^.

There are no extension problems, sine eah diagonal

p + q = ∗

^{ontains at}

most one non-zero group, so

H ^∗ (BC n ; F p ) = E _∞

^{,as desired.}

Theorem 2.14. Let

p

^{be a prime number and}

n ∈ N

^{suh that}

p | n

^{. As}

F p [u]

^{-modules,the following holds:}

(i)

^Suppose

p ∤ r + 1

^{. Then}

H ^∗ G(HP ^r ) ⁽ⁿ⁾

hS ¹ ; F p

∼ = F p [u]

1, y, y ² , . . . , y ^{r − 1} , τ, τ y, . . . , τ y ^{r − 1} . (ii)

^Suppose

p | r + 1

^{. Then}

H ^∗ G(HP ^r ) ⁽ⁿ⁾

hS ¹ ; F p ∼ = F p [u]

1, y, y ² , . . . , y ^r , σ, σy, . . . , σy ^r .

where

y

^{has degree}

4

^,

τ

^{has degree}

4r + 3

^and

σ

^{has degree}

4r − 1

^.

Proof. In the beginning, the proofs of the two ases are the same. Consider

the spetral sequene forthe brationfromProp. 1.6

G(r) −→ ES ¹ × S ¹ G(r) ⁽ⁿ⁾ −→ BS ¹ .

⁽²²⁾

Aording to our omputation of the ohomology of the bre in Theorem

2.1, neither the bre nor the base has anything in ohomology of degree 1.

This means that

H ¹ ((G(r) ⁽ⁿ⁾ ) _hS ¹ ) = 0

^{. We an use this when onsidering}

the spetral sequene forthe other brationfromProp. 1.6:

BC n −→ ES ¹ × S ¹ G(r) ⁽ⁿ⁾ −→ ∆(r).

Aording toProp. 2.13,

E ₂ ^q,s = H ^q (∆(r); H ^s (BC n ; F p )

^{looksas follows:}

3 ue uex uex ² , uet . . .

2 u ux ux ² , ut

1 e ex ex ² , et . . .

0 1 x x ² , t

0 1 2 3 4 · · ·

(23)

Let us denote the two lower rows of the

E 2

^{page by}

F

^{. Then the next two}

rows (rows 2and 3)onsists of

uF

^{,the next two are}

u ² F

^{, et. Consider the}

dierential

d ₂

^{as a map}

d ₂ : eH ^∗ (∆(r)) −→ H ^∗ (∆(r))

^{from row 1 to row}

0. Then, using the derivation property of the dierentials we see that

d 2

^is

multipliationwith

d 2 (e)

^{. When passingfrom the}

E 2

^tothe

E 3

^page,

F

^will

be replaed by two rows, Cok

d 2

^{and Ker}

d 2

^,

uF

^{willbereplaed by}

u

^Cok

d 2

and

u

^Ker

d 2

^{, et.}

So to determine the

E 3

^{page, we need to nd}

d 2 (e)

^{. As noted, the total}

spaehas

H ¹ = 0

^,so

d ^0,1 ₂ : E ₂ ^0,1 −→ E ₂ ^2,0

^{must bean injetivemap, henean}

isomorphism. Thisfores

d 2 (e) =

^unit

· x

^{;wemightaswellsay}

d 2 (e) = x

^{. So}

d 2

^ismultipliationby

x

^{,and we must determine Cok}

(x)

^{and Ker}

(x)

^{. Using}

Theorem 2.4, we see that

Cok

(x) ∼ = F p [x, t]/ h x, Q _r , Q _r+1 i ∼ = F p [t]/ h Q _r (0, t), Q _r+1 (0, t) i .

⁽²⁴⁾

Now by Lemma 2.7,

Q r (0, t) = (r + 1)t ^r

^and

Q r+1 (0, t) = (r + 2)t ^r+1

^{. This}

iswhere we must distinguish between the two ases.

Butletusrst investigatethe kernel. I havetriedtodiagramthe

dimen-sions of

F p [x, t]/ h Q _r , Q _r+1 i

^{using Remark 2.10, with boldfae indiatingthe}

degrees where, for dimension reasons, the kernel must be non-trivial. The

degrees are in the top row:

0 2 4 6 8 · · · ^{4r 4r+2 4r+4 4r+6 4r+8 4r+10 4r+12} · · ·

1 1 2 2 3 · · · r r r − 1 r − 1 r − 2 r − 2 r − 3 · · ·

Thepatternis(hopefully)lear: Theremustbeapartofthekernelindegrees

4(r + i) − 2

^for

i = 1, ..., r

^{. In partiular, the dimension is at least}

r

^{. Now,}

for the rest of the proof, we need tohandle the twoases separately.

Case

(i)

^:

p ∤ r + 1

^{. In this ase,}

r + 1

^{is a unit in}

F p

^{, so (24) beomes}

Cok

(x) ∼ = F p [t]/ h t ^r i

^{. In partiular, the dimension of Cok}

(x)

^is

r

^{, generated}

by

1, t, . . . , t ^{r − 1}

^.

Sine

dim

^Ker

(x) = dim

^Cok

(x) = r

^{,we have determined above that the}

kernel is indegrees

4(r + i) − 2

^for

i = 1, ..., r

^{. In eah degree, the kernel is}

one-dimensional,say generated by

ϕ i

^indegree

4(r + i) − 2

^{. Sowean write}

down the

E 3

^page:

3 uϕ ₁ uϕ ₂ · · · uϕ _r

2 u ut ut ² · · · ut ^{r − 1}

1 ϕ 1 ϕ 2 · · · ϕ r

0 1 t t ² · · · t ^{r − 1}

0 2 4 6 8 . . . 4r − 4 4r − 2 4r 4r+2 4r+4 4r+6 . . . 8r − 2

Beause there are no further dierentials on

t

^and

u

^{, and the} dierentials satisfy the derivation property, we see that the spetral sequene ollapses

from the

E ₃

^{page. Now let usompare this to the spetral sequene for the}

bration

G(r) −→ ES ¹ × S ¹ G(r) ⁽ⁿ⁾ −→ BS ¹

^{from (22) onsidered in the}

beginning, whihalso onverges to

H ^∗ ((G(r) ⁽ⁿ⁾ ) hS ¹ ; F p )

^{. Sine}

H ^∗ (G(r); F p ) ∼ = F p [y, τ ]/

y ^r = 0, τ ² = 0 ,

where

y

^{has degree}

4

^and

τ

^{has degree}

4r + 3

^{,we get the}

E 2

^page,

E ₂ ^{∗ , ∗} ∼ = F p [y, τ]/

y ^r = 0, τ ² = 0

⊗ F p [u]

Comparing this to the

E ₃

^{page above, we see that we have in eah ase}

2r

^{generators whih are multiplied by}

1

^,

u

^,

u ²

^{, et. This means, sine the}

rst spetral sequene ollapses, that this seond one must also ollapse.

Consequently we an read o that

H ^∗ (G(r) ⁽ⁿ⁾ _hS 1 ; F p )

^{as an}

F p [u]

^{-module is}

generated by

1, y, y ² , . . . , y ^{r − 1} , τ, τ y, . . . , τ y ^{r − 1}

Case

(ii)

^:

p | r + 1

^{. In this ase,}

r + 1

^{is zero in}

F p

^{, but}

r + 2

^{is aunit, so}

(24) beomes:

Cok

(x) ∼ = F p [t]/t ^r+1 .

In partiular, the dimension of Cok

(x)

^is

r + 1

^{,generated by}

1, t, . . . , t ^r

^.

Consequently,

dim

^Ker

(x) = r+1

^{,soweneedtondanadditionalelement}

inthe kernel. By Lemma2.7,

Q r

^{is the polynomial}

Q r = (r + 1)t ^r −

r + 2 r − 1

t ^{r − 1} x ² + · · · ± x ^2r ,

so

x

^divides

Q r

ⁱⁿ

F p [x, t]

^{. This means we have an element}

ϕ 0 = Q r /x

ⁱⁿ

degree

4r − 2

^{whih is in the kernel of}

x

^{. So together with the elements}

ϕ 1 , . . . , ϕ r

^{frombefore, we have found generators of the kernel.}

As in Case

(i)

^{, we see that the spetral sequene ollapses from the}

E 3

page. Comparingwiththe

E 2

^{pageofthe bration(22),andusingthatsine}

p | r + 1

^,

H ^∗ (G(r); F p ) ∼ = F p [y, σ]/

y ^r+1 = 0, σ ² = 0 ,

weonlude asabovethat

H ^∗ (G(r) ⁽ⁿ⁾ _hS 1 ; F p )

^asan

F p [u]

^{moduleisgenerated}

by

1, y, y ² , . . . , y ^r , σ, σy, . . . , σy ^r .

Corollary 2.15. For the Serre spetral sequene of the bration

G(HP ^r ) −→ G(HP ^r ) ⁽ⁿ⁾ _hS 1 −→ BS ¹

the following holds: If

p | n

^{, it ollapses from the}

E 2

^{page. If}

p ∤ n

^the

inlusion of the bre indues a surjetive map on even degree ohomology

H ^{2 ∗} (G(HP ^r ) ⁽ⁿ⁾ _hS 1 ; F p ) −→ H ^{2 ∗} (G(HP ^r ); F p )

Proof. Thease

p | n

^{follows diretly fromtheproof ofTheorem 2.14 above.}

Forthe ase

p ∤ n

^{, wemust hek thatthe lasses}

y ^j

^{fromTheorem 2.1are in}

the image of the inlusion of the bre. To do this, we onsider the

E 2

^page

of the spetral sequene, and must show that the lasses

y ^j

^{survive to}

E _∞

^.

Sine the dierentialsare derivations,

d _s (y ^j ) = jy ^{j − 1} d _s (y)

^{, and so it sues}

to show

y

^{survives. Clearly it does, sine any dierentialstarting at}

y

^ends

intotal degree

5

^{,and there are no}non-triviallasses in total degree

5

^.

3

K

^{-theory of spaes of geodesis in}

CP ^r

Let

G(r) = G(CP ^r )

^{be the spae of simple, losed,} parametrized geodesis in

CP ^r

^{, and let}

∆(r) = S ¹ \ G(r)

^{be the quotient spae under the rotation}

ation. In this hapter we obtain

K

^{-theoreti analogues of the results for}

ohomologyfromthe previous hapter.

By

K

^{-theory we mean omplex}

K

^{-theory, i.e.}

K ⁰ (X)

^{for aCW-omplex}

X

^{isthe groupompletionof the semi-groupofomplexvetor bundleswith}

base spae

X

^{. Dene}

K ^∗ (X)

^{for a general spae}

X

^{as follows: Chose any}

CW omplex

Y

^{weakly equivalent to}

X

^{, put}

K(X) = K(Y )

^{. This is well}

dened, sine two hoies of

Y

^{will be homotopy} equivalent, and

K

^-theory

is homotopy invariant. We most often employ the

Z/2Z

^{-grading from}

Bott-periodiity,writing

K ^∗ (X) = K ⁰ (X) ⊕ K ¹ (X)

^.

3.1 The unparametrized geodesis

Reall the model for

∆(r)

^{from the end of setion 1.2. We had}

γ 2

^{, the}

standard 2-dimensional bundle over the Grassmannian Gr

2 (C ^r+1 )

^and

p : P(γ 2 ) −→

^Gr

² (C ^r+1 )

^{the assoiated projetive bundle. Thenwe had a}

om-posite map (6),whih isan

S ¹

-equivariant dieomorphism

ϕ : ∆(r) −→ P(γ 2 )

Take the standard line bundle

γ 1

^over

P(γ 2 )

^{. The pullbak}

ϕ ^∗ (γ 1 )

^of

γ 1

under

ϕ

^{is a line bundle we willdenote}

X

^{. We onsider also the onjugate}

line bundle

γ ₁ ^⊥

^to

γ 1

^over

P(γ 2 )

^{, i.e.}

γ 1 ⊕ γ ₁ ^⊥ = p ^∗ γ 2

^{. The pullbak}

ϕ ^∗ (γ ₁ ^⊥ )

of this bundle to

∆(r)

^{we will denote}

Y

^{. In}

K ⁰ (∆(r))

^{we dene the lasses}

x = [X] − 1

^and

y = [Y ] − 1

^.

Theorem 3.1. Let

x, y ∈ K ⁰ (∆(r))

^{be the lasses dened above. Then}

K ⁰ (∆(r)) ∼ = Z[x, y ]/ h Q _r , Q _r+1 i ,

K ¹ (∆(r)) = 0,

where

Q s

^for

s ∈ N

^{is the} homogeneouspolynomial in

x, y

^{given by}

Q _s (x, y ) =

X s j=0

x ^j y ^{s − j} .

Note that thesepolynomialsare not the same isin the ohomologyase,

Proof. We apply the Atiyah-Hirzebruh spetralsequene, Theorem 1.3

H ^∗ (∆(r); K ^∗ ( ∗ )) ⇒ K ^∗ (∆(r)).

⁽²⁵⁾

Sineweknowthe ohomologyof

∆(r)

^from [Bökstedt-Ottosen ℄,

H ^∗ (∆(r)) ∼ = Z[x 1 , x ₂ ]/ h Q _r , Q _r+1 i ,

and

x 1 , x 2

^{have degree 2, we see that all} dierentials in (25) are trivial, so that

E _∞ = E ₂ ∼ = Z[x 1 , x ₂ ]/ h Q _r , Q _r+1 i ⊗ Z[β, β ^{− 1} ],

where

β

^{denotes the Bott element. This shows that}

K ¹ (∆(r)) = 0

^{, and}

K ⁰ (∆(r))

^{is free abelianof the same rankas}

H ^∗ (∆(r))

^.

Weuse the Chern harater,

h

: K ⁰ (X) −→ H ^∗ (X; Q),

whihisaring homomorphism. Byonstrution,

x 1 = c 1 (X)

^and

x 2 = c 1 (Y )

are the rst Chern lasses of

X

^and

Y

^{, f.} [Bökstedt-Ottosen℄ Thm. 3.2, so sine

X, Y

^{are linebundles, we get}

h

(x) =

^h

(X) − 1 = exp(c 1 (X)) − 1 = exp(x 1 ) − 1,

h

(y) =

^h

(Y ) − 1 = exp(c 1 (Y )) − 1 = exp(x 2 ) − 1.

ThereisarelationbetweentheChernharaterhandtheAtiyah-Hirzebruh

spetral sequene, by [Atiyah-Hirzebruh℄ Cor. 2.5. We see that h

(x ⁱ y ^j ) =

h

(x) ⁱ

^h

(y) ^j = x ⁱ ₁ x ^j ₂ +

^{higher terms, where higher terms means terms in}

higher ltration, whih in this ase is equivalent to higher total degree in

x 1

^,

x 2

^{. By}

(iii)

^{in the orollary, this shows that the ring} homomorphism

Z[x, y ] −→ K ^∗ (∆(r))

^issurjetive.

This means we an use

x, y

^{as polynomial generators for}

K ^∗ (X)

^{, and it}

remains todetermine the relations. Againwe use the Chern harater, this

time aftertensoring with

Q

^:

h

: K ⁰ (X) ⊗ Q −→ H ^∗ (X, Q)

whih is then a ring isomorphism. We now want to prove that h

(x)

^and

h

(y)

^{satisfy the relations}

Q r , Q r+1

^{. Ifwe anprovethis, we are done: Sine}

theChernharaterisanisomorphismaftertensoringwith

Q

^{,andthegroups}

are torsion-free, there an beno furtherrelationsin

K ⁰ (S(τ)/S ¹ )

^{,sine this}

has the same rank as

H ^∗ (S(τ)/S ¹ ) ∼ = Z[x 1 , x 2 ]/ h Q r , Q r+1 i

^.

Soweneedtoprovethat

Q s (exp(x 1 ) − 1, exp(x 2 ) − 1) = 0

^if

Q s (x 1 , x 2 ) = 0

for

s = r, r + 1

^{. Realling that the ideals}

h Q r , Q r+1 i

^and

Q r , x ^r+1 ₁ , x ^r+1 ₂

oinide, we rst get that

(exp(x _i ) − 1) ^r+1 = x ^r+1 _i (1 +

^higherterms

) = 0

^.

Consider the quotient map

R = Q[x 1 , x 2 ]/

x ^r+1 ₁ , x ^r+1 ₂

−→ Q[x 1 , x 2 ]/ h Q r , Q r+1 i = S,

whih has kernel

I = h Q _r i

^{. Given a power series without onstant term,}

g(z) = a 1 z + a 2 z ² + · · ·

^{, we an dene}

g _∗ : R −→ R

^by

x i 7→ g(x i )

^for

i = 1, 2

^{. In our ase,}

g(z) = exp(z) − 1

^{. If we an prove that}

g _∗ I ⊆ I

^{, the}

map

g _∗

^{will bewell-dened as a map}

S −→ S

^{, as shown below:}

0 ^// I ^//

g ∗

R ^//

g ∗

S ^//

0 0 ^// I ^// R ^// S ^// 0

Wewillshow

I =

^Ker

(x 1 − x 2 )

^{. Consider a} homogeneouspolynomial

f ∈ R

of degree

m

^{. It sues totake}

m ≥ r

^,forif

f

^{had lowerdegree, itouldnot}

be in

I = h Q r i

^{, sine}

Q r

^{has degree}

r

^{. Then, using}

x ^r+1 ₁ = x ^r+1 ₂ = 0

^{, wean}

write

f = X r i=m − r

c i x ⁱ ₁ x ^m ₂ ^{− i} ⇒ (x 1 − x 2 )f = X r i=m − r+1

(c i − 1 − c i )x ⁱ ₁ x ^m ₂ ^{− i} .

By[Bökstedt-Ottosen ℄ Lemma 3.4,

f ∈ I

^{if and only if}

c m − r = . . . = c r

^{, and}

we onlude

I =

^Ker

(x 1 − x 2 )

^{. This implies}

g _∗ I ⊆

^Ker

(g _∗ x 1 − g _∗ x 2 )

^{. So we}

alulate

g _∗ x 1 − g _∗ x 2 = X

i ≥ 1

a i (x ⁱ ₁ − x ⁱ ₂ ) = (x 1 − x 2 ) X

i

a i i − 1

X

k=0

x ^k ₁ x ⁱ ₂ ^{− k − 1}

! .

This shows

g _∗ I ⊆

^Ker

(g _∗ x ₁ − g _∗ x ₂ ) ⊆

^Ker

(x ₁ − x ₂ ) = I

^{,as desired.}

Remark 3.2. Let

M = K ^∗ (∆(r)) = Z[x, y ]/ h Q r , Q r+1 i

^{. We often use}

ltration arguments, so let us x the notation now. Let

M _j ⊆ M

^be

the group generated by monomials in

x, y

^{of total degree at least}

j

^{, i.e.}

M j = Z[x, y ] _≥ j / h Q r , Q r+1 i

^{. This makes sense sine}

Q r

^,

Q r+1

^are

homoge-neous. Then

0 = M 2r ⊆ M 2r − 1 ⊆ · · · ⊆ M 1 ⊆ M 0 = M

^{is a ltration of}

M

^.

In document Afhandling (Sider 30-38)