Equivariant K -theory of spaes of geodesis

Theorem 3.3 (Atiyah). Let

G

^{be a ompat Lie group. Then}

(i) K ⁰ (BG) ∼ = R(G) [ _I

^,

(ii) K ¹ (BG) = 0

^.

I will nowuse this theorem to determine

K ^∗ (BS ¹ )

^and

K ^∗ (BC n )

^.

Lemma 3.4. Let

T : S ¹ ֒ → C ^∗

^{be the natural} 1-dimensional representation of

S ¹

^{, and let}

t = [T ] − 1 ∈ K ⁰ (BS ¹ )

^{. Then}

R(S ¹ ) = Z[T, T ^{− 1} ], I = h T − 1 i , K ⁰ (BS ¹ ) ∼ = R(S \ ¹ ) _I = Z[[t]].

Proof. Firstnotethatarepresentation

ρ : S ¹ −→ GL n (C)

^{anbeonjugated}

to

ρ : S ¹ −→ U(n)

^{, by hoosing an inner produt on}

C ⁿ

^{(all of whih are}

onjugate) whih is

S ¹

-invariant. So it sues to look at representations

ρ : S ¹ −→ U (n)

^{. Now}

ρ(t) ∈ U (n)

^(for

t ∈ [0, 2π]

^{) is} diagonizable,

ρ(t) ∼

diag

(e ^{iθ 1} , . . . , e ^{iθ n} )

^{. This also} diagonalizes

ρ(kt)

^,

k ≥ 1

^{, so if we hoose}

t

rationallyindependent of

π

^{, this} diagonalizationworks fora dense subset of

S ¹

^{. Soby ontinuity we an} diagonalize

ρ(t)

^{for all}

t

simultaneously, and so

ρ

^{isgivenbydiag}

(ρ 1 (t), . . . , ρ n (t))

^,where

ρ k : S ¹ −→ S ¹

^isahomomorphism.

This means

ρ k (z) = z ^{m k} , m k ∈ Z

^{. Using the natural} representation

T : z 7→

z

^{,anditsinverse}

T ^{− 1} : z 7→ z ^{− 1}

^,weanreformulatethisbysayingthatevery representation of

S ¹

^{has the form}

P N

i= − N n i T ⁱ

^,

n i ≥ 0

^{. The} Groethendiek onstrution yields

R(S ¹ ) = ( _N

X

i= − N

n i T ⁱ | n i ∈ Z )

= Z[T, T ^{− 1} ].

Now tothe augmentationideal. By denition

I = ( _N

X

i= − N

n i T ⁱ | X N i= − N

n i = 0 )

.

Clearly,

T − 1 ∈ I

^{, and also,}

P N

i= − N n i T ⁱ ∈ I

^{is divisible by}

T − 1

^{, beause}

the sum of the oeients is zero. So

I = h T − 1 i

^{. Now}

I ^k =

(T − 1) ^k

,

and

R(S ¹ )/I ^k

^{hasgenerators}

1, T − 1, (T − 1) ² , . . . , (T − 1) ^{k − 1}

^. Consequently, putting

t = [T ] − 1

^{,we get}

K ⁰ (BS ¹ ) ∼ = R(S \ ¹ ) _I = Z[[t]].

Lemma3.5. Let

n ∈ N

^{be anumber withprime} fatorisation

n = Q

p | n p ^i(p)

^.

Then

K ⁰ (BC n ) ∼ = Z ⊕ M

p | n

(ˆ Z p ) ^{p i(p) − 1} ,

where

Z ˆ p

^{denotes the}

p

^{-adi integers.}

Proof. Let

W

^{be the natural} 1-dimensional representation of

C n ⊆ C ^∗

^{. As}

in the proof of Lemma 3.4 above, we only need look at representations

ρ : C n −→ U (m)

^and diagonalize, so that

ρ =

^diag

(ρ 1 , . . . , ρ m )

^{. Here eah}

ρ j : C n −→ S ¹

^{is a group} homomorphism, and so is a power of

W

^{, with}

the relation

W ⁿ = 1

^. Consequently

R(C _n ) = Z[W ]/ h W ⁿ − 1 i

^{. The}

aug-mentation ideal is

I = h W − 1 i

^{for the same reason as before, and we must}

omputetheinverse limit

lim

←− k

R(C n )/I ^k

^{. This wepropose todointwosteps:}

First assume

n = p ⁱ

^{. Then}

C _p ⁱ

^{is a}

p

^{-group, and aording to [Atiyah2℄}

the

I

^{-adi and}

p

^{-adi topologieson}

I = I(C _p ⁱ )

^are equivalent,so that

K ⁰ (BC p ⁱ ) ∼ = R(C \ p ⁱ ) _I = Z ⊕ I \ (C p ⁱ ) _I ∼ = Z ⊕ I \ (C p ⁱ ) _p .

To alulate this, let

w = W − 1

^{, and note that}

I(C _p ⁱ ) = h w i

^{in the ring}

Z[w]/ D

(w + 1) ^{p i} = 1 E

,and so

I(C p ⁱ ) ∼ = Z ^{p i − 1}

^{. Thus}

I \ (C p ⁱ ) _p ∼ = (ˆ Z p ) ^{p i − 1}

^.

Now take any

n ∈ N

^{. Observe that}

C _p ^i(p)

^{, where}

n = p ^i(p) m

^with

gcd(p, m) = 1

^{, are exatly the Sylow}

p

^{subgroups of}

C n

^{. Then by [Atiyah2℄}

Prop. 4.10, there isaninjetivemap

K ⁰ (BC n ) −→ M

p | n

K ⁰ (BC _p ^i(p) ),

and inpartiular

I(C \ n ) _{I(C n )} −→ M

p | n

I(C \ _p ^i(p) ) _I(C

pi (p) )

is injetive. By using that

C n ∼ = Q

p | n C _p ^i(p)

^{by the Chinese Remainder}

The-orem, itis easilyseen that this map is an isomorphism,sothat

K ⁰ (BC n ) ∼ = Z ⊕ M

p | n

I(C \ _p ^i(p) ) ∼ = Z ⊕ M

p | n

(ˆ Z p ) ^{p i(p) − 1} ,

by the result for

p ⁱ

^above.

Withthese results, letusrst take alook atthe

K ^∗ (BS ¹ )

^-module

stru-ture on

K ^∗ (X _hS ¹ )

^{, where}

X

^isan

S ¹

^{-spae,as desribed in Setion1.4. F}

ol-lowing the notation in Lemma 3.4, we have the anonial representation

T

of

S ¹

^{, whih by (27) gives a bundle over}

BS ¹

^{, whih we also all}

T

^{. On}

K

-theory,

T

^{denes alassin}

K ^∗ (BS ¹ )

^,and

K ^∗ (BS ¹ ) = Z[[t]]

^,where

t = T − 1

^.

Usingthe projetion pr

1 : X hS ¹ −→ BS ¹

^{,weget lasses pr}

^∗ ₁ (T )

^{and pr}

^∗ ₁ (t)

ⁱⁿ

K ^∗ (X _hS ¹ )

^{. We willsuppress the map pr}

₁

^{from the notation, and simplyall}

these lasses

T

^and

t

^again.

Weannowdeterminethe

K ^∗ (BS ¹ )

^{modulestrutureon}

∆(r) ≃ G(r) _hS ¹

^:

Lemma 3.6. The

K ^∗ (BS ¹ ) = Z[[t]]

^{module struture on}

K(∆(r))

^{is given}

by

t 7→ (x − y)/(y + 1)

^{. In partiular,}

t ^2r

^{ats as}

0

^.

Proof. We use the results from ohomology, where the

H ^∗ (BS ¹ ) = Z[u]

module struture on

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

^{is given by}

u 7→

x 1 − x 2

^,f. [Bökstedt-Ottosen℄Cor. 3.7. Reallthat

x = [X] − 1

^,

y = [Y ] − 1

^,

where

x 1 = c 1 (X)

^and

x 2 = c 1 (Y )

^{aretherstChernlasses. Also}

u = c 1 (T )

^.

The rst Chern lass gives a group isomorphism fromomplex linebundles

over

∆(r)

^to

H ² (∆(r))

^{, sosine}

c 1 (T ⊗ Y ) = c 1 (T ) + c 1 (Y ) = u + x 2 = x 1 = c 1 (X ).

we get

T ⊗ Y = X

^{. Then we alulate}

(T − 1) ⊗ (Y − 1) = T ⊗ Y − Y − T + 1 = (X − 1) − (Y − 1) − (T − 1)

Isolating

T − 1

^gives

(T − 1) = ((X − 1) − (Y − 1)) ⊗ Y ^{− 1} .

In

K ^∗ (∆(r))

^{this equality gives}

t = (x − y)(y + 1) ^{− 1}

^{, as desired. Sine in}

K(∆(r)) ∼ = Z[x, y]/ h Q r , Q r+1 i

^{all non-zero elements have a total degree in}

x, y

^{whih is less than}

2r

^{, we see that}

t ^2r = (x − y) ^2r (y + 1) ^{− 2r} = 0

^.

Now we prove the main Theorem of this setion, but rst we introdue

a bit of notation: We write

K _hS ^∗ 1 (X)

^for

K ^∗ (ES ¹ × S ¹ X)

^{, when}

X

^{is an}

S ¹

^{-spae. Reall the diagram(26)}

K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ ) ^oo K ^∗ (∆(r))

K ^∗ (BS ¹ )

OO

K ^∗ (BS ¹ )

B P ⁿ

oo OO

This gives amap

K ^∗ (BS ¹ ) ⁽ⁿ⁾ ⊗ K ^∗ (BS ¹ ) K ^∗ (∆(r)) −→ K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ )

where the

K ^∗ (BS ¹ ) ⁽ⁿ⁾

^{denotes that the map}

B P ⁿ

^{should be applied in the}

tensor produt, as the diagramindiates.

Theorem 3.7. Let

n ∈ N

^{. Then the map}

K ^∗ (BS ¹ ) ⁽ⁿ⁾ ⊗ R(S ¹ ) K ^∗ (∆(r)) −→ K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ )

isan isomorphism of rings. In partiular,

K _hS ¹ 1 (G(r) ⁽ⁿ⁾ ) = 0

^.

Tox the notationand avoidlong, umbersome expressions, put

R = R(S ¹ ) = Z[U, U ^{− 1} ], R ˆ = K ⁰ (BS ¹ ) = Z[[u]], u = U − 1.

S = R(S ¹ ) = Z[T, T ^{− 1} ], S ˆ = K ⁰ (BS ¹ ) = Z[[t]], t = T − 1.

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

Here

S

^isan

R

^{-module by the map}

U 7→ T ⁿ

^{, and likewise}

S ˆ

^{is an}

R ˆ

^-module

by

u 7→ (t +1) ⁿ − 1

^{. ByLemma3.6,}

M

^isan

R ˆ

^-moduleby

u 7→ (x − y)/(1+y)

^,

and thusan

R

^{-module by}

U 7→ (x − y)/(1 + y) + 1

^.

The Theorem says that

S ˆ ⁽ⁿ⁾ ⊗ ^R M ∼ = K hS ¹ (G(r) ⁽ⁿ⁾ )

^{. The reason for}

restriting to

R

^{instead of}

R ˆ

^{is given by the following lemma, whih also}

shows that for the isomorphism, this restrition does not matter.

Lemma 3.8.

S ˆ

^{is a at}

R

^{-module, and}

S ˆ ⊗ R ^ˆ N ∼ = ˆ S ⊗ ^R N.

forany nitely generated

R ˆ

^-module

N

^where

u ^m

^{ats as0 on}

N

^{for some}

m

^.

In partiular this holds for the ltration modules

M j

^{from Remark 3.2, for}

M = M _2r+1

^{, and for the quotients}

M _j /M _j+1

^.

Proof. Clearly,

S

^{is a free}

R

^{-module (with basis}

{ 1, U, . . . , U ^{n − 1} }

^{), so}

S

^is

at over

R

^{. Sine}

S

^isNoetherian,

S ˆ

^{isat over}

S

^,see [Atiyah-MaDonald℄, Prop. 10.14. By the natural isomorphism,for any

R

^-module

M

^,

S ˆ ⊗ R M ∼ = ˆ S ⊗ S S ⊗ R M,

wesee that

S ˆ

^{isat over}

R

^.

Take

N

^{as inthe lemma. Then the ompletion by the ideal}

I = h u i ⊆ R ˆ

gives

N ˆ = lim _←

k

N/u ^k N = N.

Also by [Atiyah-MaDonald℄, Prop. 10.13, sine

R

^{is Noetherian and}

N

is nitely generated,

N ˆ ∼ = ˆ R ⊗ ^R N

^{. Combining these two fats yields the}

isomorphism

S ˆ ⊗ R ^ˆ N ∼ = ˆ S ⊗ R ^ˆ N ˆ ∼ = ˆ S ⊗ R ^ˆ ( ˆ R ⊗ ^R N ) ∼ = ˆ S ⊗ ^R N.

Now onsider the

R ˆ

^-module

M j

^{. Sine}

u

^{ats as}

(x − y)/(1 + y)

^{, and}

M j

onsistsofpolynomialsdegreeatleast

j

^,

u ^2r+1

^{atsaszero. Forthequotient}

M j /M j+1

^,

u

^{itself ats as zero. So the} requirements of

N

^{holds for these}

modules.

We will use the ltration

M j

^of

M

^{to prove the Theorem, so we need a}

Lemma whih proves the Theorem in the ase

M = Z

^:

Lemma 3.9. The followingmap is an isomorphism:

K ^∗ (BS ¹ ) ⊗ R(S ¹ ) Z −→ K ^∗ (BC n ).

Proof. Let

A = S ⊗ ^R Z

^and

B = ˆ S ⊗ ^R Z

^{, and let}

A −→ B

^{be the map}

indued by the ompletion

S −→ S ˆ

^{. We now deneanother map}

A −→ R(C _n ) = Z[W ]/ h W ⁿ − 1 i , T 7→ W.

This is learly an isomorphism,and preserves the augmentation ideal.

Con-sider the diagram:

B

A

oo ^{∼ =} //

R(C _n )

ˆ

B ^oo A ˆ ^{∼ = //} K ⁰ (BC n )

Herethe vertialarrowsdenoteompletionwithrespettotheaugmentation

ideals; respetively

tB

^,

(T − 1)A

^{, and}

h W − 1 i )

^{. To prove the Lemma, we}

must show

B ∼ = ˆ A

^{. Firstnotethat}

A ˆ −→ B ˆ

^isanisomorphism,sineforany

k

^{,the map given by}

T 7→ t + 1

^{, is an}isomorphism:

A/(T − 1) ^k = Z[T ]/

T ⁿ − 1, (T − 1) ^k

−→ Z[t]/

(t + 1) ⁿ − 1, t ^k

= B/t ^k B.

Next we show that

B −→ B ˆ

^{is an}isomorphism. To show this, onsider the exat sequene given by multipliationby

u − 1 ∈ R

^,

0 ^// R ^{u − 1 //} R ^// Z ^// 0 .

Sine

S ˆ

^{is at over}

R

^{, we obtain anew exat sequene,}

0 ^// S ˆ ⊗ ^R R ^{1 ⊗ (u − 1) //} S ˆ ⊗ ^R R ^// S ˆ ⊗ ^R Z ^// 0 ,

whih, afterapplying the natural isomorphism,beomes

0 ^// S ˆ ^(t+1)

n − 1

// ˆ S ^// ˆ S ⊗ R Z ^// 0 .

⁽²⁸⁾

Completing this with respet to the ideal

h t i

^{, whih is an exat funtor, we}

obtainyet another exat sequene

0 ^// lim

←

S/ ˆ

t ^k (t+1) ⁿ − 1

// lim

←

S/ ˆ t ^k

// lim

← ( ˆ S ⊗ ^R Z)/

t ^k

// 0 .

Reall

S ˆ = Z[[t]]

^{. After applying the} isomorphism

lim _← S/ ˆ

t ^k ∼ = ˆ S

^{, we get}

the exat sequene,

0 ^// S ˆ ^(t+1)

n − 1

// ˆ S ^// lim _← ( ˆ S ⊗ R Z)/

t ^k

// 0 .

⁽²⁹⁾

Comparing(28) and (29), we see that

B ∼ = ˆ B

^{. Asalready noted, this means}

that

A ˆ ∼ = B

^{, and this provesthe result.}

Now we an provethe mainTheorem 3.7:

Proof of Theorem3.7. Firstwe laim that the map

K ^∗ (BS ¹ ) ⊗ ^Z K _hS ^∗ 1 (G(r)) −→ K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ )

⁽³⁰⁾

issurjetive. Tosee this, werst note that the map

K ^∗ (BC n ) −→ K ^∗ (BS ¹ )

issurjetive. Thisfollowsfromthefatthatthe mapof representationrings,

R(C n ) −→ R(S ¹ )

^{is surjetive, sine any} representation of

C n

^{an be}

ex-tended toarepresentationof

S ¹

^{. Nowtoprove} surjetivity of (30), weuse a ltrationargumentin the spetral sequene

H ^∗ (∆(r); K ^∗ (BC _n )) ⇒ K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ ).

This ollapses, sine everything sits in even degrees. As in the proof of

Theorem 3.1, we now use Cor. 2.5 of [Atiyah-Hirzebruh℄, so let

A

^denote

the image of

K ^∗ (BS ¹ ) ⊗ Z K _hS ^∗ 1 (G(r))

ⁱⁿ

K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ )

^{. In ltration degree}

0 we have

K ^∗ (BC _n )

^{. As already shown}

K ^∗ (BS ¹ )

^{is surjetive onto this,}

so the lowest ltration an be hit. Anything else in

H ^∗ (∆(r); K ^∗ (BC n ))

^is

generatedby monomials

x ⁱ ₁ x ^j ₂

^{, andwe have}

x ⁱ y ^j ∈ A

^withh

(x ⁱ y ^j ) = x ⁱ ₁ x ^j ₂ +

higherterms. This shows that

A = K _hS ^∗ 1 (G(r) ⁽ⁿ⁾ )

^{,so (30) is surjetive.}

Nowweshowthatthemapisinjetive. Wewillusealtrationargument,

where we lter

M = K ⁰ (S(τ )/S ¹ )

^{as in Remark 3.2. We look at the exat}

sequene,

0 −→ M i+1 −→ M i −→ M i /M i+1 −→ 0.

As

S ˆ

^{isat over}

R

^{by Lemma3.8, weget the exat sequene}

0 −→ S ˆ ⊗ R M i+1 −→ S ˆ ⊗ R M i −→ S ˆ ⊗ R M i /M i+1 −→ 0.

⁽³¹⁾

Werstapplythisto

K

^-theorywith

F p = Z/pZ

^{oeients. Forthe eld}

F p

^{,wehavebytheUniversalCoeientTheorem,}

K ^∗ (X; F p ) ∼ = K ^∗ (X) ⊗ F p

^.

Clearly the ltration

M _i ^′ = M i ⊗ F p

^{works for}

F p

^{oeients, so we an}

use the result above. But sine

F p

^{is a eld, the exat sequene (31) splits,}

so we an do a ounting argument quite easily. Observe that

M _i ^′ /M _i+1 ^′ = F p [x, y] i / h Q r , Q r − 1 i = (F p ) ^{n i}

^{, where}

n i ∈ N

^{. By Lemma3.9, we know}

S ˆ ⊗ ^R M _i ^′ /M _i+1 ^′ ∼ = (K ⁰ (BC n ; F p )) ^{n i} .

⁽³²⁾

and in addition,

K ⁰ (BC n ; F p )

^{is a nite numberof opies of}

F p

^{, so it makes}

sense to ount them. Also

M _2r ^′ _{− 1} = F p

^{, so}

S ˆ ⊗ R M _2r ^′ _{− 1} ∼ = K ⁰ (BC n ; F p )

^{. So}

indutively, sine

M ⊗ F p

^{is a graded ring with a total of}

r(r + 1)

^{opies of}

F p

^{, then}

S ˆ ⊗ R M ⊗ F p ∼ = (K ⁰ (BC n ; F p )) ^r(r+1) .

Weompare this with

K ^∗ (G(r) ⁽ⁿ⁾ ; F p )

^{via the spetralsequene for the}

ver-tial brationin Prop. 1.6:

E 2 = H ^∗ (∆(r); K ^∗ (BC n ; F p )) ⇒ K ^∗ (G(r) ⁽ⁿ⁾ ; F p ).

Weseeeverythingsitsinevendegreesin

E 2

^{,sothereareno}dierentials,and, workingoveraeld

F p

^{,weansimplyountthedimensionof}

K ⁰ (G(r) ⁽ⁿ⁾ ; F p )

as the sum of the dimensions of

E ₂ ^m,n

^{on the diagonal}

m + n = 0

^{. Sine}

H ^∗ (∆(r); F p ) ∼ = F p [x, y]/ h Q r , Q r+1 i

^{also has a total of}

r(r + 1)

^{opies of}

F p

^,

again by Lemma 3.5, we get,

K ⁰ (G(r) ⁿ ; F p ) ∼ = (K ⁰ (BC n ; F p )) ^r(r+1) .

Sothe map of

F p

^{-vetor spaes}

S ˆ ⊗ ^R M ⊗ Z p = K ⁰ (BS ¹ ) ⊗ R(S ¹ ) K ⁰ (S(τ)/S ¹ ; Z p ) −→ K ⁰ (S(τ) ^(p) _hS 1 ; Z p )

isasurjetion between spaesof the samedimension, and isthusan

isomor-phism, and this holds for every prime number

p

^.

Nowwe ompare

Z

^{- and}

F p

^{-oeients (for aprime}

p

^{)by the diagram}

K ⁰ (∆(r)) ^// K ⁰ (G(r) ⁽ⁿ⁾ ; F p )

S ˆ ⊗ R M ^//

ϕ

OO

S ˆ ⊗ R M ⊗ F p

∼ =

OO

(33)

Assume

a ∈ S ˆ ⊗ R F

^{isin the kernel of}

ϕ

^{. Then, by the diagram,}

a

^redued

mod

p

^{iszero, so}

a = p · a 1

^{for some}

a 1

^{. But then, sine}

K ⁰ (∆(r))

^{is torsion}

free,

a 1 ∈ ker(ϕ)

^{, so}

a 1 = p · a 2

^{, et.} Consequently, if

a ∈ ker(ϕ)

^{, then}

a

is divisible by

p

^{innitely often. Reall that this holds for any prime}

p

^{, and}

thusalso for

n

^{, so}

a

^{is innitelyoften divisible by}

n

^.

Now take a look atthe ltration again

0 −→ S ˆ ⊗ R M _{i − 1} −→ S ˆ ⊗ R M _i −→ S ˆ ⊗ R M _i /M _{i − 1} −→ 0.

⁽³⁴⁾

If

a ∈ S ˆ ⊗ ^R M i

^{isdivisible by}

n

^{innitelyoften, then the image in}

S ˆ ⊗ R M i /M i − 1 ∼ = Z ^N ⊕ M

p | n

(ˆ Z p ) ^{N p}

is zero (the isomorphism is Lemma 3.5 and Lemma 3.9). So

a

^{omes from}

a ^′

ⁱⁿ

S ˆ ⊗ R M i − 1

^and

a ^′

^{is also innitely oftendivisible by}

n

^{. Soindutively}

a

^{omes from}

a 0 ∈ S ˆ ⊗ R F 0 ∼ = Z ⊕ L

p | n (ˆ Z p ) ^{p i − 1}

^{, and}

a 0

^{is divisible by}

n

innitelyoften, and so

a 0 = 0

^{, whih implies}

a = 0

^.

This shows that the kernel of

ϕ

^{is zero,and thusthe map}

ϕ : K ⁰ (BS ¹ ) ⊗ R(S ¹ ) K ⁰ (S(τ )/S ¹ ) −→ K ⁰ (S(τ) ^(p) _hS 1 )

⁽³⁵⁾

isan isomorphism.

4 The free loop spae and Morse theory

Now we turn to study the free loop spae

L(FP ^r )

^{, where as usual}

F = C

^or

F = H

^{. First a denition:}

Denition 4.1. Let

X

^{be a topologialspae. The spae}

LX = { f : [0, 1] −→ X | f(0) = f (1), f

^isontinuous

} ,

with the ompat-open topology,is alled the free loopspae of

X

^.

We are going to use Morse theory to study

LM

^{for a smooth manifold}

M

^{, where we will take}

M = FP ^r

^{. It is a fat that it does not hange the}

homotopy typeof

LM

^{if we requireall}

f ∈ LM

^tobe dierentiable,or even smooth, sowe do that.

Now let us onsider how one ould do Morse theory on the free loop

spae

LM

^{as well the spae of homotopy orbits}

LM hS ¹

^{, where}

M

^denotes

a ompat

n

-dimensional manifold. For details, I refer to [Klilngenberg1℄, and [Bökstedt-Ottosen ℄, espeially hapters 7 and 8.

LM

^{is not a}

nite-dimensionalmanifold,but one an makeamodelof

LM

^{whihisaso-alled}

Hilbert manifold, f. [Klilngenberg1℄ Ÿ1.2, meaning there are harts on

LM

making it loally homeomorphi to a Hilbert spae. The tangent spae of

a loop

f ∈ LM

^{is the spae}

Γ(f )

^{of vetor elds along}

f

^{. Let}

h· , ·i

^denote

the Riemannian metri on

M

^{. Now the tangent spae}

T f LM

^{arries the}

struture of a Hilbert spae via

h ξ, η i c = Z

S ¹

h ξ(t), η(t) i + c h∇ ξ(t), ∇ η(t) i

dt,

⁽³⁶⁾

where

ξ, η ∈ T f LM

^{are vetor elds along}

f

ⁱⁿ

LM

^{, and}

∇

^{denotes the}

ovariant derivative along

f

^{. The onstant}

c ∈ R

^{makes the inner produt}

vary. This isneessary toensurethat the

n

^{-folditerationmap,}

P n

^{, beomes}

anisometry

P n ^∗ = D f ( P n ) : T f LM −→ T _P n f LM, P n ^∗ (ξ(z)) = ξ(z ⁿ )

sine

hP n ^∗ ξ, P n ^∗ η i 1 = h ξ, η i n ²

^,see [Bökstedt-Ottosen ℄ Ÿ7.

Weare going todo Morse theory via the energy funtion

E : LM −→ R, f 7→

Z

S ¹ | f ^′ (t) | ² dt.

Foreah

a ∈ R

^{, weset}

F (a) = E ^{− 1} (] − ∞ , a]) ⊆ LM

^{. The ritial points of}

E

^{are the losed geodesis on}

M

^{. We shall assume that the ritial points}

are olleted on ompat submanifolds, eah of whih satisfy the Bott

non-degeneray ondition. This strong ondition is needed for the Morse theory

mahinery,anditissatisedfor

M = FP ^r

^{,andmoregenerallyforsymmetri}

spaes, [Ziller℄. Callthe ritial values

0 = λ ₀ < λ ₁ < . . .

^{, and onsider the}

ltration

F (λ 0 ) ⊆ F (λ 1 ) ⊆ · · · ⊆ LM.

⁽³⁷⁾

This ltration is equivariant with respet to the

S ¹

^{ation. This means it}

induesa ltrationof

LM hS ¹

^,

F (λ ₀ ) _hS ¹ ⊆ F (λ ₁ ) _hS ¹ ⊆ · · · ⊆ LM _hS ¹ .

⁽³⁸⁾

Thenon-degeneray ondition ensurethat eahritialsubmanifold

N (λ) = E ^{− 1} (λ)

^isnite-dimensional,and the tangent bundle

T (LM ) | ^N(λ) ⊆ T (LM )

splits

S ¹

-equivariantly:

T (LM) | N (λ) ∼ = µ ⁻ (λ) ⊕ µ ⁰ (λ) ⊕ µ ⁺ (λ),

intothe bundles of negative, zero-,and positivediretions, respetively, and

the negativebundle

µ ⁻ (λ)

^is nite-dimensional. To ease the notation, write

F ⁿ = F (λ n )

^and

µ ⁻ _n = µ ⁻ (λ n )

^{. The main result of Morse theory in this}

setting is proved by Klingenberg in [Klilngenberg1℄, Ÿ2.4: There is an

S ¹

-equivariant homotopy equivalene

F ⁿ )/ F ⁿ − 1 ≃ T h(µ ⁻ _n ).

⁽³⁹⁾

We want a similar result for

(LM) _hS ¹

^{, so we onsider the quotients of the}

ltration(38):

ES ¹ × S ¹ F n /ES ¹ × S ¹ F n − 1 ∼ = ES ₊ ¹ ∧ S ¹ F n / F n − 1 ,

where

ES ₊ ¹ ∧ S ¹ X = (ES ₊ ¹ ∧ X)/S ¹

^{is the smash produt modded out by}

the diagonal

S ¹

^{ation. The obvious map dened on} representatives is a homeomorphism. Thus by the Morse theorem in(39),

ES ¹ × S ¹ F ⁿ /ES ¹ × S ¹ F ⁿ − 1 ≃ ES ₊ ¹ ∧ S ¹ T h(µ ⁻ _n ).

We an use [Bökstedt-Ottosen ℄ Lemma 5.1tond that

( F ⁿ ) hS ¹ /( F ⁿ − 1 ) hS ¹ ≃ ES ₊ ¹ ∧ ^{S 1} T h(µ ⁻ _n ) ∼ = T h((µ ⁻ _n ) hS ¹ )

⁽⁴⁰⁾

where for an

S ¹

^{-vetor bundle}

ξ

^{given by a projetion map}

p : E −→ B

^we

denoteby

ξ hS ¹

^{thebundlewithprojetionid}

× p : ES ¹ × S ¹ E −→ ES ¹ × S ¹ B

^.

This means wealso have a Morse theorem for the

S ¹

-equivariantltration.

4.1 The negative bundle

In [Bökstedt-Ottosen ℄ Lemma 5.1, it is shown that the negative bundle

(µ ⁻ _n ) _hS ¹

^{is an oriented vetor bundle if}

µ ⁻ _n

^{is. But to use the Thom}

iso-morphismin

K

^{-theoryweneedtoknowthatthenegativebundleisomplex,}

ormore preisely

Proposition 4.2. The negative bundle

µ ⁻ _n

^{for the energyltration of}

LCP ^r

an be written as

ε ⊕ ν

^{, where}

ε

^{is a trivial real}

S ¹

^{-line bundle, and}

ν

^{is a}

omplex

S ¹

^{vetor bundle.} Consequently, the negative bundle

(µ ⁻ _n ) hS ¹

^{for the}

energyltration of

LCP ^r _hS 1

^{an also be written as}

ε ⊕ ν _hS ¹

^.

Proof. ThereisaHermitianinnerprodut

h· , ·i ^C

^on

T CP ^r

^,andthe

Rieman-nian metri is

h· , ·i =

^Re

( h· , ·i C )

^{. The tangent spae}

T f LCP ^r

^{is a omplex}

vetor spae,and itarries the struture of a Hilbert spae via

h ξ, η i = Z

S ¹

( h ξ(t), η(t) i + h∇ ξ(t), ∇ η(t) i ) dt,

where

ξ, η ∈ T f LCP ^r

^{are vetor elds along}

f

ⁱⁿ

LCP ^r

^{, and}

∇

^{denotes the}

ovariant derivativealong

f

^{. Sine}

h· , ·i =

^Re

( h· , ·i C )

^{, weget}

h zξ, zη i = h ξ, η i

^for

z ∈ S ¹ .

⁽⁴¹⁾

If

f

^{is a ritial point of the energy funtional}

E

^{(a geodesi), then the}

tangent spae of

LCP ^r

^{splits as}

T f LCP ^r = Γ(Rf ^′ ) ⊕ Γ(Rif ^′ ) ⊕ Γ((f ^′ ) ^⊥ )

⁽⁴²⁾

where e.g.

Γ(Rf ^′ ) ⊆ Γ(f)

^{denotes the vetor elds}

ξ

^along

f

^with

ξ(t) ∈ Rf ^′ (t) ⊆ T _f(t) CP ^r

^{. We an use the inner produt to represent the Hessian}

H = D ² E

^of

E

^{by a linear operator}

A = A f

^on

T f LCP ^r

^{, by requiring}

h Aξ 1 , ξ 2 i = H(ξ 1 , ξ 2 )

^{. Then we get by (41) that}

zAz ¯ = A

^for

z ∈ S ¹

^{, whih}

implies that

A

^{isomplex linear.}

Aording to Klingenberg, [Klilngenberg1℄ Thm. 2.4.2,

A f =

^id

− (1 − ∇ ² ) ^{− 1} ◦ ( ˜ K f + 1),

where

K ˜ f (ξ)(t) = R(ξ(t), f ^′ (t))f ^′ (t)

= π ² (f ^′ (t) h ξ(t), f ^′ (t) i − 2f ^′ (t) h f ^′ (t), ξ(t) i + ξ(t) h f ^′ (t), f ^′ (t) i )

Notethefator

π ²

^{;itappearsbeauseourmetrion}

CP ^r

^{issaledsothatthe}

irumferene is 1, not

π

^{. This gives us the following eigenvalue equation}

Aξ = λξ

^:

(λ − 1) ∇ ² ξ = ( ˜ K f + λ)ξ

⁽⁴³⁾

Thenegativebundleonsistsofsolutionstothis equationwith

λ < 0

^{. Notie}

thatbytheformulafor

A

^{,itpreservesthe}deomposition(42),sineovariant derivativeommuteswiththeomplexstrutureon

T CP ^r

^{. Thusweansolve}

(43) inthe three spaes separately.

(i) ξ ∈ Γ(Rf ^′ )

^{: Then}

ξ(t) = g(t)f ^′ (t)

^where

g : [0, 1] −→ R

^{is a smooth}

funtionwith

g(0) = g(1)

^{. Then}

K ˜ f (t) = 0

^{,andequation(43)beomes}

(λ − 1)g ^′′ = λg ⇔ g ^′′ = _λ ^λ _{− 1} g ⇒ g = 0

sine

λ < 0

^and

g

^{mustbeperiodi. Sowehaveno}non-trivialsolutions.

(ii) ξ ∈ Γ((f ^′ ) ^⊥ )

^{: Sine}

(f ^′ ) ^⊥

^{is a omplex vetor spae, and}

A

^{is omplex}

linear as noted, we see that

Aξ = λξ

^implies

A(iξ) = λ(iξ)

^{. So this}

spaeof solutions has a omplex struture.

(iii) ξ ∈ Γ(Rif ^′ )

^{: Then}

ξ(t) = g (t)if ^′ (t)

^{, where}

g : [0, 1] −→ R

^{is asmooth}

funtionwith

g(0) = g(1)

^{. Then}

K ˜ f (t) = 4π ² k f ^′ (t) k ² ξ(t) = 4π ² n ² ξ(t)

^,

sine

f

^{is ageodesi of length}

n

^{. The equation (43) then beomes}

(λ − 1)g ^′′ = (4π ² n ² + λ)g ⇔ g ^′′ = 4π ² n ² + λ

λ − 1 g

To get a periodi solution

g

^{, we must have}

^{4π 2} _λ ⁿ ₋ ² ₁ ^+λ ≤ 0

^{, i.e.}

λ ≥

− 4π ² n ²

^{. For}

λ = − 4π ² n ²

^{we must have}

g

^{onstant, and this gives the}

trivialreal linebundle

ε

^{. If}

− 4π ² n ² < λ < 0

^{we have the solution set}

spanned over

R

^by

g ₁ ^K (t) = cos(K · 2πt),

^and

g ₂ ^K (t) = sin(K · 2πt), t ∈ [0, 1]

where

K = s

− 4π ² n ² + λ

2π(λ − 1) ,

^and

K ∈ N,

sine the funtions must be periodi with period 1. This happens if

and onlyif

λ = 4π ² (K ² − n ² ) 4π ² K ² + 1 ,

so for a xed

n

^{we get solutions with}

λ < 0

^for

K = 1, . . . , n − 1

^.

Thisspae ofsolutionsan be given aomplexstruture

J

^{by rotating}

t 7→ t − _4K ¹

^,where

t ∈ [0, 1]

^,i.e.

J(g ₁ ^K ) = g ₂ ^K , J (g ₂ ^K ) = − g ^K ₁ .

and extendinglinearly. Clearly

J

^satises

J ² = −

^id.

This gives the bundle

ν

^{, whih is learly an}

S ¹

^{bundle, with the}

S ¹

^ation

given by rotation.

Nowletussee thattheresultfor

µ ⁻ _n

^{impliesthatfor}

(µ ⁻ _n ) hS ¹

^{. Thebundle}

(µ ⁻ _n ) _hS ¹

^{isdened sothatthe pullbakof}

(µ ⁻ _n ) _hS ¹

^{agreeswith pr}

^∗ (µ ⁻ _n )

^inthe

followingdiagram,

µ ⁻ _n

pr

∗ (µ ⁻ _n )

//

oo (µ ⁻ _n ) hS ¹

G n (r) ES ¹ × G _n (r) ^//

^pr

^oo ES ¹ × S ¹ G _n (r)

where

G n (r)

^{denotesthespaeof}

n

^{-timesiteratedgeodesis. Sine}

µ ⁻ _n = ε ⊕ ν

isadeompositionin

S ¹

^-bundles,weautomatiallygetthedeompositionfor

(µ ⁻ _n ) hS ¹

^.

In document Afhandling (Sider 38-51)