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### Transcript of Lower bounds for Gromov width of coadjoint orbits in …milena/UIUC.pdf · Lower bounds for Gromov...

• Lower bounds for Gromov width of

coadjoint orbits in SO(n).

Milena Pabiniak

UIUC December 6, 2011

• Problem suggested by

prof Yael Karshon,

extension of PhD work of her student

Masrour Zoghi

Key points:

Hamiltonian torus action symplectic embeddings of balls

Action of the Gelfand-Tsetlin torus (Cetlin, Zetlin).

1

• Let (M,) be a 2N-dimensional symplectic

manifold.

Gromov Non-Squeezing Theorem being symplectomorphism is much more

restrictive then just being volume preserv-

ing.

The Gromov width of M is the supremum of the set of as such

that a ball of capacity a

B2Na ={z CN

Ni=1

|zi|2 < a},

can be symplectically embedded in (M,).

2

• G- compact connected Lie group

Coadjoint action: Gy g

For matrix groups, coadjoint action is by conjugation.

T G choice of maximal torus(t)+ choice of positive Weyl chamber

coadjoint points in positive Weyl chamber

orbits11 (t)+

Fact: For any (t)+, the coadjoint orbit through , O, isa symplectic manifold with Kostant-Kirillov symplectic form .

3

• Example: G = U(n)

u(n) = u(n) = n n Hermitian matrices,coadjoint action is conjugation

T =

eit1

eit2. . .

eitn

, t+ =

a1a2

. . .an

; a1 a2 . . . an

Coadjoint orbits = Hermitian matrices with the same eigenvalues

4

• Example: G = SO(2n+ 1), coadjoint action is conjugationso(2n+ 1) = (2n+ 1) (2n+ 1) skew symmetric matricesLet

R() =

(cos() sin()sin() cos()

), L(a) =

(0 aa 0

)Then

TSO(2n+1) =

R(1)R(2)

. . .R(n)

1

; j S1

(tSO(2n+1))+ =

L(1)L(2)

. . .L(n)

0

; j R, 1 . . . n 0

Coadjoint orbits = matrices with char. pol. tnj=1(t

2 + 2j ).

5

• Example: G = SO(2n), coadjoint action is conjugationso(2n) = (2n) (2n) skew symmetric matrices

TSO(2n) =

R(1)R(2)

. . .R(n)

; j S1

(tSO(2n))+ =

L(1)L(2)

. . .L(n)

; j R, 1 . . . n1 |n|

6

• Theorem 1. Let = (1 > . . . > n) int t+, (i.e. regular),M := O - SO(2n+ 1) coadjoint orbit through .The Gromov width of M is at least the minimum

min{1 2, . . . , n1 n, 2n},

what in the language of coroots is

min{, ; a coroot}.

Method:

- construct a proper, centered, Hamiltonian T -space,

- use it to construct explicit embeddings of symplectic balls;

7

• The root system of SO(2n+ 1) consists of vectorsej, j = 1, . . . , n, of squared length 1,(ej ek), j 6= k, of squared length 2.

Therefore this root system for SO(n) is non-simply laced.Note that

(ej),

= 2

ej,

ej, ej

= 2j,and

(ej ek),

= 2

ej ek,

ej ek, ej ek

= j kTherefore for in our chosen positive Weyl chamber

min{, ; a coroot} = min{1 2, . . . , n1 n, 2n}.

8

• Why do we care about such lower bound?

1. (Zoghi) For regular, indecomposable (i.e. with some integral-

ity conditions) U(n) coadjoint orbits their Gromov width is given

by min{, ; a coroot}.

2. (P.) For a class of not regular U(n) coadjoint orbits the above

formula is a lower bound of their Gromov width.

3. (Zoghi) For any compact connected Lie group G, an up-

per bound of the Gromov width of a regular, indecomposable

coadjoint G orbit is given by the above formula.

Corollary 2. (P., Zoghi) The Gromov width of regular, indecom-

posable SO(n) coadjoint orbits is min{, ; a coroot}.

4. (Caviedes) is working on the upper bounds for non-regular

monotone U(n) orbits.

9

• Action is Hamiltonian: there exists a T -invariant momentum

map : M t, such that

(M) = d , t,

where M is the vector field on M corresponding to t.

This sign convention for p MT , the isotropy weights ofT y TpM are pointing out of the momentum map image.(S1)2 y C2 gives

NOT

10

• Let T t be an open convex set which contains (M).The quadruple (M,,, T ) is a proper Hamiltonian T-manifoldif is proper as a map to T .

We will identify Lie(S1) with R using the convention that theexponential map exp : R =Lie(S1) S1 is given by t e2it,that is S1 = R/Z.

11

• A proper Hamiltonian T -manifold (M,,, T ) is centered about

a point T if

KT ctd XMK , (X).

Not centered:

Centered:

12

• Hamiltonian T action on M is called toric if dimT = 12 dimM.

Example 3.M - compact symplectic toric manifold

: M t - moment mapThen:

:= (M) is a convex polytope,

and for any , F face of

F

1(rel-int F )

is the largest subset of M that is centered about .

13

• Proposition 4. (Karshon, Tolman) Let:

(M2n, ,, T ) - a proper Hamiltonian T -manifold,centered about T and

1({}) = {p} a single fixed point.

Then

M is equivariantly symplectomorphic to{z Cn | +

|zj|2j T

},

where 1, . . . ,n are the isotropy weights at p.

14

• Example: Isotropy weights at : 1,2

2

1 51

22

1

2

1(shaded region T ) is equivariantly symplectomorphic to

{z C2|+ (|z1|21 + |z2|22) T }Notice that

z B2 = {z C2(|z1|2+|z2|2) < 2} +(|z1|21+|z2|22) T

B2 M embedds symplectically15

• G = SO(2n+ 1), T -maximal torus of G, dimT = n

Fix t+ regular, O-coadjoint orbit through , dimO = n2

T y O coadjoint (conjugation).

Centered region for this action is too small.

For example, for the SO(5) orbit through = (6,1)

min{, ; a coroot} = min{5,7,12,2} = 2,

while the centered region is

16

• ()

(p)

E1

E2

2

2

1

5

= e1 + e2

e1

= e2

() 17

• Therefore we will use the Gelfand-Tsetlin action.

First define the Gelfand-Tsetlin functions for a group G of

rank k.

Consider a sequence of subgroups

G = Gk Gk1 . . . G1,

with maximal tori T = Tk Tk1 . . . T1.

Inclusion Gj G an action of Gj on the G-coadjoint orbit O.

This action is Hamiltonian with momentum map

j : g gj

18

• Every Gj orbit intersects the (chosen) positive Weyl chamber

(tGj)+ exactly once.

This defines a continuous (but not everywhere smooth) map

sj : gj (tGj)

+.

Let (j) denote the composition sj j:

O j

//

(j) ##HH

HHHH

HHHH

gjsj

(tGj)+

The functions {(j)}, j = 1, . . . , k1, form the Gelfand-Tsetlinsystem denoted by : O RN .

19

• Example: G = U(n) U(n 1) . . . U(1)

- maximal tori: diagonal matrices,- t: diagonal Hermitian matrices,- positive Weyl chambers: eigenvalues in non-increasing order.

Then for a Hermitian matrix A,

j(A) is its j j top left submatrix and

(j)(A) = ((j)1 (A) . . . (j)1 (A)) R

j

is a sequence of eigenvalues of j(A) ordered in a non-increasingway.

Due to this ordering, the function (j) is not smooth on thewhole orbit. The singularities may occur when eigenvalues coin-cide.

20

• Gelfand-Tsetlin system for SO(2n+ 1).

SO(2n+ 1) SO(n) . . . SO(2).

For any k = 2, . . . ,2n, SO(k) injects into SO(2n+ 1) by

SO(k) 3 B 7(B 00 I

).

SO(k) also acts on O by a subaction of a coadjoint action.This action is Hamiltonian with a momentum map

k : O so(k),

k(A) k k top left submatirx of A.

21

• Then

(k)1 (A)

(k)2 (A) . . .

(k)

bk2c(A)

are such that

k(A) SO(k)

L((k)1 (A)) . . .

L((k)bk2c

(A))

0

if k oddor

k(A) SO(k)

L((k)1 (A)) . . .

L((k)bk2c

(A))

if k even .

22

• Why not smooth everywhere?

Due to ordering. The singularities may occur when generalized

eigenvalues coincide.

Proposition 5. The functions (k) are smooth at the preimage

of the interior of the positive Weyl chamber,

USO(k) := ((k))1(int (tSO(k))

+).

23

• Torus action induced by the Gelfand-Tsetlin system

On USO(k), (k) is inducing a smooth action of TSO(k).

For t TSO(k) and A O this new action is

t A =(B1 tB

I

)A

(B1 t1B

I

)

where B SO(k) is such that

Bk(A)B1 (tSO(k))+.

Proposition 6. (k) is a momentum map for the Hamiltonian

action of the torus TSO(k) on USO(k).

24

• Putting together actions of all TSO(k), we obtain the action of

the Gelfand-Tsetlin torus TGTs

TGTs = TSO(2n) TSO(2n1) . . . TSO(2)on the set

U :=2nk=2

USO(k).

Momentum map for this action is

= ((2n),(2n1), . . . ,(2)) : O tGTs = Rn2.

25

• Image of the momentum map

Let {x(k)j |1 k 2n, 1 j bk2c} be basis of R

n2.

Proposition 7. The image of the Gelfand-Tsetlin functions

: O Rn2

is the polytope, which we will denote by P, definedby the following set of inequalities x

(2k)1 x

(2k1)1 x

(2k)2 x

(2k1)2 . . . x

(2k)k1 x

(2k1)k1 |x

(2k)k |,

x(2k+1)1 x

(2k)1 x

(2k+1)2 x

(2k)2 . . . x

(2k+1)k |x

(2k)k |,

for all k = 1, . . . , n, where x(2n+1)j = j.

Moreover, the dimension of the polytope P is n2, what is half ofthe dimension of O.

26

• Graphically,

. . .

. . .. . .

. . .

. . .

1 2 n

x(2n)1 x

(2n)2 x

(2n)n1 |x

(2n)n |

n1

x(2n1)1 x

(2n1)2 x

(2n1)n1

Every coordinate is between its top right and top left neighbors.

27

• Where do these inequalities come from?

Any A O can be SO(2n+ 1) conjugated toL(1)

. . .L(n)

0

and U(2n+ 1) conjugated to

i1. . .

in0in

. . .i1

28

• Therefore the Hermitian matrix 1iA can be U(2n+1) conjugated

to

diag(1, . . . , n,0,n, . . . ,1)

and similarly (1iA)2n can be U(2n) conjugated to

diag((2n)1 , . . . , |(2n)n |,|(2n)n |, . . . ,(2n)1 ).

Min-max principle intertwining inequalities on the eigenvalues:

j (2n)j j+1, n |

(2n)n | 0

29

• Going the other way:

Lemma 8. For any a1 b1 a2 . . . ak bk ak+1 R

x1, . . . , xk C, xk+1 R

such that the Hermitian matrix

A :=

b1 0 x1

. . . ...0 bk xkx1 . . . xk xk+1

,has eigenvalues a1, . . . , ak+1.

30

• For our example of SO(5) orbit through = (6,1) have

6 (4)1 1 |(4)2 |,

(4)1

(3)1 |

(4)2 |,

(3)1 |

(2)2 |.

31

• Note:

- is in U as all generalized eigenvalues for are distinct.

- is fixed under the action of Gelfand-Tsetlin torus

- () is a vertex of P as all Gelfand-Tsetlin functions are equalto their upper bounds.

Moreover:

( face S of P, () S) 1( rel-intS) U.

Therefore the region

T :=

S; ()S1(rel-int S) U.

is centered around ().

32

• Need: weights of the Gelfand-Tsetlin action on TO.

First: identify the edges of P starting from ():

pick one of these inequalities defining P that is an equationon ()

consider the set of points in P satisfying all the equationsthat () does except possibly this chosen one. Call this set

E.

33

• Back to example of SO(5) orbit through = (6,1):

At () = (6,1,6,6) all Gelfand-Tsetlin functions are equal totheir upper bounds.

Can choose the equation (4)1 = (3)1 .

Then the set E case consists of points satisfying

6 1= =

x(4)1 x

(4)2

x

(3)1

=

x(2)1

34

• That is,

(x(4)1 , x(4)2 , x

(3)1 , x

(2)1 ) = (6,1, s, s) R

4

where

s [1,6].

Note that the edge E connects () = (6,1,6,6) and (6,1,1,1),

so

~E = (0,0,5,5).

35

• The preimage 1(E) consists of matrices A of the form

A :=

0 s 0 a 0s 0 0 b 00 0 0 c 0a b c 0 00 0 0 0 0

,

where a, b, c R are such that

(A)4 SO(4)

0 66 0

0 11 0

36

• In particular, the characteristic polynomial of (A)4,

t4 + t2(a2 + b2 + c2 + s2) + c2s2,

must be equal to t4 + t2(6 + 1) + 6.

This gives S1 worth of solutions for any s (1,6), and uniquesolution for s = 1 or 6.

Therefore 1(E) is S2.

37

• Calculating weights of the action of the Gelfand Tsetlin torus

TGTs = TSO(4) TSO(3) TSO(2).

First of TSO(4). Let

R =

(R(1)

R(2)

) TSO(4).

Then

R A =(B1RB

1

) ((A)4 0

0 0

) (B1R1B

1

)

=

(B1R (B(A)4B1) R1B 0

0 0

)=

((A)4

0

)= A

The action is trivial.

38

• Now let

R =

(R(1)

1

) TSO(3), W =

(a b c0 0 0

).

Then

RA =(R

I2

) ((A)3 WTW 0

) (R1

I2

)=

((A)3 RWTW R1 0

).

As

RWT =

R(1)(ab

)0

c 0

The action has weight 1.

Similarly TSO(2) acts with weight 1.

39

• Therefore the Gelfand-Tsetlin torus acts on TO with weight

= (0,0,1,1).

Recall that ~E = (0,0,5,5), so

~E = (6 1) = (e1 e2), (6,1) = 2e1 e2,

e1 e2, e1 e2.

40

• Now, for example, choose the equation

(4)2 = 1.

Then the set E case consists of points

6 1= x

(4)1 x

(4)2

=

x(3)1

=

x(2)1

41

• That is,

(x(4)1 , x(4)2 , x

(3)1 , x

(2)1 ) = (6, l,6,6) R

4

where

l [1,1].

Note that the edge E connects () = (6,1,6,6) and (6,1,1,1),so

~E = (0,2,0,0).

42

• The preimage 1(E) consists of matrices A of the form

A :=

0 6 0 0 y16 0 0 0 y20 0 0 l y30 0 l 0 y4y1 y2 y3 y4 0

,

where Y = (y1, y2, y3, y4) R4 is such that A O that is

A SO(5)

0 66 0

0 11 0

0

43

• Compare characteristic polynomials and get

y1 = y2 = 0,

y23 + y24 = 6

2 + 12 62 l2 = 12 l2.

This gives S1 worth of solutions for any l (1,1), and uniquesolution for l = 1.

Therefore 1(E) is S2.

44

• Calculating weights of the action of the Gelfand Tsetlin torus

TGTs = TSO(4) TSO(3) TSO(2).First of TSO(4). Let

R =

(R(1)

R(2)

) TSO(4).

Then

RA =(R

1

) ((A)4 Y TY 0

) (R1

1

)=

((A)4 RY TY R1 0

),

RY T =

00

R(2)

(y3y4

)

The action of TSO(4) has weight (0,1).

45

• Now let

R =

(R(1)

1

) TSO(3).

Then R A is R(1) 1 11

0 6 0 0 06 0 0 0 00 0 0 l y30 0 l 0 y40 0 y3 y4 0

R(1)

1

11

1

=

(A)3

(R(1)

1

) 0 00 0l y3

(0 0 l0 0 y3

) (R(1)1

1

)0

= A.

The action is trivial.Similarly, the action of TSO(2) is trivial.

46

• Therefore the Gelfand-Tsetlin torus acts on TO with weight

= (0,1,0,0).

Recall that ~E = (0,2,0,0), so

~E = 2 = 2(e2), (6,1) = 2e2, e2, e2

.

47

• Lemma 9. Every edge E of P starting from () is at least

r := min{, ; a coroot}

multiple of E where (E) is the weight of the action along thisedge.

48

• Denote the weights by 1, . . . ,n2.

The centered region T is equivariantly symplectomorphic to

W :={z Cn | +

|zj|2j T

}.

Due to Lemma 9, for any z in a ball of capacity r,

Br = {z Cn2 n2i=1 |zi|2 < r}, have

+ n2i=1

|zi|2i T .

Thus:

Br W Br symplectically embeds into T O. r is the lower bound for Gromov width of O.

49