§4. The moment map 21

The classification of coadjoint orbits reduces in this case to the problem of

classification of symmetric matrices up to transformations

S →

gtSg,

g ∈ Sp(2n, R).

♦

Remark 4. The problems arising in Examples 7 and 8 are particular

cases of the following general problem: classification of a pair (S, A) where

S is a symmetric matrix and A is an antisymmetric matrix with respect to

simultaneous linear transformations:

(S, A) →

(gtSg, gtAg),

g ∈ GL(n, R).

♥

Example

8.∗

Let M be a compact smooth simply connected 3-dimen-

sional

manifold5

with a given volume form vol. Let G = Diff(M, vol) be the

group of volume preserving diffeomorphisms of M.

The role of the Lie algebra g = Lie(G) is played by the space Vect(M, vol)

of all divergence-free vector fields on M. We recall that the divergence of a

vector field ξ with respect to a volume form vol is a function div ξ on M

such that Lξ(vol) = div ξ · vol. Here Lξ is the Lie derivative along the field

ξ. Using the identity (see formula (16) in Appendix II.2.3)

Lξ = d ◦ iξ + iξ ◦ d

we obtain that iξvol = dθξ where θξ is some 1-form on M defined modulo

exact forms (differentials of functions). Now any smooth map K :

S1

→ M

defines a linear functional FK on g:

(24) FK (ξ) =

S1

K∗(θξ).

(It is clear that adding a differential of a function to θξ does not change

the value of the integral.) Moreover, the functional FK does not change if

we reparametrize

S1

so that the orientation is preserved. In other words, it

depends only on the oriented curve

K(S1).

We see that the classification of coadjoint orbits in this particular case

contains as a subproblem the classification of oriented knots in M up to a

volume preserving isotopy. ♦

Example

9.∗

Let G =

Diff+(S1)

denote the group of orientation pre-

serving diffeomorphisms of the circle, and let G be its simply connected

5The

famous Poincar´ e conjecture claims that such a manifold is diffeomorphic to

S3

but it

is still unknown. We use only the equalities H2(M) = H1(M) = {0}.