10 1. Geometry of Coadjoint Orbits

Then [ξ1, ξ2](e, 0) = ([X1, X2], K(X1)F2 − K(X2)F1) and we also have

ξ1θ(ξ2) = ξ1 F, X2 = F1, X2 , ξ2θ(ξ1) = ξ2 F, X1 = F2, X1 ,

θ([ξ1, ξ2]) = F, [X1, X2] .

The desired formula (12) follows from these computations.

The group G×G acts on G by left and right shifts: (g1, g2)·g = g1g g2

−1.

This action extends to a Hamiltonian action on T

∗G.

Actually, the extension

is again given by left and right shifts on elements G viewed as elements of

the group T ∗G. Using the above identification we can write it in the form

(13) (g1, g2) · (g, F ) =

(

g1g g2

−1,

K(g2)F

)

.

Recall (see Appendix III.1.1) that to any X ∈ g there correspond two vector

fields on G:

– the infinitesimal right shift X, which is a unique left-invariant field

satisfying X(e) = X, and

– the infinitesimal left shift X, which is a unique right-invariant field

satisfying X(e) = X.

Exercise 1. Using the above identifications, write explicitly

a) the vector fields X and X on G;

b) their Hamiltonian lifts

˜

X

∗

and

ˆ

X

∗

on T

∗(G).

Answer:

(14)

a) X(g) = X; X(g) = Ad

g−1X;

b)

X∗(g,

F ) = (X, 0),

X∗(g,

F ) =

(

Ad

g−1X,

K∗(X)F

)

.

Hint. Use formula (13).

To construct the reduction of T

∗G

with respect to the left, right, or

two-sided action of G (see Appendix II.3.4), we have to know the moment

map.

Lemma 4. For the action of G × G on T

∗G

the moment map μ : T

∗G

→

g∗ ⊕ g∗ is given by the formula

(15) μ (g, F ) = (F ⊕ K(g)F ).