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 ).
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