96
F. Barbaresco
X and Y are two left invariant vectors fields on G. X
and Y
right invariant vectors
fields coinciding with X and Y at the point e, neutral element of G. If T =
X
, Y
is tight invariant vectors fields which coincide with −[X, Y ] on e, then:
[X, J T ] = J [X, [X, Y ]] − [X, J [X, Y ]] at point e
(5.5)
At point e, we have the equality:
4α(X, Y ) =
2n
i=1
ξ i ([J [X, Y ], X i ] − J [[X, Y ], X i ])
(5.6)
As the form α is invariant on the left by G, this equality is verified for all points.
For any endomorphism of the space g such that b ⊂ b, we denote by T r b the
trace of the restriction of to b and by T r g/b the trace of the endomorphism of
g/b deduced from by passage to the quotient, with T r = T r b + T r g/b . We
have:
T r g/b =
2n
i=1
ξ i (X i )
(5.7)
Whatever X ∈ g and s ∈ B, we have J (Xs) − (J X)s ∈ b. If ad(Y ) is the
endomorphism of g defined by ad(Y ).Z = [Y, Z ], we have (J ad(Y ) − ad(Y )J )g ⊂
b for all Y ∈ b. We can deduce, for all X ∈ g, the endomorphism ad(J X) − J ad(X )
leaves steady the subspace b. Koszul defines a linear form on the space g by
defining:
(X ) = T r g/b (ad(J X) − J ad(X )), ∀X ∈ g
(5.8)
Koszul has finally obtained the following fundamental theorem:
Theorem of Koszul [77] The Kähler form of the Hermitian canonical form
has for image by p
∗ the differential of the form −
1
4
(X ) with (X ) =
T r g/b (ad(J X) − J ad(X )), ∀X ∈ g
Koszul note that the form is independent of the choice of the tensor J. It is
determined by the invariant complex structure of G/B. The form is right invariant
by B. For all s ∈ B, note the endomorphism r (s) : X → Xs of g . Since J (Xs) =
(J X)s mod b and that T r g/b ad(Y ) = 0, we have:
(Xs) = T r g/b (ad((J X)s) − J ad(Xs)), ∀X ∈ g, ∀Y ∈ b
(5.9)
(Xs) = T r g/b
r (s)ad(J X)r (s)
−1
− Jr(s)ad(X )r (s)
−1
(5.10)