TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...
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3.1. Application of Z × Z2-graded Lie superalgebras to quadratic Lie superalgebras<br />
Example 3.1.9. Let X ∈ Vx be a homogeneous element in V of degree x and define the endomorphism<br />
ιX of E(V ) by<br />
ιX(A)(X1,...,Xa−1) = (−1) xb A(X,X1,...,Xa−1), ∀ A ∈ E (a,b) (V ), X1,...,Xa−1 ∈ V.<br />
Then ιX is a super-derivation of E(V ) of degree (-1,x) [BP89]. In particular<br />
ιX(A ∧ A ′ ) = ιX(A) ∧ A ′ + (−1) −a+xb A ∧ ιX(A ′ ), ∀ A ∈ E (a,b) (V ), A ′ ∈ E(V ).<br />
Lemma 3.1.10. Let X 0 ∈ V 0 and X 1 ∈ V 1 then for all Ω ⊗ F ∈ A ω ⊗ S f , one has:<br />
(1) ιX 0 (Ω ⊗ F) = ιX 0 (Ω) ⊗ F,<br />
(2) ιX 1 (Ω ⊗ F) = (−1) ω Ω ⊗ ιX 1 (F).<br />
Proof. We have:<br />
(1) ιX 0 (Ω ⊗ F) = ιX 0 ((Ω ⊗ 1) ∧ (1 ⊗ F)) = ιX 0 (Ω ⊗ 1) ∧ (1 ⊗ F) + (−1) −ω (Ω ⊗ 1) ∧ ιX 0 (1 ⊗<br />
F) = ιX 0 (Ω) ⊗ F.<br />
(2) ιX 1 (Ω ⊗ F) = ιX 1 ((Ω ⊗ 1) ∧ (1 ⊗ F)) = ιX 1 (Ω ⊗ 1) ∧ (1 ⊗ F) + (−1) −ω (Ω ⊗ 1) ∧ ιX 1 (1 ⊗<br />
F) = (−1) ω Ω ⊗ ιX 1 (F).<br />
Remark 3.1.11.<br />
(1) If Ω ∈ A ω then ιX(Ω)(X1,...,Xω−1) = Ω(X,X1,...,Xω−1), for all X,X1,...,Xω−1 ∈ V 0.<br />
(2) Let X be an element of the canonical basis B of V1 and p ∈ V ∗<br />
1<br />
Corollary II.1.52 in [Gié04] one has:<br />
Moreover,<br />
ιX(p n )(X n−1 ) = (−1) n p n (X n ) = (−1) n (−1) n(n−1)/2 n!.<br />
be its dual form. By<br />
∂ pn<br />
∂ p (X n−1 ) = n(p n−1 )(X n−1 ) = (−1) (n−1)(n−2)/2 n!. It implies that<br />
ιX(p n )(X n−1 ∂ pn<br />
) = −<br />
∂ p (X n−1 ).<br />
Since each F ∈ S f is regarded as a polynomial in the variable p and by linearizing so<br />
one has the following property: let X ∈ V1 and p ∈ V ∗ be its dual form then<br />
1<br />
{A,A ′ m<br />
ω+ f +1<br />
} = (−1) ∑<br />
j=1<br />
ι X j<br />
0<br />
ιX(F) = − ∂F<br />
, ∀ F ∈ S .<br />
∂ p<br />
Proposition 3.1.12. Fix an orthonormal basis {X 1 0 ,...,X m 0 } of V0 and a canonical basis B =<br />
{X 1 1 ,...,X n,Y 1,...,Y<br />
n<br />
1 1 1 } of V1. Then the super Z × Z2-Poisson bracket on E(V ) is given by:<br />
<br />
ιX k(A)<br />
∧ ιY k(A<br />
1<br />
1<br />
′ ) − ιY k(A)<br />
∧ ιX k(A<br />
1<br />
1<br />
′ <br />
)<br />
for all A ∈ A ω ⊗ S f and A ′ ∈ E(V ).<br />
(A) ∧ ι j(A<br />
X<br />
0<br />
′ n<br />
ω<br />
) + (−1) ∑<br />
k=1<br />
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