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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3.5. Singular quadratic Lie superalgebras of type S1 with non-Abelian even part<br />
3.5 Singular quadratic Lie superalgebras of type S1 with non-<br />
Abelian even part<br />
Let g be a singular quadratic Lie superalgebra of type S1 such that g 0 is non-Abelian. If<br />
[g 1,g 1] = {0} then g is an orthogonal direct sum of a singular quadratic Lie algebra of type S1<br />
and a vector space. There is nothing to do. Therefore, we can assume that [g 1,g 1] = {0}. Fix<br />
α ∈ VI and choose Ω0 ∈ A 2 (g 0), Ω1 ∈ S 2 (g 1) such that<br />
I = α ∧ Ω0 + α ⊗ Ω1.<br />
Let X 0 = φ −1 (α) then X 0 ∈ Z(g) and B(X 0,X 0) = 0. We define the maps C0 : g 0 → g 0,<br />
C1 : g 1 → g 1 by Ω0(X,Y ) = B(C0(X),Y ), for all X,Y ∈ g 0 and Ω1(X,Y ) = B(C1(X),Y ), for all<br />
X,Y ∈ g 1. Let C : g → g defined by C(X +Y ) = C0(X) +C1(Y ), for all X ∈ g 0, Y ∈ g 1.<br />
Proposition 3.5.1. For all X,Y ∈ g, the Lie super-bracket of g is defined by:<br />
[X,Y ] = B(X 0,X)C(Y ) − B(X 0,Y )C(X) + B(C(X),Y )X 0.<br />
In particular, if X,Y ∈ g 0, Z,T ∈ g 1 then<br />
(1) [X,Y ] = B(X 0,X)C0(Y ) − B(X 0,Y )C0(X) + B(C0(X),Y )X 0,<br />
(2) [X,Z] = B(X 0,X)C1(Z),<br />
(3) [Z,T ] = B(C1(Z),T )X 0<br />
Proof. For all X,Y,T ∈ g 0, since B([Y,Z],T ) = α ∧ Ω0(X,Y,Z) then one has (1) as in Chapter<br />
2. For all X ∈ g 0, Y,Z ∈ g 1<br />
B([X,Y ],Z) = α ⊗ Ω1(X,Y,Z) = α(X)Ω1(Y,Z) = B(X 0,X)B(C1(Y ),Z).<br />
Hence we obtain (2) and (3).<br />
Now, we show that g 0 is solvable. Consider the quadratic Lie algebra g 0 with 3-form I0 =<br />
α ∧ Ω0. Write Ω0 = ∑i< j ai jαi ∧ α j, with ai j ∈ C. Set Xi = φ −1 (αi) then<br />
Recall the space WI0 in Chapter 2 as follows:<br />
C0 = ∑ ai j(αi ⊗ Xj − α j ⊗ Xi).<br />
i< j<br />
WI0 = {ιX∧Y (I0) | X,Y ∈ g 0}.<br />
Since WI0 = φ([g 0,g 0]) one has Im(C0) ⊂ [g 0,g 0]. In Section 3.2, it is known that {α,I0} = 0<br />
and then [X 0,g 0] = 0. As a sequence, B(X 0,[g 0,g 0]). That deduces B(X 0,Im(C0)) = 0. Therefore<br />
[[g 0,g 0],[g 0,g 0]] = [Im(C0),Im(C0)] ⊂ CX 0 and then g 0 is solvable.<br />
Since B is non-degenerate then we can choose Y 0 ∈ g 0 isotropic such that B(X 0,Y 0) = 1 as in<br />
Chapter 2 to obtain a straightforward consequence as follows:<br />
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