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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3.3. Elementary quadratic Lie superalgebras<br />
Proposition 3.3.3. Let g be a reduced elementary quadratic Lie superalgebra then g is iisomorphic<br />
to one of the following Lie superalgebras:<br />
(1) gi where gi, 3 ≤ i ≤ 6, are reduced singular quadratic Lie algebras of type S3 given in<br />
Proposition 2.2.29.<br />
(2) g s 4,1 = (CX 0 ⊕ CY 0 ) ⊕ (CX 1 ⊕ CZ 1) such that the bilinear form B is defined by<br />
B(X 0 ,Y 0 ) = B(X 1,Z 1) = 1,<br />
the other are zero and the Lie super-bracket is defined by [Z 1,Z 1] = −2X 0 , [Y 0 ,Z 1] = −2X 1,<br />
the other are trivial.<br />
(3) g s 4,2 = (CX 0 ⊕ CY 0 ) ⊕ (CX 1 ⊕ CY 1) such that the bilinear form B is defined by<br />
B(X 0 ,Y 0 ) = B(X 1,Y 1) = 1,<br />
the other are zero and the Lie super-bracket is defined by [X 1,Y 1] = X 0 , [Y 0 ,X 1] = X 1,<br />
[Y 0 ,Y 1] = −Y 1, the other are trivial.<br />
(4) g s 6 = (CX 0 ⊕ CY 0 ) ⊕ (CX 1 ⊕ CY 1 ⊕ CZ 1 ⊕ CT 1) such that the bilinear form B is defined by<br />
Proof.<br />
B(X 0 ,Y 0 ) = B(X 1,Z 1) = B(Y 1,T 1) = 1,<br />
the other are zero and the Lie super-bracket is defined by [Z 1,T 1] = −X 0 , [Y 0 ,Z 1] = −Y 1,<br />
[Y 0 ,T 1] = −X 1, the other are trivial.<br />
(1) This statement corresponds to the case where I1 = 0 and I = I0 a decomposable 3-form in<br />
A 3 (g 0 ). In this case, I(g 0,g 1,g 1) = B([g 0,g 1],g 1) = 0. It implies [g 0,g 1] = [g 1,g 1] = {0}<br />
since B is non-degenerate and then g 1 ⊂ Z(g). On the other hand, g is reduced so Z(g) ⊂<br />
[g,g] ⊂ g 0. Therefore, g 1 = {0} and we obtain (1).<br />
Assume that I = α ⊗ pq ∈ A 1 (g 0 ) ⊗ S 2 (g 1). By the previous lemma, g 0 = CX 0 ⊕ CY 0<br />
where X 0 = φ −1 (α), B(X 0 ,X 0 ) = B(Y 0 ,Y 0 ) = 0, B(X 0 ,Y 0 ) = 1. Let X 1 = φ −1 (p), Y 1 =<br />
φ −1 (q) and U = span{X 1,Y 1}.<br />
(2) If dim(U) = 1 then Y 1 = kX 1 with some non-zero k ∈ C. Therefore, q = kp and I =<br />
kα ⊗ p 2 . Replace with<br />
X 0 = kX 0 , Y 0 = Y 0<br />
k<br />
and Z 1 is an element in g 1 such that B(X 1,Z 1) = 1 then g = (CX 0 ⊕ CY 0 ) ⊕ (CX 1 ⊕ CZ 1)<br />
and I = α ⊗ p 2 .<br />
Now, let X ∈ g 0,Y,Z ∈ g 1. By using (1.7) and (1.8) of [BP89], one has:<br />
B(X,[Y,Z]) = −2α(X)p(Y )p(Z) = −2B(X 0 ,X)B(X 1,Y )B(X 1,Z).<br />
Since B|g 0 ×g 0 is non-degenerate then:<br />
[Y,Z] = −2B(X 1,Y )B(X 1,Z)X 0 , ∀ Y,Z ∈ g 1.<br />
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