TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...

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