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

3.5. Singular quadratic Lie superalgebras of type S1 with non-Abelian even part
so D(X) = µX. For Y 0, D([Y 0,X]) = DC(X) = µC(X) for all X ∈ g. But also, D([Y 0,X]) =
[D(Y 0),X] = µC(X) + ϕ(Y 0)C(X), hence ϕ(Y 0) = 0. As a consequence, D(Y 0) = µY 0 + Z(Y 0).
Now, we prove that D(X 0) = µX 0. Indeed, since D is even and [g 1,g 1] = CX 0 then one has
D(X 0) ⊂ D([g 1,g 1]) = [D(g 1),g 1] ⊂ [g 1,g 1] = CX 0.
It implies that, D(X 0) = aX 0. Combined with B(D(Y 0),X 0) = B(Y 0,D(X 0)), we obtain µ = a.
Let X ∈ q, B(D(X 0),X) = µB(X 0,X) = 0. Moreover, B(D(X 0),X) = B(X 0,D(X)), so ϕ(X) =
0.
Since C(g) is generated by invertible centromorphisms then the necessary condition of
Lemma is finished. The sufficiency is obvious.
We turn now the proposition. By the previous lemma, the bilinear form B ′ defines an associated
invertible centromorphism D = µ Id+Z for some non-zero µ ∈ C and Z : g → Z(g)
satisfying Z| [g,g] = 0. For all X,Y,Z ∈ g, one has:
I ′ (X,Y,Z) = B ′ ([X,Y ],Z) = B(D([X,Y ]),Z) = B([D(X),Y ],Z) = µB([X,Y ],Z).
That means I ′ = µI and then dup B ′(g) = dup B (g) = 1.
Finally, if dup B (g) = 0, then g cannot be of type S3 or S1 with respect to B ′ , so dup B ′(g) = 0.
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