Lie algebras with unique invariant bilinear symmetric form

A complex, finite dimensional Lie algebra admitting a (up to multiplication by a non-zero scalar) unique, invariant, symmetric, non-degenerate bilinear form is 1-dimensional or simple. Cf 'Lie algebras with unique invariant scalar product' June 20 2019.

I would like to drop the non-degeneracy:

Can you characterize all complex Lie algebras admitting a (up to multiplication) unique, invariant, symmetric bilinear form?

One easy example is the 2-dimensional Lie algebra [x,y]=y.

• Well, it characterizes the Lie algebra $\{0\}$ (assuming you mean "unique up to nonzero scalar multiplication", and also if you mean unique at all).
– YCor
Jul 15, 2019 at 16:39
• Maybe you mean to replace "non-degenerate" with "non-zero" (and keep "up to nonzero scalar multiplication"), in which case the question will be less trivial, but the answer will be a bit disappointing. First, say "so-simple" to mean simple or 1-dimensional abelian. Then you get the Lie algebras $\mathfrak{g}$ with a single so-simple quotient $\mathfrak{g}/\mathfrak{n}$ (and hence $\mathfrak{g}$ either perfect [solvable-by-simple] or solvable with 1-dim abelianization), with no other quotient having a non-degenerate symmetric bilinear form. This is a huge class.
– YCor
Jul 15, 2019 at 16:43
• So well, among those Lie algebras with a single so-simple quotient, I'd say that "generically" all Lie algebras satisfy your property. For instance, all solvable 3-dimensional with 1-dim abelianization do, and all 4-dim do, with the exception of the oscillator algebra. For $\mathfrak{s}$ simple and $V$ irreducible $\mathfrak{s}$-module, the semidirect product $\mathfrak{s}\ltimes V$ satisfies your property iff $V$ is not the adjoint representation.
– YCor
Jul 15, 2019 at 17:49
• Small modifications to a question with an unsatisfactory answer sometimes indicate that there is an unspecified bigger question lurking behind the scenes. Is that true? If so, what is it? Jul 15, 2019 at 20:11