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.

  • 1
    $\begingroup$ Well, it characterizes the Lie algebra $\{0\}$ (assuming you mean "unique up to nonzero scalar multiplication", and also if you mean unique at all). $\endgroup$
    – YCor
    Jul 15, 2019 at 16:39
  • 2
    $\begingroup$ 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. $\endgroup$
    – YCor
    Jul 15, 2019 at 16:43
  • 2
    $\begingroup$ 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. $\endgroup$
    – YCor
    Jul 15, 2019 at 17:49
  • $\begingroup$ 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? $\endgroup$
    – LSpice
    Jul 15, 2019 at 20:11


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