Timeline for Lie algebras with unique invariant bilinear symmetric form
Current License: CC BY-SA 4.0
7 events
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| Jul 15, 2019 at 20:11 | comment | added | LSpice | 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 19:00 | history | edited | Jim Humphreys | CC BY-SA 4.0 |
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| Jul 15, 2019 at 17:49 | comment | added | YCor | 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. | |
| Jul 15, 2019 at 16:43 | comment | added | YCor | 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. | |
| Jul 15, 2019 at 16:39 | comment | added | YCor | Well, it characterizes the Lie algebra $\{0\}$ (assuming you mean "unique up to nonzero scalar multiplication", and also if you mean unique at all). | |
| Jul 15, 2019 at 16:39 | history | edited | Nate Eldredge | CC BY-SA 4.0 |
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| Jul 15, 2019 at 16:36 | history | asked | Thomas Schucker | CC BY-SA 4.0 |