Timeline for Lie algebras with unique invariant scalar product
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 15, 2019 at 17:27 | vote | accept | Thomas Schucker | ||
| Jun 26, 2019 at 21:07 | comment | added | YCor | In general, for given $\mathfrak{g}$ let $K$ be the intersection of all kernels of all [symmetric] invariant bilinear forms. Then these are those for which $\mathfrak{g}/K$ is simple or 1-dimensional. Remains to understand $K$, and this goes along with understanding Lie algebras admitting a non-degenerate symmetric bilinear form. | |
| Jun 26, 2019 at 17:40 | comment | added | Thomas Schucker | Now I would like to drop the condition of 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. | |
| Jun 26, 2019 at 17:30 | comment | added | Thomas Schucker | Thanks you InfiniteLooper and YCor. | |
| Jun 26, 2019 at 15:19 | comment | added | YCor | Actually I can think of an $n$-dimensional nilpotent Lie algebra (for all odd $n\ge 5$) for which this dimension is $5$ ($4$ for symmetric ones), while chains have length $n$. | |
| Jun 26, 2019 at 15:16 | comment | added | YCor | By the way, no, I don't prove the inequality you claim. Indeed first it would require a normal composition series (not only subnormal). Second would it require all the quotients to admit a non-degenerate invariant form, which is usually not the case. | |
| Jun 26, 2019 at 15:13 | comment | added | YCor | If the abelianization has dimension $n$ this dimension is $\ge n^2$ ($\ge n(n+1)/2$ for the symmetric part). | |
| Jun 26, 2019 at 15:06 | comment | added | InfiniteLooper | Using this approach you show that the dimension of such bilinear form is greater than the length of maximal chains of composition in the Lie algebra. Can you show the equality ? | |
| Jun 26, 2019 at 14:07 | comment | added | YCor | Also the result (and proof) works over an arbitrary field of characteristic zero, when in the statement "simple" is replaced with "absolutely simple". (For a simple, non-absolutely-simple Lie algebra, the dimension of the space of invariant symmetric bilinear forms is $\ge 2$.) | |
| Jun 26, 2019 at 14:05 | comment | added | YCor | Here's a slight variant of the proof, not using Levi. If $\mathfrak{g}$ is not simple or $\le 1$-dimensional, then it has a quotient that is simple or $\le 1$-dimensional. Since $\mathfrak{g}$ has a non-degenerate invariant bilinear form $b_0$ and this quotient too, it also admits a degenerate non-zero invariant bilinear form $b_1$. Hence the space of invariant bilinear forms on $\mathfrak{g}$ has dimension $\ge 2$. So it contains 2 non-proportional non-degenerate elements (say $b_0+tb_1$ for all but finitely many scalars $t$). | |
| Jun 26, 2019 at 13:55 | history | answered | InfiniteLooper | CC BY-SA 4.0 |