Skip to main content
15 events
when toggle format what by license comment
Sep 6, 2021 at 9:02 vote accept WunderNatur
Sep 5, 2021 at 8:27 answer added Paul Levy timeline score: 4
Sep 2, 2021 at 7:02 comment added YCor @LSpice Yes: any Lie subalgebra of $gl_n$ whose action is completely reducible (over a field of char 0) is reductive (=semisimple $\times$ abelian). A similar result holds for linear algebraic groups (with "reductive" in the usual sense of algebraic groups).
Sep 1, 2021 at 23:42 comment added LSpice @YCor, I know reductivity as a condition on all representations. Under what circumstances can you check it by looking at a single representation? (Is the relevant point that the representation is faithful?)
Sep 1, 2021 at 7:30 comment added YCor Other remark: The standard representation is clearly (absolutely) irreducible, so the generated Lie algebra $g_n$ is semisimple (well, reductive, but with center contained in scalar matrices, and since $g_n$ consists of trace zero matrices, the center is trivial).
Sep 1, 2021 at 6:56 comment added WunderNatur @YCor I checked from 3 to 9.
Sep 1, 2021 at 6:36 comment added YCor Here's a roadmap to a solution. (a) compute the space $B_n$ of invariant quadratic (for $n$ odd) or alternating forms by $U_n$ and $L_n$. Then $B_n$ should be 1-dimensional and generated by a non-degenerate form (otherwise the conclusion fails), say $q$. So, the Lie subalgebra $g_n$ generated by $\{U_n,L_n\}$ preserves $q$, i.e., is contained in so(q), resp sp(q). (b) Show the latter inclusion is an equality, e.g., computing enough iterated brackets to obtain a good lower bound on the dimension of $g_n$?
Sep 1, 2021 at 6:31 comment added YCor You checked numerically with which values of $n$?
Sep 1, 2021 at 5:16 comment added WunderNatur @abx Thanks for the suggestion! I've made the change in the text.
Sep 1, 2021 at 5:15 history edited WunderNatur CC BY-SA 4.0
added 28 characters in body
Sep 1, 2021 at 5:13 comment added abx Oh, I see. Maybe you could say that $L_n$ and $U_n$ generate a Lie algebra isomorphic to $\mathfrak{so}(n,\mathbb{C})$ or $\mathfrak{sp}(n,\mathbb{C})$.
Sep 1, 2021 at 4:31 comment added WunderNatur @abx That should be part of the question. It should be some weird bilinear form which I could not determine.
Sep 1, 2021 at 4:29 comment added abx For which bilinear forms? For the standard quadratic form $\mathfrak{so}(n,\mathbb{C})$ is the space of skew-symmetric matrices, which does not contain $L_n$ or $U_n$.
Sep 1, 2021 at 1:39 history edited WunderNatur CC BY-SA 4.0
edited title
Sep 1, 2021 at 0:57 history asked WunderNatur CC BY-SA 4.0