Timeline for Centralizers of semisimple subgroups
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 19, 2022 at 16:30 | comment | added | unknownymous | This makes sense. Thank you! | |
| Feb 5, 2022 at 23:08 | comment | added | Paul Levy | Sorry, scratch that. Easy explanation: let $x\in{\rm Lie}(K)$ and $y\in{\mathfrak u}$. Then there's a Borel subalgebra ${\mathfrak b}$ of ${\rm Lie}(K)$ containing $x$, and then ${\mathfrak u}$ is contained in the nilradical of ${\mathfrak b}$, and we may assume ${\mathfrak b}$ consists of upper triangular matrices, then it's easy to see ${\rm tr}(xy)=0$. | |
| Feb 4, 2022 at 16:28 | comment | added | unknownymous | I'm not seeing why this helps | |
| Feb 2, 2022 at 23:04 | comment | added | Paul Levy | Levi decomposition ${\rm Lie}(K)={\mathfrak l}\ltimes({\mathfrak s}\oplus{\mathfrak u})$ where ${\mathfrak l}$ is semisimple, ${\mathfrak s}$ is a torus and ${\mathfrak s}\oplus{\mathfrak u}$ is the radical. Now ${\mathfrak l}$ is perfect and we can assume ${\mathfrak s}$ consists of diagonal matrices. | |
| Jan 26, 2022 at 13:16 | comment | added | unknownymous | Maybe I am missing something simple, but I can't follow the last step. We have $\mathfrak{u}$ an ideal of $Lie(K)$ consisting of upper triangular matrices. Why does this contradict the non-degeneracy of the trace form? | |
| Sep 6, 2021 at 22:17 | comment | added | Paul Levy | For a semisimple element of $H$, the centraliser is a pseudo-Levi subgroup (subset of the affine Dynkin diagram). For an arbitrary subgroup of $H$, I think almost anything can occur, e.g. there is an $F_4$ in $E_8$ with centraliser equal to $G_2$ (and vice versa). The centralisers of ${\rm SL} _2$s are the "reductive parts" of centralisers of nilpotent elements, so can be found e.g. in Collingwood-McGovern ("Nilpotent Orbits...") or Carter's "Finite Groups of Lie type" (with some subtleties around component groups). | |
| Sep 6, 2021 at 20:52 | comment | added | unknownymous | Could I ask a follow-up question? Given the case $\rho : \mathbb{C}^{*} \to H$ (or more generally given a real semisimple element of the Lie algebra), the centralizer is a Levi subgroup of a parabolic associated to $\rho$. Is there a generalization of this in the setting of my question? | |
| Sep 6, 2021 at 20:47 | history | bounty ended | unknownymous | ||
| Sep 6, 2021 at 20:47 | vote | accept | unknownymous | ||
| Sep 6, 2021 at 20:47 | comment | added | unknownymous | That makes sense. Thank you! Although I am not sure that closed connected subgroups of $G$ are in $1-1$ correspondence with Lie subalgebras of $\mathfrak{g}$ in the case that $G$ is only reductive. But this doesn't seem to affect the argument. | |
| Sep 6, 2021 at 13:45 | comment | added | Paul Levy | Apologies, yes I did flip $G$ and $H$. I have included some more detail - I'm not sure what else you think is unclear. | |
| Sep 6, 2021 at 13:42 | history | edited | Paul Levy | CC BY-SA 4.0 |
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| Sep 5, 2021 at 21:58 | comment | added | unknownymous | Could you please give some more detail? (Also, in your answer, are your $G$ and $H$ flipped from the notation in the question?) | |
| Sep 5, 2021 at 21:39 | comment | added | Paul Levy | Ok, I want the form to be a trace form - I'm fairly sure ${\rm Lie}({\mathbb G}_a)$ can't have a non-degenerate form coming from a rational representation for ${\mathbb G}_a$. | |
| Sep 5, 2021 at 21:37 | history | edited | Paul Levy | CC BY-SA 4.0 |
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| Sep 5, 2021 at 18:35 | comment | added | Will Sawin | I think another step is needed, because the Lie algebra of $\mathbb G_a$ also has a nondegenerate form. | |
| Sep 5, 2021 at 14:29 | history | edited | Paul Levy | CC BY-SA 4.0 |
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| Sep 5, 2021 at 14:27 | comment | added | Paul Levy | Good point - I guess I just mean that there exists a $G$-equivariant symmetric bilinear form on ${\mathfrak g}$. Any such form (it doesn't have to be the Killing form) will work for this argument. | |
| Sep 5, 2021 at 13:25 | comment | added | LSpice | What is the Killing form on a reductive (as opposed to semisimple) group? Do you restrict from some embedding into $\operatorname{SL}(V)$? | |
| Sep 5, 2021 at 13:03 | history | answered | Paul Levy | CC BY-SA 4.0 |