Timeline for Properties of stabilizers of adjoint action general linear group
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2021 at 13:29 | vote | accept | Tommaso Scognamiglio | ||
| Apr 16, 2021 at 11:27 | answer | added | Jef | timeline score: 5 | |
| Apr 16, 2021 at 11:05 | comment | added | Tommaso Scognamiglio | Actually if each Jordan block has different size the group seems more pleasant. In the situation where one has different Jordan blocks of the same size,I wasn't really able to find relevant normal subgroups even in the case of a $4 \times 4$ matrix with two blocks of size $2$. Do you have any idea whether the guess is true or not? | |
| Apr 16, 2021 at 10:23 | comment | added | Jef | I agree, but at least you only need to understand their reductive quotients. This is because an extension of special groups is special and a unipotent group over $\mathbb{C}$ is special. So if $x$ is nilpotent, the unipotent radical $R_u(C(x))$ is special and we just need to know that $C(x)/R_u(C(x))$ is special. | |
| Apr 16, 2021 at 10:16 | comment | added | Tommaso Scognamiglio | I've tried to do that but the centralizer of a nilpotent matrix seems not to easy to compute actually | |
| Apr 15, 2021 at 19:41 | comment | added | Jef | You might as well take $x$ to be any $n\times n$-matrix, and then using the Jordan decomposition you can reduce the question to looking at centralizers of nilpotent elements in Levi subgroups of $GL_n(\mathbb{C})$, i.e. nilpotent elements in $GL_{n_1} \times \dots \times GL_{n_k}$. It therefore suffices to consider nilpotent elements in $GL_n$, by looking at each factor individually. | |
| Apr 14, 2021 at 8:58 | history | edited | Tommaso Scognamiglio | CC BY-SA 4.0 |
added 71 characters in body; edited tags
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| Apr 13, 2021 at 17:59 | history | asked | Tommaso Scognamiglio | CC BY-SA 4.0 |