Timeline for An upper bound on the dimension of a subalgebra of $\mathfrak{so}(p,q)$ with non-trivial centre
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2022 at 14:58 | vote | accept | MathQuest | ||
| Mar 6, 2022 at 11:04 | answer | added | YCor | timeline score: 3 | |
| Mar 6, 2022 at 10:46 | comment | added | YCor | Oops, the bound should be codimension $2n-4$ regardless of $p,q$. I'll try to post an answer with details. | |
| Mar 6, 2022 at 9:57 | comment | added | MathQuest | Thanks for the comment. Indeed, that is where I am coming from. I am looking for the maximal dimension of centralizers of non-trivial elements. Maximal subgroups are only a crude upper bound, not structurally related to the question at hand. | |
| Mar 6, 2022 at 9:46 | comment | added | YCor | This is the same as asking the maximal dimension of centralizers of nontrivial elements, and shouldn't require the classification of maximal subgroups. For example, for $\mathfrak{so}(n)$ it seems to be $\lfloor n/2\rfloor^2$. For the split case $\mathfrak{so}(p,q)$ with $p+q=n$, $p\le q\le p+1$, it should be $n(n-1)/2 - (2n-4)$. This corresponds to the dimension of the centralizer of a root space. | |
| Mar 5, 2022 at 20:12 | history | edited | YCor | CC BY-SA 4.0 |
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| S Mar 5, 2022 at 19:16 | review | First questions | |||
| Mar 5, 2022 at 19:22 | |||||
| S Mar 5, 2022 at 19:16 | history | asked | MathQuest | CC BY-SA 4.0 |