Timeline for Choice of generators to make the centralisers connected
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2022 at 0:29 | comment | added | user488802 | I appreciate all your input and time! Merry Christmas in advance! | |
| Dec 18, 2022 at 0:24 | comment | added | LSpice | My repeated (1 2 3) claim about intersections being possibly disconnected is false (I forgot to think specifically in $\operatorname{PGL}$), but I don't know about the question in general. I am thinking about it. | |
| Dec 17, 2022 at 22:41 | comment | added | LSpice | Yes, if your group is contained in a torus, then you may arrange (over $\mathbb C$) that that torus is whichever (maximal) torus you prefer; for example, the (image of the) diagonal torus. However, it still does not force the intersection of the identity components of the various centralisers to be connected. | |
| Dec 17, 2022 at 21:53 | history | edited | user488802 | CC BY-SA 4.0 |
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| Dec 17, 2022 at 21:49 | comment | added | user488802 | The settings I am working in are in fact conjugacy classes of toral elementary abelian 2-subgroups in $G$. So are you saying I cannot assume each such class has a representative in the $T$? I thought all the maximal tori are conjugate. Got a bit confused. Thank you! | |
| Dec 17, 2022 at 21:30 | comment | added | LSpice | Re, no. First, while every individual involution can be conjugated into a fixed torus, they cannot all be conjugated simultaneously into a common torus (not even necessarily if they commute). But maybe you meant to assume this. Even if so, then it need not be true that the intersection of the connected centralisers is connected. | |
| Dec 17, 2022 at 21:27 | comment | added | user488802 | Thank you! So we agree that my question makes sense given that I assumed all the toral groups have been conjugated into $T$? | |
| Dec 17, 2022 at 21:23 | comment | added | LSpice | Yes, if all your elements come from a common torus, then you can describe centralisers and connected centralisers easily. It's clear in complete generality that $H \mathrel{:=} \bigcap_{i \in I} C_G(e_i)^\circ$ is between $C_G(e_i : i \in I)$ and its identity component, but there is no reason in general that $H$ should be connected. | |
| Dec 17, 2022 at 20:01 | comment | added | LSpice | I do not follow your claim about elementary Abelian 2-groups. In answer to another question of yours, I showed that $C_G(e_1, e_2)^\circ$ equals $C_G(e_1)^\circ \cap C_G(e_2)^\circ$ if $e_1$ and $e_2$ lie in a common torus. But I do not know that the claim is true if $e_1$ and $e_2$ do not lie in a common torus, and I know that it can fail if you replace two involutions by three, even if they lie in a common torus. So how do you deduce your claim about elementary Abelian 2-groups? | |
| Dec 17, 2022 at 9:04 | history | edited | user488802 | CC BY-SA 4.0 |
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| Dec 17, 2022 at 8:56 | history | edited | user488802 | CC BY-SA 4.0 |
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| Dec 17, 2022 at 8:37 | history | asked | user488802 | CC BY-SA 4.0 |