Timeline for Connectedness of stabilizer of regular element
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2019 at 19:12 | answer | added | LSpice | timeline score: 1 | |
| Oct 3, 2019 at 16:18 | comment | added | LSpice | (For "semisimple group" in my comment read "simply connected group".) So I should also have said that I disagree with your comment: an element (at least on the group level, if not, in characteristic 0, on the Lie-algebra level) can have connected centraliser in the simply connected, but not the adjoint, group. | |
| Oct 3, 2019 at 16:05 | comment | added | LSpice | By the way, your argument in the semisimple case doesn't work: it is true that the image in the adjoint group of the centraliser in the semisimple group is connected, but the centraliser in the adjoint group can, a priori, be strictly larger. (This happens with @YCor's example of $g = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$, whose centraliser in $\operatorname{PGL}_2(\mathbb C)$ is bigger than the image of its centraliser in $\operatorname{SL}_2(\mathbb C)$; but that's on the group side.) The conclusion is still true (in characteristic 0), just not the proof. | |
| Oct 3, 2019 at 15:55 | comment | added | AThomas | @PraphullaKoushik: the adjoint group $G$ is the unique Lie group with Lie algebra $\mathfrak{g}$ which is centerless. If the centralizer is connected in a covering space, of course it is also connected in the adjoint group, but the converse is of course not true. | |
| Oct 3, 2019 at 14:43 | comment | added | LSpice | The centraliser of an arbitrary element is the centraliser in the centraliser of the semisimple part, of the (regular there) nilpotent part. | |
| Oct 3, 2019 at 14:18 | comment | added | YCor | At least this is true in $\mathrm{PGL}_n(\mathbf{C})$, without assuming regular. Indeed, for $x\in\mathfrak{sl}_n(\mathbf{C})$, the centralizer $C_x$ of $x$ in $\mathrm{GL}_n(\mathbf{C})$ is connected (centralizers in $\mathrm{GL}_n(\mathbf{C})$ are connected, and this applies to $x+t$ for suitable scalar $t$). Hence the centralizer $C_x/\mathbf{C}^*$ of $x$ in $\mathrm{PGL}_n(\mathbf{C})$ is connected. (On the other hand, in $\mathrm{PGL}_2(\mathbf{C})$ there are regular elements inside the group itself, with non-connected centralizer.) | |
| Oct 3, 2019 at 14:08 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Oct 3, 2019 at 13:34 | comment | added | Praphulla Koushik | I don't know what is the adjoint group.. can I assume $G$ is a Lie group and $\mathfrak{g}$ be it's Lie algebra?? | |
| Oct 3, 2019 at 13:31 | history | asked | AThomas | CC BY-SA 4.0 |