Timeline for What is the orbit of the standard conformal structure on $S^2$ under $\operatorname{SL}(3,\mathbb{R})$?
Current License: CC BY-SA 4.0
15 events
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Nov 3, 2020 at 17:06 | history | edited | Malkoun | CC BY-SA 4.0 |
edited title
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S Nov 3, 2020 at 14:19 | history | suggested | gmvh |
Added top-level tag and "group-actions" tag
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Nov 3, 2020 at 8:44 | review | Suggested edits | |||
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Nov 3, 2020 at 5:38 | answer | added | Malkoun | timeline score: 1 | |
Nov 3, 2020 at 4:18 | comment | added | Malkoun | I edited the question, taking into account your comments. Thank you! | |
Nov 3, 2020 at 4:16 | history | edited | Malkoun | CC BY-SA 4.0 |
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Nov 3, 2020 at 4:15 | comment | added | Ian Agol | $PSL_2(\mathbb{C})$ is isogenous to $O(3,1;\mathbb{R})$ by the hermitian action you describe. But I don't think that there's a 3-dim. rep. | |
Nov 3, 2020 at 4:13 | comment | added | Ian Agol | Why then is $gxg^*$ trace-free too? | |
Nov 3, 2020 at 4:13 | comment | added | Malkoun | @LSpice, you are right. | |
Nov 3, 2020 at 4:11 | comment | added | LSpice | Why does that preserve $\mathbb R^3$? It seems to me that we have $\operatorname{diag}(2, 1/2)\cdot\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & -1/4 \end{pmatrix}$, which is not trace free. (Maybe I don't know what $*$ means in this context.) | |
Nov 3, 2020 at 4:03 | comment | added | LSpice | @IanAgol, I'm not sure it is a subgroup, but I don't think that's an obstruction; $\operatorname{PGL}(3, \mathbb R)$ contains $\operatorname{GL}(2, \mathbb R)$, which has such subgroups. | |
Nov 3, 2020 at 3:57 | history | edited | Malkoun | CC BY-SA 4.0 |
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Nov 3, 2020 at 3:54 | comment | added | LSpice | Shouldn't you be looking at the quotient $\operatorname{GL}_+(3, \mathbb R)/\mathbb R_+^\times \cong \operatorname{SL}(3, \mathbb R)$, since scalars act trivially on the sphere at infinity? | |
Nov 3, 2020 at 3:53 | history | edited | LSpice | CC BY-SA 4.0 |
\DeclareMathOperator
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Nov 3, 2020 at 3:44 | history | asked | Malkoun | CC BY-SA 4.0 |