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Nov 3, 2020 at 17:06 history edited Malkoun CC BY-SA 4.0
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S Nov 3, 2020 at 14:19 history suggested gmvh
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Nov 3, 2020 at 8:44 review Suggested edits
S Nov 3, 2020 at 14:19
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
Nov 3, 2020 at 3:44 history asked Malkoun CC BY-SA 4.0