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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}\DeclareMathOperator\PSL{PSL}$Consider the group $\GL_+(3,\mathbb{R})$ acting on $\mathbb{R}^3$. It induces an action of $\GL_+(3,\mathbb{R})/\mathbb{R}_+$, which is isomorphic to $SL(3,\mathbb{R})$, on the "sphere at infinity". Moreover, the group of Moebius transformations $\PSL(2,\mathbb{C})$ of the sphere at infinity is a $6$-dimensional group (over $\mathbb{R}$). What is the $\SL(3,\mathbb{R})$-orbit of the standard conformal structure $c$ on the $2$-sphere at infinity?

Could someone perhaps describe this homogeneous space more concretely? (I took LSpice's and IanAgol's comments into consideration and re-edited my post!).

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    $\begingroup$ @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. $\endgroup$
    – LSpice
    Commented Nov 3, 2020 at 4:03
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    $\begingroup$ 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.) $\endgroup$
    – LSpice
    Commented Nov 3, 2020 at 4:11
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    $\begingroup$ @LSpice, you are right. $\endgroup$
    – Malkoun
    Commented Nov 3, 2020 at 4:13
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    $\begingroup$ $PSL_2(\mathbb{C})$ 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. $\endgroup$
    – Ian Agol
    Commented Nov 3, 2020 at 4:15
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    $\begingroup$ I edited the question, taking into account your comments. Thank you! $\endgroup$
    – Malkoun
    Commented Nov 3, 2020 at 4:18

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$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SL{SL}$It is clear that $\SO(3,\mathbb{R}) \subset \SL(3,\mathbb{R})$ acts isometrically, and thus by Moebius transformations, on the sphere at infinity.

Moreover, any $g \in \SL(3,\mathbb{R})$ can be written, by the polar decomposition, as

$$g = pq$$

where $q \in \SO(3)$ and $p$ is a real symmetric positive-definite $3$-by-$3$ matrix of determinant $1$. So the action of $g$ on the conformal structure $c$ is the same as the action of $p$ on the conformal structure $c$.

I claim that if $p \neq \mathbf{1}$, then $p^*(c) \neq c$, i.e. $p$ (and therefore $g$) does not induce a Moebius transformation on the sphere at infinity.

To see that, note that one can always find an orthonormal basis of $\mathbb{R}^3$, with respect to which $p$ becomes diagonal, say $p$ becomes $\operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)$. If $p \neq \mathbf{1}$, then there exist two diagonal elements which are not equal, say $\lambda_1 \neq \lambda_2$, WLOG. Consider then the point $N = (0,0,1)^T \in S^2_\infty$ (in the new coordinates). Then the Euclidean metric $g(N) = dx^2 + dy^2$ at the point $N$ gets pulled back by $p$ to $\lambda_1^2 dx^2 + \lambda_2^2 dy^2$ which is not a positive multiple of $g(N)$. This shows that the action of $p \neq \mathbf{1}$ is not conformal.

We thus conclude that the subgroup of $\SL(3,\mathbb{R})$ which acts by Moebius transformations on the sphere at infinity is actually $\SO(3,\mathbb{R})$, which actually acts isometrically on the $S^2_\infty$ with the round metric.

The orbit of $c$ is thus simply $\SL(3,\mathbb{R})/\SO(3,\mathbb{R})$.

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  • $\begingroup$ I thought your question was about the nature of the homogeneous space, not the identification of the orbit as a homogeneous space? (Of course, as you point out, the polar decomposition in a sense is a description of the homogeneous space.) $\endgroup$
    – LSpice
    Commented Nov 3, 2020 at 19:51
  • $\begingroup$ well, this kind of answers my question. By the way, I have an easier argument for why only rotations are conformal in this case. Roughly speaking, thinking of the Minkowski picture, you are splitting spacetime into space and time in a specific way, thus breaking the symmetry of the group of Moebius transformations, and not allowing for "boosts". I hope it is clear enough... $\endgroup$
    – Malkoun
    Commented Nov 3, 2020 at 20:16

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