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Let $\textrm{SO}_2(\textbf{R})$ act on $\textrm{GL}_2(\textbf{R})$ by conjugation. Does the quotient $\textrm{GL}_2(\textbf{R})/\textrm{SO}_2(\textbf{R})$ exist as a manifold?

I asked the same question in MO and got a negative answer for the quotient $\textrm{GL}_2(\textbf{R})/\textrm{SL}_2(\textbf{R})$. Thus I really would like to know if the quotient $\textrm{GL}_2(\textbf{R})/\textrm{SO}_2(\textbf{R})$ exist as a manifold?

https://mathoverflow.net/questions/249724/quotient-of-textrmgl2-textbfr-by-the-conjugate-action-of-textrmsl

Let $X$ be a manifold and $R$ be an equivalence relation of $X$. There is a criterion for the quotient $X/R$ to exist in Serre's book "Lie algebras and Lie groups", which says the following. Let $Gr\subset X\times X$ be the graph of $R$. Then the quotient $X/R$ exists as a manifold if and only if

(1) $Gr\subset X\times X$ is a closed sub manifold (i.e., the inclusion $Gr\rightarrow X\times X$ is a closed embedding), and

(2) the projection map $pr_1: Gr\rightarrow X$ is a submersion.

But I do not know how to check the corresponding maps are immersion and submersion or not in this case.

Anybody can help? Any comment, suggestion will be appreciated.

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    $\begingroup$ Did you check whether the action is proper and free? $\endgroup$
    – Math Lover
    Sep 12, 2016 at 18:34
  • $\begingroup$ @VictorBarg If my understanding is correct, free means that the stabilizer of any point is trivial. But the stabilizer of 1 is the whole $SL_2(\textbf{R})$. $\endgroup$
    – Q. Zhang
    Sep 12, 2016 at 18:36
  • $\begingroup$ Just to clarify, is it $SO(2)$ or $SL_2(\mathbb{R})$ that's acting? If it's $SO(2)$, then the action is proper by compactness. $\endgroup$
    – PeterJL
    Sep 13, 2016 at 3:53
  • $\begingroup$ I changed the problem from SL2 to SO2. It is proper now $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 5:24

1 Answer 1

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The question with $\mathrm{SL}_2$ is clearly negative since orbits are not all closed.

For $\mathrm{SO}(2)$, since the acting group is compact, the quotient is Hausdorff so this is more reasonable. Actually, you get a manifold with boundary, which can be completely described:

Every matrix in $M_2(\mathbf{R})$ decomposes as: $$M=\begin{pmatrix} a & b \\ c & d\end{pmatrix}=\frac{a+d}2\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}+\frac{b-c}2\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}+\frac12\begin{pmatrix} a-d & b+c \\ b+c & d-a\end{pmatrix}.$$

The conjugation by the matrix $R_t=\begin{pmatrix} \cos t& -\sin t \\ \sin t & \cos t\end{pmatrix}$ acts trivially on the first two coordinates and acts as the matrix $R_{2t}$ on the pair $(a-d,b+c)$. As a consequence, the map $$\phi: M\mapsto (a+d,b-c,(a-d)^2+(b+c)^2)$$ is a faithful invariant, i.e. induces an injection from the quotient $M_2(\mathbf{R})/\mathrm{SO}(2)_{\mathrm{conj}}$ into $\mathbf{R}^3$, and more precisely into $\mathbf{R}^2\times\mathbf{R}_+$, where $\mathbf{R}_+$ denotes non-negative reals. This map is actually surjective, by an immediate verification (use the coordinates $(a+d,a-d,b+c,b-c)$).

If $\phi(M)=(x,y,z)$, we see that $\det(M)=-(z-x^2-y^2)/4$. In particular, $\phi(\mathrm{GL}_2(\mathbf{R})/\mathrm{SO}(2)_{\mathrm{conj}})$ is the open subset of $\mathbf{R}^2\times\mathbf{R}_+$ consisting of those $(x,y,z)$ with $x^2+y^2\neq z$. It has two components: $\{x^2+y^2<z\}$ which is a manifold with no boundary and corresponds to $\phi(\mathrm{GL}_2(\mathbf{R})^-/\mathrm{SO}(2)_{\mathrm{conj}})$, and $\{x^2+y^2>z\ge 0\}$, which corresponds to $\phi(\mathrm{GL}_2(\mathbf{R})^+/\mathrm{SO}(2)_{\mathrm{conj}})$, and contains the boundary $\{z=0\}$, which itself is the (injective) image of $\mathbf{R}_{>0}\mathrm{SO}(2)$ (the group of positive homotheties) in the quotient.

Added: we also see that the quotient $\phi(\mathrm{SL}_2(\mathbf{R})/\mathrm{SO}(2)_{\mathrm{conj}})$ is the surface with boundary $\{(x,y,z):z=x^2+y^2-4,z\ge 0\}$.

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  • $\begingroup$ Thanks for this nice answer. I thought we should use the criterion I quoted above from Serre's book, but I really do not know how to check the corresponding maps are immersive and submersive. $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 22:04
  • $\begingroup$ @QingZhang But in Serre's book I think that a manifold with nonempty boundary is not considered as a manifold. So this shouldn't satisfy the criterion. $\endgroup$
    – YCor
    Sep 13, 2016 at 22:07
  • $\begingroup$ That's correct. But we can consider $(GL_2(R)-R_{>0}SO(2))/SO_2(R)$ in this example according your solution. If there is a way to check whether the given maps are immersive and submersive, I guess we can also compute the boundary explicitly using that method. Given a compact Lie group $G$ acts on a manifold $X$, is it true that there exists an open $G$-invariant subset $X_o\subset X$ such that $X_o/G$ always exists as a manifold? $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 22:12
  • $\begingroup$ I don't know (I guess you want $X_o$ nonempty, or maybe even dense). It's a natural question. $\endgroup$
    – YCor
    Sep 13, 2016 at 22:22
  • $\begingroup$ Yes. I think we expect $X-X_o$ has lower dimension. I really think this should be standard but I don't know where I can find this. In the usual textbook on manifolds, there is a quotient theorem which requires the action is free and proper. The criterion in Serre's book (which is named after Godement, and that result is later published in Bourbaki's elements of math) is not very common in the standard textbook. There are not examples in Serre's book and thus I really would like to see how to apply that criterion. That criterion looks very powerful. $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 22:29

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