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I do not know if this is the right place to ask the following question. If it is not, I will delete it. I asked a similar question in math stack exchange and get a nice answer by @YCor.

https://math.stackexchange.com/questions/1924189/quotient-of-textrmgl2-textbfr-by-the-conjugate-action-of-textrmso

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

I thank @ThiKu for his nice answer in this case.

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  • $\begingroup$ The question for $SO(2)$ is now answered in MathSE (link above) $\endgroup$
    – YCor
    Sep 13, 2016 at 20:39
  • $\begingroup$ @YCor Thanks for your nice answer, which makes me think whether I should delete this post. But since I accept and answer, it seems that I cannot delete this one. $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 22:06
  • $\begingroup$ One option would be that you edit to only leave the question about quotient by $SL_2$, and that enough people vote to move it to MathSE. $\endgroup$
    – YCor
    Sep 13, 2016 at 22:09

1 Answer 1

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No. By the Jordan normal form theorem every matrix in $GL(2,{\bf R})$ is conjugate to

  • either a diagonal matrix $$\left(\begin{array}{cc}\lambda_1&0\\ 0&\lambda_2\end{array}\right)$$ (clearly the set of these classes is homeomorphic to $({\bf R}\setminus 0)^2/({\bf Z}/2{\bf Z})$ with ${\bf Z}/2{\bf Z}$ acting by permutation of the two coordinates.)

  • or to a multiple of a triangular matrix $$\left(\begin{array}{cc}\pm 1&n\\ 0&\pm 1\end{array}\right)$$

  • or to a multiple of a matrix in $SO(2)$.

However, the triangular matrices with eigenvalues $1$ are all in one of three orbits because of $$\left(\begin{array}{cc}\lambda_1&0\\ 0&\lambda_1^{-1}\end{array}\right)\left(\begin{array}{cc}1&n\\ 0&1\end{array}\right)\left(\begin{array}{cc}\lambda_1^{-1}&0\\ 0&\lambda_1\end{array}\right)=\left(\begin{array}{cc}1&\lambda_1^2n\\ 0&1\end{array}\right).$$ The three orbits correspond to the cases $n>0, n=0, n<0$. Note that the orbits for $n>0$ and $n<0$ can not be separated from the identity matrix (the case $n=0$) by any open neighborhood of the identity, because $\lambda_1^2n$ tends to zero for a suitable sequence of $\lambda_1$'s.

So the quotient space is non-Hausdorff, it contains of a copy of $({\bf R}\setminus 0)^2/({\bf Z}/2{\bf Z})$ and another two points which can not be separated from $(1,1)$.

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  • $\begingroup$ I am sorry. I cannot understand some points. For example, how do you get every matrix in $\textrm{GL}_2(\textbf{R})$ is conjugate to the given form under the conjugation by $\textrm{SL}_2(\textbf{R})$? I don't know how to prove this even under the conjugation action of $\textrm{GL}_2(\textbf{R})$ because $\textbf{R}$ is not algebraically closed. In your second point, you showed that all the unipotent matrices are in the same orbit under the action of $\textrm{GL}_2(\textbf{R})$, but for $\textbf{SL_2}(\textbf{R})$-conjugacy class, I guess it's different. $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 1:27
  • $\begingroup$ It doesn't really make a difference whether you conjugate by GL(2,R) or SL(2,R) because every matrix in GL^+(2,R) is a multiple of a matrix in SL(2,R), and conjugation by a multiple of the identity acts trivially. So the only difference is that you have an additional Z/2Z-action coming from matrices with negative determinant. In any case, I have now edited the answer to consider only the SL(2,R)-action. $\endgroup$
    – ThiKu
    Sep 13, 2016 at 1:38
  • $\begingroup$ I am sorry. I am still confused at some points. What do you mean by "every matrix in $\textrm{GL}_2(\textbf{R})$" is a multiple of a matrix in $\textrm{SL}_2(\textbf{R})$? For example, how do you write diag(-1,1) as "a multiple of a matrix in $\textrm{SL}_2(\textbf{R})$? $\endgroup$
    – Q. Zhang
    Sep 13, 2016 at 1:46
  • $\begingroup$ For the normal form: I kind of ignore matrices with non-real eigenvalues because already those with real eigenvalues yield a non-Hausdorff space, so the same must be true for the larger space. $\endgroup$
    – ThiKu
    Sep 13, 2016 at 1:46
  • $\begingroup$ For your last question: I meant matrices in GL^+(2,R), i.e., those with positive determinant. $\endgroup$
    – ThiKu
    Sep 13, 2016 at 1:48

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