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.



Quotient of $\textrm{GL}(2,\textbf{R})$ by the conjugate action of $\textrm{SL}(2,\textbf{R})$
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 $(1,1)$.


