Orthogonal group acts on symmetric matrices Let $O_n(\mathbb{R})$ act on $Sym_n(\mathbb{R})$, the symmetric matrices with real entries, via $S \mapsto A^T SA$ for $A \in O_2(\mathbb{R})$ and $Sym_n(\mathbb{R})$. What is the space of orbits $O_n(\mathbb{R})/Sym_n(\mathbb{R})$ as a set (and what is a basis for the topology)?
I know that we can diagonalize a symmetric matrix $A$ with $Q\in O_n(\mathbb{R})$ such that $QSQ^{-1}$ is diagnonal but I don't know how to continue. Thanks a lot for your help!
 A: Every orbit contains exactly one diagonal matrix $$\begin{pmatrix} a & 0 \\ 0 & b\end{pmatrix},$$ with $a\geq b,$ so the quotient space is exactly the set of pairs $(x, y)\in \mathbb{R}^2,$ with $x\geq y.$ Notice that the answer would be different for quotient by $SO(2),$ since then you can't sort, because the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ has determinant $-1.$
A: Each orbit contains a diagonal matrix of the form
$$
\pmatrix{\lambda_1\\&\lambda_2}
$$
with $\lambda_1 \geq \lambda_2$.  This gives us a homeomorphism between $Sym_2(\mathbb{R})/O_2(\Bbb R)$ and $\Bbb R/ [(x,y) \sim (y,x)]$.  That is, $Sym_2(\mathbb{R})/O_2(\Bbb R)$ is homeomorphic to the set of unordered real pairs.

We have the following basis for $\Bbb R/ [(x,y) \sim (y,x)]$: for every $(x,y)$ with $x \neq y$, take every neighborhood of $(x,y)$ in $\Bbb R$ which is small enough to miss the line $y = x$.  For every point $(x,x)$, consider the usual neighborhoods, but exclude any points $(x,y)$ for which $x < y$.
