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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!

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    Do you mean the set of orbits $Sym_2(\mathbb{R})/O_2(\mathbb{R})$? – 2017-01-17
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    yes I do, added. – 2017-01-17

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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 $x < y$.

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    Thank you, does this statement generalize to unordered $n$-tuples of the reals? – 2017-01-17
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    Yes, it generalizes exactly in that way. Interestingly, the resulting space is non-orientable. In the $2 \times 2$ case, I think you have something homeomorphic to an open Mobius strip. – 2017-01-17
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    Can you give a basis of the topology of the orbit space? – 2017-01-17
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    And: Can one say that the that symmetric matrices of the same eigenvalues are identified in this orbit space? – 2017-01-17
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    To that last bit: yes, that's exactly what we're saying – 2017-01-17
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    It just comes to my mind: Is it possible to just identify the matrices with same signature or would this require an action by honest regular matrices (that are not necessarily orthogonal)? – 2017-01-17
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    If a regular matrix means the same thing as an [invertible matrix](https://en.wikipedia.org/wiki/Invertible_matrix), then that's exactly right. – 2017-01-18
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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 $-1.$

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    Can one generalize this result for arbitrary $n$? – 2017-01-17
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    Thanks for your answer and the very useful remark on sorting with $\begin{pmatrix}0&1\\1&0\end{pmatrix}\in O(n)$. – 2017-01-17