Topology of space of symmetric matrices with fixed number of positive and negative eigenvalues Let $M$ be real non-singular symmetric $n \times n$ matrix with $p$ positive and $n-p$ negative eigenvalues. What is the topology of the space of such matrices?
For a trivial case $n=1$ the matrix is an ordinary number (which is also its eigenvalue), there are two 1-dimensional spaces with a trivial topology, one for negative and one for positive eigenvalue.
For $n=2$, if eigenvalues have the same sign, space of such matrices is 3-dimensional space of trivial topology, but if they have different signs, then space of such matrices is isomorphic to $\mathbb R^2 \times S^1$.
How to find the topology for a general case of $p$ positive and $n-p$ negative eigenvalues?

Here are my comments as an answer (with few more details):
Suppose you fix the eigenvalues and set $q=n−p$. Then, in view of the orthogonal diagonalization of quadratic forms, what you get is the homogeneous space $$O(n)/(O(n)\cap O(p,q))=O(n)/(O(p)\times O(q))$$
which is the Grassmannian $G_p(R^n)$ of $p$-dimensional subspaces in $R^n$. There is a huge literature describing topology of Grassmannians. If you do not want to fix the eigenvalues, only their sign, you get a certain $R^n$-bundle over the above Grassmannian. The homotopy type, of course, will be the same. 
See Books on topology and geometry of Grassmannians for references on topology of Grassmannians. In addition to the answers there, Hatcher has a good discussion of Grassmannians in his "Algebraic Topology". 
