# 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?

• Suppose you fix the eigenvalues and set $q=n-p$. Then what you get is the homogeneous space $O(n)/(O(n)\cap O(p,q))=O(n)/(O(p)\times O(q))$ (assuming $p\ne q$), which is a certain real flag-manifold. There is a huge literature describing topology of flag manifolds. If you do not want to fix the eigenvalues, you get a certain $R^n$-bundle over the above flag manifold. – Moishe Kohan Apr 30 '14 at 17:42
• @studiosus I want only the signs of the eigenvalues to be fixed. – Danijel Apr 30 '14 at 17:51
• As I said, the spaces will be homotopy equivalent to each other whether you fix only the signs or the actual eigenvalues. – Moishe Kohan Apr 30 '14 at 17:53
• @studiosus Thanks, that helped! Which literature do you recommend for introduction to the topology of flag manifolds? – Danijel Apr 30 '14 at 17:58
• I was stupid and should have said "Grassmannian", see my answer for the details. – Moishe Kohan Apr 30 '14 at 18:59

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.