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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?

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  • $\begingroup$ 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. $\endgroup$ – Moishe Kohan Apr 30 '14 at 17:42
  • $\begingroup$ @studiosus I want only the signs of the eigenvalues to be fixed. $\endgroup$ – Danijel Apr 30 '14 at 17:51
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    $\begingroup$ As I said, the spaces will be homotopy equivalent to each other whether you fix only the signs or the actual eigenvalues. $\endgroup$ – Moishe Kohan Apr 30 '14 at 17:53
  • $\begingroup$ @studiosus Thanks, that helped! Which literature do you recommend for introduction to the topology of flag manifolds? $\endgroup$ – Danijel Apr 30 '14 at 17:58
  • $\begingroup$ I was stupid and should have said "Grassmannian", see my answer for the details. $\endgroup$ – Moishe Kohan Apr 30 '14 at 18:59
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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".

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