Signature is constant on a rank stratum of symmetric matrices

Idea. Signs of nonzero eigenvalues are stable, and with the rank constant, zero eigenvalues cannot be created or destroyed, so no eigenvalue can cross $0$. Conversely, inside one signature all matrices can be deformed into a standard one.

Part 1: lower semicontinuity of $p$ and $q$. By Sylvester's law of inertia (quoted), $p(S)$ is the largest dimension of a subspace $V$ on which $x^TSx>0$ for all $x\ne0$ (take the span of eigenvectors with positive eigenvalues; and no larger subspace works). Let $V$ be such a subspace of dimension $p(S)$. The unit sphere of $V$ is compact and $x\mapsto x^TSx$ is positive there, so its minimum is some $c>0$. If $\|S'-S\|<c$ then for unit $x\in V$,
$$x^TS'x\ge x^TSx-\|S'-S\|>0,$$
so $V$ also works for $S'$ and $p(S')\ge p(S)$. The same argument applied to $-S$ gives $q(S')\ge q(S)$.

If now $S'$ has the same rank $r$ as $S$ and is close to $S$, we have $p(S')+q(S')=r=p(S)+q(S)$ together with $p(S')\ge p(S)$ and $q(S')\ge q(S)$; hence $p(S')=p(S)$ and $q(S')=q(S)$. So $S\mapsto(p,q)$ is locally constant on the rank-$r$ set, and therefore constant along continuous paths (a locally constant function on an interval is constant, by connectedness of the interval).

Part 2: each stratum is path connected. Let $\mathcal S_{p,q}$ ($p+q=r$) be the set of symmetric matrices of signature $(p,q)$. Take $S\in\mathcal S_{p,q}$ and write $S=Q\Lambda Q^T$ with $Q\in O(n)$ and $\Lambda=\mathrm{diag}(\lambda_i)$. Let $D=\mathrm{diag}(d_i)$ with $d_i=\operatorname{sign}\lambda_i\in\{1,-1,0\}$.

*(a) Normalise the eigenvalues.* For $t\in[0,1]$, put $S_t=Q\bigl((1-t)\Lambda+tD\bigr)Q^T$. Each diagonal entry $(1-t)\lambda_i+td_i$ is a convex combination of two numbers of the same sign (or both $0$), so it keeps the sign of $\lambda_i$. Thus $S_t\in\mathcal S_{p,q}$ and $S_0=S$, $S_1=QDQ^T$.

*(b) Reduce to $Q\in SO(n)$.* If $\det Q=-1$ let $F=\mathrm{diag}(-1,1,\dots,1)$. The diagonal matrices $F$ and $D$ commute, so $(QF)D(QF)^T=QFDF^TQ^T=QDQ^T$ and $\det(QF)=1$. So we may assume $Q\in SO(n)$.

*(c) Connect to $D$.* $SO(n)$ is path connected (quoted; it is shown in the connectedness chapter). Take a path $Q_t$ in $SO(n)$ from $Q$ to $I$. Then $t\mapsto Q_tDQ_t^T$ is a path in $\mathcal S_{p,q}$ (orthogonal conjugation preserves eigenvalues) from $QDQ^T$ to $D$.

So every point of $\mathcal S_{p,q}$ is joined by a path in $\mathcal S_{p,q}$ to the fixed matrix $D_{p,q}=\mathrm{diag}(I_p,-I_q,0)$. Hence each $\mathcal S_{p,q}$ is path connected.

Conclusion. The rank-$r$ set is the disjoint union of the sets $\mathcal S_{p,q}$ with $p+q=r$ ($p=0,1,\dots,r$, so $r+1$ sets), each path connected, and by Part 1 no path leaves a given $\mathcal S_{p,q}$. Therefore these are exactly the path components, and there are $r+1$ of them. (Each is also open in the rank-$r$ set by local constancy of the signature, so they are also the connected components.)

Remark. For $r=n$ this says that the nondegenerate symmetric matrices form $n+1$ components, indexed by $p=0,\dots,n$; the component $p=n$ is the open cone of positive definite matrices, which is convex. For $r=0$ the set is $\{0\}$, one component. Note that the problem would not be true if we allowed the rank to vary: the positive definite matrices and the negative definite ones are joined through singular matrices.

Lesson. A discrete invariant that is lower semicontinuous and whose total is fixed (here $p+q=r$) is locally constant.