I am getting discriminant to be $(1/4)(s-3)^2(-s^4 + 6s^3 - 24s^2 +56s - 60)$. The remaining polynomial of degree $4$ is negative for all $s$.

Consider the polynomial of degree three with roots $x,y,z$. The information given implies that it is equal to $f(t)=t^3-st^2 +(s^2/2 -3s/2 + 3)t -1$. Three roots are real iff the discriminant is nonnegative. This gives you a polynomial inequality on $s$, which you presumably can check.

Decent question, wrong forum.

If the probabilities $p_i$ are very small, then in more than half the cases the total number will be even $0$. You have to at least make some assumptions on $p_i$.

I consider it highly unlikely that the critical points are always nondegenerate. Intuitively, you can imagine a finite number of such points for a generic choice of $q$, which could then "collide" for special choices of $q$. Even taking something as simple as regular $n$-gon and all $q=1$ might give a degenerate critical point at the origin (but I haven't done the calculation).

Just a reference, we have a proof.

@YCor Thank you, this makes more sense than trying to somehow think of the boundary.

Is my intuition correct that the boundary of any small neighborhood of the singular point would have a nontrivial fundamental group, and this should not happen for topological manifolds? Or am I being naive here?

Good question. I think the proof that we have works for $S$ complex if you allow $P$ to be unitary. But we only need it for $S$ real and then can do it for $P$ orthogonal.

The square of such matrix would not be diagonalizable, let alone real symmetric.

I found the relevant page online, it's not the right result. It is basically trivial to prove the statement without the orthogonality assumption on $P$. And it is not particularly hard to prove it with the orthogonality condition, but so far I didn't see a reference.

@Suvrit No, I haven't looked at this book. Will take a look at it on Tuesday at the departmental library.

At least in characteristics zero, Lefshetz hyperplanes theorem, together with $h^{1,0}\neq 0$ for abelian varieties shows that $n=2$ is the only case.

Just as a comment: in the case of $S$ being an involution, the statement is pretty much equivalent to the principal angles between subspaces story.

I don't know, it could be an exercise in some textbook, although it could be just a touch too tricky for that.

I don't know, it may be a difficult problem. Have you tried to code it in, say, dimension three or four and just see what you get?