This might be ridiculously obvious, but…
For each $n \in \mathbb{N}$, let $M_n$ denote the manifold of $n \times n$ matrices with real entries. It is well known that the $n$-dimensional determinant function $d_n:M_n \to \mathbb{R}$ is a Morse function if and only if $n = 2$ since the zero matrix is a degenerate critical point of $d_n$ whenever $n > 2$. (See the simple homework exercise in this pdf for a proof).
My question is as follows:
For a given $n$, is there a characterization of the sub-manifolds of $M_n$ for which $d_n$ is a Morse function?
Just to be clear, by Morse function I mean that $d_n$ must have:
- Isolated critical points (not necessarily finitely many), and
- Non-degenerate Hessian at each critical point.
For general subspaces of $M_n$ it seems that the critical set of $d_n$ would not be isolated (one can easily construct examples of critical subspaces), so if there is a Morse-Bott theory for submanifolds of $M_n$ that would be good to know as well.
Best Answer
I'm not sure exactly what a characterization would be. Is there a characterization of the submanifolds of $\mathbb R^n$ for which $x_1$ is a Morse function?
Critical points of the submanifold will come in two forms - critical points of $M_n$ that lie on the submanifold, and noncritical points of $M_n$ that become critical in the submanifold. I don't know how to handle the second type but there is a very nice characterization of the critical points of $M_n$ - they are the matrices of rank at most $n-2$.
The matrices of rank $n-2$ have a Hessian of rank $4$, so any submanifold that includes those points must be no more than $4$-dimensional. In addition, the submanifold must also intersect transversely the submanifold of matrices of rank exactly $n-2$ (and there is another condition to ensure the Hessian is degenerate if the dimension is less than $4$)
The matrices of rank $n-3$ and lower have a Hessian that is entirely zero, so any positive-dimensional submanifold on which the function is Morse must avoid those matrices entirely.
The easiest way to see that these facts are true is to apply a linear transformation to put the matrix into Smith normal form, so that you only have to consider one matrix of each rank - the one that is zero outside the diagonal and has ones, then zeroes, on the diagonal.