Let $\operatorname{PSD}_n$ be the cone of $n\times n$ semidefinite positive matrices. For any $X\in \operatorname{PSD}_n$, define $$f(X)=\log(\det(X)).$$ Then $f$ is a concave function on $\operatorname{PSD}_n$. This fact has some significance in convex optimization.
Is there an analogous result for the permanent? In particular, if we define $$g(X)=\log(\operatorname{Perm}(X)),$$ can we identify some non-trivial set $M$ of matrices over which $g$ is concave (or convex for that matter)?
(I'm hoping for some set larger than "diagonal matrices".)
EDIT: In the comments below, Mark L. Stone included a reference to a very interesting theorem; here's the relevant part.
Let $H_n$ be the set of $n\times n$ Hermitian matrices. For $X,Y\in H_n$, we say that $X\preccurlyeq Y$ iff $Y-X$ is semidefinite positive. Let $S_n$ be the group of permutations on $n$ objects, and $G$ a subgroup of $S_n$.
We define a function $d:H_n \rightarrow R$ for an irreducible character $\chi$ of $G$ by: $$d(X)=\sum_{\sigma\in G} \chi(\sigma) \prod_{i=1}^{n} X_{\sigma(i),i}$$
(Note that the determinant and the permanent are instances of this function, as are, I think, the immanants.)
Theorem: If $X,Y\in H_n$, $0 \preccurlyeq Y \preccurlyeq X$ and $0\leq \lambda \leq 1$, then $d$ is convex, i.e., $$ d(\lambda X + (1-\lambda)Y) \leq \lambda d(X) + (1-\lambda)d(Y) $$
In particular, that holds for the permanent. However, this result doesn't quite answer my original question, in that there is no fixed set $M$ that works— it requires a relationship between $X$ and $Y$. (Is there some way to lift this result with a matrix exponential to get $M=\operatorname{PSD}_n$?)
