Let $PSD_n$ be the set of $n\times n$ semidefinite positive matrices.  For any $X\in PSD_n$, define $$f(X)=\log(\det(X)).$$  Then $f$ is a concave function over $PSD_n$.  This fact has some significance in convex optimization.

Is there an analogous result for the permanent?  In particular, if we define
$$g(X)=\log(Perm(X)),$$ 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".)

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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][1].)

Theorem: If $X,Y\in H_n$, $0 \preccurlyeq Y \preccurlyeq X$ and $0\leq \lambda \leq 1$, then $d$ is convex, i.e.,
$$ d(\lambda X + (1-\lambda)Y) \leq \lambda d(A) + (1-\lambda)d(B) $$

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 $M=PSD_n$?)


  [1]: https://en.wikipedia.org/wiki/Immanant