Geometric Interpretation of the Permanent

Question

The Permanent of a square matrix M = ($m_{i,j}$) is defined as follows:

$Perm(M) = \sum_{\sigma\in S_n}\prod_{i=1}^{n} m_{i,\sigma(i)}$

The Permanent is quite similar to the Determinant of a square matrix, defined as follows:

$Det(M) = \sum_{\sigma\in S_n}sign(\sigma)\prod_{i=1}^{n} m_{i,\sigma(i)}$

The Determinant has an intuitive geometric interpretation. Is anything similar known about the Permanent? If not, why does the signed sum of the permutations lend itself to a geometric interpretation and the unsigned sum does not?

Answer 1

There is no obvious geometric interpretation of a permanent, which has been remarked upon in Wikipedia and other places, for example, see the comments of this posting:

https://terrytao.wordpress.com/2008/04/16/on-the-permanent-of-a-random-bernoulli-matrix/

This is one reason why certain (interesting) results have been proven for determinantal (or fermion) point processes, but not perminental (or boson) point processes.

Answer 2

In the article by Terrance Tao linked in Keeler's answer, the first commenter, "nobody", points out that

As for a geometric interpretation of the permanent, we can think of the determinant of a vector space $\mathbb{R}^n$ (which is just a vector bundle over a point) as a section of the top (n-th) exterior power of that vector bundle; equally we can regard the permanent as a section of the n-th symmetric tensor bundle – and given a Young diagram, we can play the same sort of game for any immanant...

I wouldn't say this is a particularly intuitive explanation (I'm not sure I even really understand it), but I believe it adds something.