Geometric interpretation of the permanent

Question

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

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

The permanent is quite similar to the determinant of a square matrix, which is defined as follows

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

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 whereas 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:

• Terence Tao, On the permanent of a random Bernoulli matrix, 16 April 2008.

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 Terence 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.