4
$\begingroup$

Suppose that $p\ge 5$ is a prime, $n$ a positive integer divisible by $p-1$, and $L<\mathbb F_p^n$ a subspace of dimension $d=n/(p-1)$. Do there exist vectors $l_1,\dotsc,l_n\in L$ such that the matrix with $l_1,\dotsc,l_n$ as its columns has a nonzero permanent? Clearly, the answer is negative if $L$ is contained in a coordinate hyperplane, or in a linear subspace like $\{(x_1,\dotsc,x_n)\in\mathbb F_p^n\colon x_1=\dotsb=x_p\}$; are there other obstructions of this sort?

Is it possible to classify those subspaces $L<\mathbb F_p^n$ for which any square matrix of order $n$ with all its column vectors in $L$ has a vanishing permanent?

Notice that if the permanent vanishes for $l_1,\dotsc,l_n$ being the elements of some particular basis of $L$, with each element repeated $p-1$ times, then in fact it vanishes for any $l_1,\dotsc, l_n\in L$. (It is this property that depends critically on the assumption $d=n/(p-1)$.)

In the case where $d=1$, the requirement that $L$ is not contained in a coordinate hyperplane is easily seen to be also sufficient. The case $d=2$ does not look that easy to me.

$\endgroup$
2
  • $\begingroup$ Your highlighted question doesn't include the hypothesis on the dimension of $L$ that is in the introductory paragraph. Do you want to continue imposing that hypothesis? $\endgroup$
    – LSpice
    Commented Apr 20, 2021 at 18:43
  • 1
    $\begingroup$ @LSpice: Absolutely; everything is about subspaces $L$ of dimension $d=n/(p-1)$. $\endgroup$
    – Seva
    Commented Apr 20, 2021 at 18:56

1 Answer 1

1
$\begingroup$

Let $v_1,\dots, v_d$ be a basis for $L$. Then the permanent of the matrix obtained from $p-1$ repetitions each of $v_1,\dots, v_d$ is a polynomial function in the entries of $v_1,\dots, v_d$. Keeping all the entries but the last one in each $v_i$ constant, we get a linear function. Since it is linear, among all $p^{nd}$ possible tuples of vectors, the number that satisfy it is at least $p^{nd-1}$.

Not all $d$-tuples of vectors in $\mathbb F_p^n$ form a basis, but $1 - O( p^{d-n-1})$ do.

So the fraction of subspaces $L$ where this determinant vanishes is at least $$ \frac{1}{p} - O( p^{d-n-1}) $$

On the other hand, the fraction that satisfy one of your conditions is smaller than that for $d>1$ - just $\frac{ d (p-1)}{p^d}$ satisfy the coordinate condition and $\frac{ \binom{d (p-1) }{ p } } { p^{ (p-1) d }}$ satisfy the $p$-fold repetition condition.

So there certainly are other obstructions. I do not expect it is possible to classify them.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .