TITLE: Counting nonzero hyperdeterminants over $\mathbb{F}_q$
QUESTION [11 upvotes]: The hyperdeterminant  $D(A)$ is a multidimensional generalization of the
determinant. It is a polynomial in the entries of a $(k_1+1)\times
(k_2+1)\times\cdots \times (k_n+1)$ array $A$. The hyperdeterminant is defined
when $2\max k_i\leq \sum k_i$. For further information see for
instance http://arxiv.org/pdf/1301.0472v1.pdf.
Assuming that $(k_1,\dots,k_n)$ satisfies the above condition, let
$N(q)$ be the number of nonzero $(k_1+1)\times\cdots\times(k_n+1)$
hyperdeterminants over the finite field $\mathbb{F}_q$. For instance,
if $n=2$ then $k_1=k_2$, and the number of $m\times m$ matrices over
$\mathbb{F}_q$ with nonzero determinant is well-known and easily seen
to equal $(q^m-1)(q^m-q)\cdots (q^m-q^{m-1})$. A few years ago some
M.I.T. graduate students computed that for $2\times 2\times 2$
hyperdeterminants we get a polynomial in $q$, though I seem to have
misplaced the formula. Can anything be said about the general case? Do
we always get a polynomial in $q$? (This seems unlikely to me.) Can
someone compute the $2\times 2\times 3$ case?

REPLY [3 votes]: I saw this question years ago and thought it was interesting. I never got around to doing anything with the question, but today I happened to come across some slides of Steven Sam with some data on the problem. I imagine the slides refer to the same M.I.T. students alluded to in the question. The slides can be found on Steven Sam's website and information on hyperdeterminants is on pages 16, 17, and 18.
The count for $2 \times 2 \times 2$ is reported to be $(q^4 - 1)(q^4 - q^3)$ and is attributed to Musiker--Yu. Attributed to Lewis--Sam are counts for $2 \times 2 \times 3$, $2 \times 3 \times 3$, and $2 \times 2 \times 4$ which are
$$q^4(q-1)^4 [2]^2 [3]\\ q^{10}(q-1)^3[2]^2[3] \\ q^4(q-1)^2[3][4](q^3+q^2-1).$$
It says "Caveat: these need to be double-checked..." which I have not done. But now maybe I (or someone else) will be inclined to check since we have an expression to compare to.