# Zeta function of $\Delta[\text{det},m]$

In Geometric complexity theory the following variety $\Delta[\text{det},m]$ is crucial.

Let $X=(x_1,\ldots,x_r)$ be a tuple of $r=m^2$ variables, so that $X$ can be thought of as an $m\times m$ variable matrix, identifying $x_i$'s with the entries of $X$ in any way. By a homogeneous symbolic determinant of size $m$ over $X=(x_1,\ldots,x_r)$, we mean the determinant of a symbolic $m \times m$ matrix, whose each entry is a homogeneous linear function over $K$ of $x_1,\ldots,x_r$. Let ${\cal X}$ be the vector space over $K$ of homogeneous polynomials of degree $m$ in the variables $x_1,\ldots,x_r$, and $P({\cal X})$ the projective space associated with ${\cal X}$. Let $\Sigma[\det,m] \subseteq P({\cal X})$ be the set of all points in $P({\cal X})$ that correspond to nonzero homogeneous polynomials in ${\cal X}$ that can be expressed as homogeneous symbolic determinants of size $m$ over $X$. Then $\Delta[\det,m] \subseteq P({\cal X})$ is the Zariski-closure $\overline{\Sigma[\det,m]}$ of $\Sigma[\det,m]$. Its dimension is $\le m^4$.

Consider $\Delta[\det,m]$ over $\overline{\mathbb{F}_q}$. Consider its zeta function $\zeta(s)$.

By Dwork's theorem $\zeta(s) = \frac{p(s)}{q(s)}$ for some polynomials $p$ and $q$.

My question: Is it true that $\deg(p)$, $\deg(q) \le 2^{poly(m)}$?

If $\Delta[\det,m]$ were smooth, it would follow from Weil's conjectures. However, $\Delta[\det,m]$ is not even normal on $\mathbb{C}$.

UPD: I am sorry I was wrong about my estimation of these degrees if $\Delta[\det,m]$ were smooth.

• what is poly(m)? – Dan Petersen Mar 5 '17 at 10:15
• @DanPetersen some polynomial in $m$. – Alexey Milovanov Mar 5 '17 at 10:18
• oh, I see. Then your proof in the case $\Delta[det, m]$ smooth should work just as well in the general case. The only part of the Weil conjecture that fails for singular $X$ is the Riemann hypothesis (i.e. the estimate for the absolute values of the roots of the factors $P_i(T)$ of the zeta function). It is still true for $X$ a singular compact variety that the zeta function is rational and the factors in the factorization have degrees given by the Betti numbers. (If $X$ is not compact then you need to work with the compactly supported Betti numbers.) – Dan Petersen Mar 5 '17 at 10:30
• Since you said you already have a proof assuming $\Delta[det,m]$ smooth, I haven't tried to work out what precise bounds you get in this way. But the key is that the degrees of these polynomials are bounded by the sum of the Betti numbers of $\Delta[det,m]$, and these are bounded by the degrees of the defining equations. One reference is a paper by Katz, "Sums of Betti numbers in arbitrary characteristic". – Dan Petersen Mar 5 '17 at 10:32
• Also posted to (and answered at) m.se, math.stackexchange.com/questions/2171324/is-delta-det-m-smooth without notice to either site. – Gerry Myerson Mar 5 '17 at 10:52