Ok, it's time for me to ask my first question on MO.

Consider the affine curve $Y+Y^q=X^{q+1}$ over the finite field $\mathbf{F}_q$. It's interesting because it has the largest number of points over $\mathbf{F}_{q^2}$ possible relative to its genus, which is $q(q-1)/2$. In other words, this curve realizes the Weil bound over $\mathbf{F}_{q^2}$. This seems to be well-known in the literature.

Now consider the following hypersurface $\mathcal{X}$ over the finite field $\mathbf{F}_q$:

$$Z+Z^q+Z^{q^2}=\det\left(\begin{matrix} 0 & X & Y \\ Y^q & 1 & X^q \\ X^{q^2} & 0 & 1\end{matrix}\right)$$

Empirical observation seems to point to the following: The compactly supported cohomology of $\mathcal{X}$ is only nonzero in degrees 2 and 4. In degree 2, the dimension of $H^2(\mathcal{X})$ is $q^2-1$ and the $q^3$-power frobenius acts as the scalar $q^3$. Which is all a fancy way of saying that for all $n$,


Thus $\mathcal{X}$ has the largest number of $\mathbf{F}_{q^3}$-points among any hypersurface with the same compactly supported Betti numbers.

Can anyone help me prove the above formula? (I can do $n=1$ alright...) Bonus points if you can also compute the automorphism group of $\mathcal{X}$. Many more bonus points if you can formulate the generalization to hypersurfaces of higher dimension!

The above hypersurface arises in the study of the bad reduction of Shimura varieties, if anyone cares to know.

EDIT: Admittedly, this is a narrow problem about a very particular surface. Therefore I'm going to accept an answer to the following question: Is there an algorithm to compute the zeta function of a hypersurface of this sort, that's quicker than counting points?

(I've already noticed that it's enough to recur over X and Y, and to test for each pair that the expression on the right lies in the image of the linear map defined by the expression on the left. But I can't think of anything faster than this.)

  • $\begingroup$ If you're serious about bonus points, you should add a bounty =p. $\endgroup$ – Harry Gindi Dec 2 '09 at 3:55
  • $\begingroup$ How do I do that? $\endgroup$ – Jared Weinstein Dec 2 '09 at 7:41
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    $\begingroup$ Jared -- when you view this question, I think you should see a little grey link labeled "start a bounty", right above "add a comment". Click that if you want to offer a bounty. See the FAQ for more details. (In my opinion, though, an elegant question like this will draw people regardless of whether you offer a bounty.) $\endgroup$ – David E Speyer Dec 3 '09 at 5:45

Here is a complete solution to the main question when $n$ and $q$ are both odd, and a partial solution for the other parities. The partial solution includes a reduction to the case $n=2$.

Let $\text{Tr}_k$ denote the trace map from $\mathbb{F}_{q^n}$ to $\mathbb{F}_{q^k}$, assuming that $k|n$.

The equation is $$z+z^q + z^{q^2} = x^{q^2+q+1} - xy^q - x^{q^2}y.$$ First, make the change of variables $y \leftarrow -yx^{q^+1},$ so that the equation becomes $$z+z^q+z^{q^2} = x^{q^2+q+1}(y^q+y+1).$$ Second, when $q$ is odd, we can clarify matters a little with the change of variables $y \leftarrow y - \frac12$ to get rid of the constant. The image of the $q$-linear map $z \mapsto z + z^q + z^{q^2}$, acting on $\mathbb{F}_{q^{3n}}$, consists of those elements $Z'$ such that $\text{Tr}_3(z') = \text{Tr}_1(z')$. Whenever such a $z'$ is reached by the right side, there are $q^2$ solutions for $Z$. On the right side, $y \mapsto y' = y^q+y$ is an $q$-linear isomorphism when $n$ is odd, while when $n$ is even its image is the locus of $\text{Tr}_2(y') = \text{Tr}_1(y')$. Meanwhile $y' \mapsto x^{q^2+q+1}y'$ is a $q$-linear isomorphism unless $x=0$.

Thus when $n$ and $q$ are both odd, there are $(q^{3n}-q^2)(q^{3n}-1)$ solutions with $x,y \ne 0$. There are $q^2(2q^{3n}-1)$ more solutions when one of them is zero.

When $q$ is odd and $n$ is even, then in principle different $x' = x^{q^2 + q + 1}$ could behave differently in the equation $z' = x'y'$. For a fixed $x'$, the set of possible $x'y'$ is a certain $q$-linear hyperplane with codimension $1$, while the set of possible $z'$ is a certain $q$-linear hyperplane with codimension $2$. In the special case that $\{x'y'\}$ contains $\{z'\}$, then there are $q^{3n+1}$ solutions for that value of $x'$. For generic non-zero values of $x'$, there are $q^{3n}$ solutions. Dualize the hyperplanes with respect to $\text{Tr}_1$. The dual of $\{y'\}$ is the line $L$ of trace 0 elements in $\mathbb{F}_{q^2}$, while the dual of $\{z'\}$ is the plane $P$ of trace 0 elements in $\mathbb{F}_{q^3}$. Meanwhile $\{x'\}$ is the subgroup $G$ of $\mathbb{F}_{q^{3n}}$ of index $q^2+q+1$. A special value of $x'$ in this subgroup is one such that $x'L \subset P$. A priori I am not sure that it never happens. What I can say is that if $x'$ is special, then it must lie in $\mathbb{F}_{q^6}$ because both $L$ and $P$ do. So you can reduce the counting problem to the case that $3n = 6$ or $n = 2$.

If $q$ is even, then the equation is $$z' = x'(y'+1),$$ where as before $z' = z+z^q+z^{q^2}$ and $y' = y + y^q$. In this case $y'$ is any element with zero trace, and the dual line $L$ is just $\mathbb{F}_q$ itself. I have not worked out exactly how it looks, but I suppose that it reduces to the case $n=1$ for similar reasons as above.

Afterthought: I don't feel like changing all of the equations, but I'm wondering now whether there a dual change of variables to put the right side in the form $y''(x^{q^2}+x)$. I think that the map $x \mapsto x^{q^2}+x$ is always non-singular when $q$ is odd.

A remark about where the trace conditions come from. If $a$ is an irreducible element of $\mathbb{F}_{q^n}$, then the map $x \mapsto x^q$ is a cyclic permutation matrix in the basis of conjugates of $a$. A map such as $z \mapsto z+z^q+z^{q^2}$ is then a sum of disjoint permutation matrices and it easy to compute its image and cokernel.

Some remarks about Jared's second, more general question: C.f. the answer to this other mathoverflow question about counting points on varieties. For fixed $q$, the equation of a hypersurface is equivalent to a general Boolean expression, and there may not be much that you can do other than count one by one. There are several strategies that work in the presence of special structure: You can use zeta function information, if you have it, to extrapolate to large values of $q$. You can count the points on a variety if you happen to know that it's linear, or maybe quadratic, or the coset space of a group. And you can use standard combinatorial counting tricks, which in algebraic geometry form amount to looking at fibrations, blowups, inclusion-exclusion for constructible sets, etc.

This particular variety decomposes a lot because it can be made jointly linear in $Y$ and $Z$, and $X$ only enters in a multiplicative form.

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    $\begingroup$ Excellent work! I can now finish the argument when $n=2$, using your criterion. (The claim is false for $q=3$, by the way.) $\endgroup$ – Jared Weinstein Dec 5 '09 at 5:16
  • $\begingroup$ Aha! Now I know why I got stuck at the end. $\endgroup$ – Greg Kuperberg Dec 5 '09 at 6:26
  • $\begingroup$ I think people like Alan Lauder in Oxford know much better ways of computing zeta functions than the naive approach. I think the idea is that they deform to characteristic zero and compute Monsky-Washnitzer or overconvergent p-adic cohomology and then compute char polys of Frobenius. Sounds like using a sledgehammer to crack a nut but my understanding is that these fancy cohomology theories are actually very concrete in terms of cocycles over coboundaries and produce computationally effective point-counting methods. $\endgroup$ – Kevin Buzzard Dec 5 '09 at 8:14
  • $\begingroup$ Maybe you're right. $\endgroup$ – Greg Kuperberg Dec 5 '09 at 18:43

I'll put up a quick proof that the curve has the claimed number of points, in case it inspires anyone. For every $X \in \mathbb{F}\_{q^2}$, the left hand side $X^q+X$ is the trace of $X$, and is thus in $\mathbb{F}\_q$. The trace is an $\mathbb{F}\_q$ linear map, so each fiber has size $q$. For those $X$ where $Tr(X)=0$, there is one root $Y=0$. For the other $X$'s, there are $q+1$ roots of $Y^{q+1} = Tr(X)$ in $\mathbb{F}\_{q^2}$. So there are $q+(q^2-q)(q+1) = q^3$ points on the affine curve.

I have no idea how to deal with that determinant though. I hope someone will have a better idea!


The equation of the surface, for fixed $x$, is $F_q$-linear in $y,z$. I'd try to use that.

  • $\begingroup$ I did notice that! I did some experiments: Fix $n\geq 1$, let $x\in\mathbf{F}_{q^{3n}}^*$, and let $N(x)$ be the number of values of $y\in\mathbf{F}_{q^{3n}}$ for which the above equation has a solution in $W\in \mathbf{F}_{q^{3n}}$. It seems that $N(x)=q^{3n-2}$ for all $x$! (Of course it's a power of $q$, because it's the zero locus of an additive polynomial in $Y$.) Establishing this would imply my guess for the zeta function. It's pretty miraculous, I think. $\endgroup$ – Jared Weinstein Dec 4 '09 at 1:00

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