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This question is a sequel to an earlier question, which asked about the zeta function of a certain affine variety over a finite field $k$. The unusual thing about this variety is that it had the maximum number of rational points over every extension of $k$, relative to the constraints imposed by the topology. Here we're going to construct a family of varieties which we suspect behaves the same way. If my hunches are correct, the result will be a means to construct irreducible characters of certain unipotent algebraic groups over finite fields, akin to the way Deligne-Lusztig varieties are used to construct irreducible characters of Chevalley groups.

Let $n\geq 3$ be prime. We define a group variety $U/\mathbf{F}_q$ as follows: for an $\mathbf{F}_q$-algebra $R$, we set

$$U(R)=\{1+x_1\tau+\dots+x_n\tau^n\biggm\vert x_i\in R\}$$

Here $\tau$ is an indeterminate satisfying $\tau x_i=x_i^q \tau$ and $\tau^{n+1}=0$. Thus $U$ is a unipotent group, abstractly isomorphic to $\mathbf{A}^n$. Let $p\colon U\to \mathbf{A}^1$ be the projection onto the coefficient of $\tau^n$. Also let $g\mapsto g^{(q)}$ be the automorphism of $U$ which raises each coordinate to the $q$th power.

Let $X\subset U$ be the hypersurface defined by $p\left(g^{(q^n)}g^{-1}\right)=0$. Then $X$ is a nonsingular affine variety of dimension $n-1$.

My question is:

Prove that the zeta function of $X$ over $\mathbf{F}_{q^n}$ is $$Z(X/\mathbf{F}_{q^n},T)=\left(1-q^{n(n-1)/2}T\right)^{-q^{n(n-1)/2}(q^n-q)} \left(1-q^{n(n-1)}T\right)^{-q}$$

The form of this zeta function is going to make more sense when we consider that $X$ has a large group of $\mathbf{F}_{q^n}$-linear automorphisms, namely $U(\mathbf{F}_{q^n})$, which acts on $X$ on the right.

Let $Z\subset U$ be the center: $Z=\{1+a_n\tau^n\}$. Call a character $\psi\colon Z(\mathbf{F}_{q^n})= \mathbf{F}_{q^n}\to\overline{\mathbf{Q}}_\ell^\times$ generic if it does not factor through the trace map $\mathbf{F}_{q^n}\to\mathbf{F}_{q}$. It can be shown that if $\psi$ is generic, then there is a unique irreducible representation $V_\psi$ of $U(\mathbf{F}_{q^n})$ whose central character is $\psi$. The dimension of $V_\psi$ is $q^{n(n-1)/2}$. (Quick construction: the subgroup $U^{(n-1)/2}\subset U$ defined by $x_1=\dots=x_{(n-1)/2}=0$ is abelian. Extend $\psi$ to a character $\tilde{\psi}$ of $U^{(n-1)/2}(\mathbf{F}_{q^n})$ any way you like, and let $V_\psi$ be the induction of $\tilde{\psi}$ to $U(\mathbf{F}_{q^n})$.)

Each non-generic character $\psi$ extends to a one-dimensional character of $U(\mathbf{F}_{q^n})$, which we also call $V_\psi$. Indeed, there is a "reduced norm" map $N\colon U(\mathbf{F}_{q^n})\to Z(\mathbf{F}_q)$ extending the trace map $T\colon Z(\mathbf{F}_{q^n})\to Z(\mathbf{F}_q)$, and if $\psi=\psi_0\circ T$ then let $V_\psi=\psi_0\circ N$. (In fact this $N$ is really a morphism $U\to Z$, albeit not a homomorphism, and an alternate definition of $X$ is $\{g\vert N(g)\in Z(\mathbf{F}_q)\}$. Thus $X$ has at least $q$ connected components.)

Our question now becomes:

Show there is an isomorphism of $U(\mathbf{F}_{q^n})$-modules $$H^*_c(X\otimes\overline{\mathbf{F}}_q,\overline{\mathbf{Q}}_\ell)\cong \bigoplus_\psi V_\psi,$$ where the sum is over all characters of $\mathbf{F}_{q^n}$ and each $V_\psi$ appears exactly once. The eigenvalue of the $q^n$-power Frobenius on $V_\psi$ equals $q^{n(n-1)/2}$ if $\psi$ is generic and $q^{n(n-1)}$ otherwise.

There will be a generalization of these statements to the case of $n$ composite, but they are more complicated; there will be contributions to $H^*_c$ of degrees strictly between $n-1$ and $2(n-1)$.

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Do you already know how to compute the cardinality of $X(\mathbf{F}_{q^n})$ ? –  François Brunault Nov 11 '10 at 10:40
    
I think I don't understand your definition of $X$, is it true that $X(\mathbf{F}_{q^n})=U(\mathbf{F}_{q^n})$ ? –  François Brunault Nov 11 '10 at 11:07
    
@François, it is indeed true that $X(\mathbf{F}_{q^n})=U(\mathbf{F}_{q^n})$, so that $\#X(\mathbf{F}_{q^n})=q^{n^2}$. The equation determining $X$ can be written explicitly as one equation in the $n$ variables. When $n=3$, this equation is $\det M=0$, where $M=\begin{pmatrix} x_1^{q^3}-x_1 & x_2^{q^3}-x_2 & x_3^{q^3}-x_3 \\ 1 & x_1^q & x_2^q \\ 0 & 1 & x_1^{q^2} \end{pmatrix}$. –  Jared Weinstein Nov 11 '10 at 16:20
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1 Answer

This is very far from a solution but just some ideas that could be possibly be used as a start.

First a simple observation on an Artin-Schreier type extension $X\rightarrow Y$, where $X:=\mathrm{Spec}\mathcal O_Y[t]/(t^q-t-f)$. Here $q=p^m$, $p$ a prime with $p\mathcal O_Y=0$ and $f\in\mathcal O_Y(Y)$. This a finite étale covering with Galois group $\mathbb F_q$. For any surjective group homomorphism $\varphi\colon\mathbb F_q\rightarrow\mathbb Z/p$ we get an induced $\mathbb Z/p$-covering $X_\varphi\rightarrow Y$. It can be described in the following way: There is an $\alpha\in\mathbb F_q^\ast$ such that $\varphi(\lambda)=\mathrm{Tr}(\alpha\lambda)$ and then $X_\alpha=\mathrm{Spec}\mathcal O_Y[t]/(t^p-t-\alpha^{-1}f)$. This is particularly relevant when we want to consider the cohomology of $X$. We get an action of $\mathbb F_q$ on $H^*(X,K)$, where $K$ is a finite extension of $\mathbb Q_\ell$ containing $p$'th roots of unity. Hence, the action on any isotypical component factors through some $\varphi$ and hence the kernel of $\varphi$ acts trivially on it. We can then use the fact that the invariants on cohomology is the cohomology of the quotient to conclude that $H^\ast(X,K)$ is the direct sum of the cohomology of $Y$ and the parts of the $H^\ast(X_\alpha,K)$ where $\mathbb Z/p$ acts non-trivially.

Assume now that we have a (smooth) affine algebraic group $G$ defined over $\mathbb F_q$ and a central subgroup $\mathbb G_a\subseteq G$ also defined over $\mathbb F_q$. We then put $H:=G/\mathbb G_a$ which is also affine which means that $G\rightarrow H$ has a section, in any case we choose a particular such section such that $G$ is the product $H\times\mathbb G_a$. The multiplication of $G$ is then given by a cocycle $\sigma\colon H\times H\rightarrow\mathbb G_a$. Using it we can describe the Lang torsor $Fg\cdot g^{-1}\colon G\rightarrow G$, where $F$ is the Frobenius map (with respect to $\mathbb F_q$), in terms of $\sigma$: Given $g=(h,z)$ an element of $H\times\mathbb G_a$ we have $Fg=(Fh,Fz)$ and $g^{-1}=(h^{-1},-z-\sigma(h,h^{-1}))$ and hence $Fg\cdot g^{-1}=(Fh\cdot h^{-1},Fz-z+\sigma(Fh,h^{-1})-\sigma(h,h^{-1}))$. In particular this means that the inverse image of $H\times\{0\}$ under the Lang torsor is defined by $\{(h,z)|Fz-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})\}$. As $Fz=z^q$ we get exactly an Artin-Schreier type covering of $H$ given by $z^q-z=\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$ and hence if we want the cohomology of this covering we need only consider the cohomology of $H$ itself and ordinary Artin-Schreier coverings $t^p-t=\alpha^{-1}(\sigma(h,h^{-1})-\sigma(Fh,h^{-1}))$ for $\alpha\in\mathbb F_q^\ast$.

This is about as far as I've come with only a few comments on the group at hand. We have in that case that $m=ns$ and $H=\mathbb G_a^{n-1}$ with $\sigma(a,b)=\sum_ia_ib_{n-i}^{r^i}$, where $r=p^s$ (I have changed notation a little so that my $r$ is the OP's $q$ and my $q$ is the OP's $q^n$). Note that somewhat curiously $\sigma$ is biadditive with respect to the addition on $\mathbb G_a^{n-1}$. Perhaps more relevantly $\sigma$ is quasi-homogeneous; we can extend $G$ to elements $x_0^{-1}+x_1\tau+\cdots+x_n\tau^n$, with $x_0$ invertible and then $G$ is a normal subgroup so that we get a $\mathbb G_m$-action on $G$. Under this action $x_i$ is homogeneous of degree $r^i-1$ and with these weights $\sigma$ is quasi-homogeneous of degree $q-1$. Unfortunately $\sigma(h,h^{-1})-\sigma(Fh,h^{-1})$ is not homogeneous as even though $\sigma(h,h^{-1})$ is homogeneous (of degree $q-1$), $\sigma(Fh,h^{-1})$ is not homogeneous. However, $\mu_{q-1}\subseteq\mathbb G_m$ acts trivially on both $\sigma(h,h^{-1})$ and $\sigma(Fh,h^{-1})$ so we get some symmetry.

Addendum: I forgot to mention that the cohomology of an Artin-Schreier extension $t^p-t=f$ is presumably best studied by computing $Rf_!$ on $\mathbb A^1$ and then see the cohomology of the Artin-Schreier extension as the fibre of its Fourier transform over the point $1$.

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Thank you for this, Torsten. The problem is still open, but your comments have helped me generalize this construction to the situation where $U$ is replaced by a more general type of unipotent group. (I knew about the action of the group of roots of unity.) I have yet to absorb the significance of Deligne-Fourier transforms to this situation. Can you see any way to at least compute the dimension of the cohomology? –  Jared Weinstein Nov 23 '10 at 19:18
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