2 added phrase "where the sum is over all characters..."

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 $$H^*_c(X,\overline{\mathbf{Q}}_\ell)\cong H^*_c(X\otimes\overline{\mathbf{F}}_q,\overline{\mathbf{Q}}_\ell)\cong \bigoplus_\psi V_\psi,$$ 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 $2(n-1)$.

1

# A family of hypersurfaces with many points

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 $$H^*_c(X,\overline{\mathbf{Q}}_\ell)\cong \bigoplus_\psi V_\psi,$$ where 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 $2(n-1)$.