2 The local systems

Fix a prime $p$ , a finite field $\mathbb{F}_{q}$ of characteristic $p$ , and a non-trivial additive character

$$\begin{eqnarray}\unicode[STIX]{x1D713}:(\mathbb{F}_{q},+)\rightarrow \unicode[STIX]{x1D707}_{p}(\mathbb{Z}[\unicode[STIX]{x1D701}_{p}]).\end{eqnarray}$$

Denote by $\unicode[STIX]{x1D712}$ a non-trivial multiplicative character

$$\begin{eqnarray}\unicode[STIX]{x1D712}:\mathbb{F}_{q}^{\times }\rightarrow \unicode[STIX]{x1D707}_{q-1}(\mathbb{Z}[\unicode[STIX]{x1D701}_{q-1}]).\end{eqnarray}$$

We extend $\unicode[STIX]{x1D712}$ to a function on all of $\mathbb{F}_{q}$ by defining $\unicode[STIX]{x1D712}(0)=0$ .

We choose a prime number $\ell \neq p$ , and an embedding of $\overline{\mathbb{Q}}$ , say viewed as the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$ , into an algebraic closure $\overline{\mathbb{Q}_{\ell }}$ of $\mathbb{Q}_{\ell }$ . This allows us to speak of the lisse, rank one $\overline{\mathbb{Q}_{\ell }}$ Artin–Schreier sheaf ${\mathcal{L}}_{\unicode[STIX]{x1D713}}$ on $\mathbb{A}^{1}/\mathbb{F}_{q}$ and of the lisse, rank one $\overline{\mathbb{Q}_{\ell }}$ Kummer sheaf ${\mathcal{L}}_{\unicode[STIX]{x1D712}}$ on $\mathbb{G}_{m}/\mathbb{F}_{q}$ . For $j:\mathbb{G}_{m}\subset \mathbb{A}^{1}$ the inclusion, the extension by zero $j_{!}{\mathcal{L}}_{\unicode[STIX]{x1D712}}$ on $\mathbb{A}^{1}$ will also be denoted ${\mathcal{L}}_{\unicode[STIX]{x1D712}}$ when no ambiguity can result. We denote by $\unicode[STIX]{x1D7D9}$ the trivial multiplicative character, and adopt the convention that ${\mathcal{L}}_{\unicode[STIX]{x1D7D9}}$ is the constant sheaf $\overline{\mathbb{Q}_{\ell }}$ on $\mathbb{A}^{1}$ . We write $\text{FT}$ for $\text{FT}_{\unicode[STIX]{x1D713}}$ , the Fourier transform using ${\mathcal{L}}_{\unicode[STIX]{x1D713}(xy)}$ as the kernel.

Given an integer $D\geqslant 2$ which is prime to $p$ , we denote by

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D):=\text{FT}({\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})}),\end{eqnarray}$$

and, for each non-trivial $\unicode[STIX]{x1D712}$ ,

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D):=\text{FT}({\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})}).\end{eqnarray}$$

These Fourier transform sheaves are lisse on $A^{1}/\mathbb{F}_{q}$ , pure of weight one, and geometrically irreducible, cf. [Reference Deligne4, Reference Katz19, Reference Laumon24]. They are cohomologically rigid, being the Fourier transforms of rank one objects, cf. [Reference Katz18, 3.0.2]. Their ranks are

$$\begin{eqnarray}\displaystyle \operatorname{rank}({\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D)) & = & \displaystyle D-1,\nonumber\\ \displaystyle \operatorname{rank}({\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D)) & = & \displaystyle D\quad \text{for}~\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D7D9}.\nonumber\end{eqnarray}$$

Their trace functions are given as follows, cf. [Reference Grothendieck11]. For $k/\mathbb{F}_{q}$ a finite extension, define

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{k/\mathbb{F}_{q}}:=\unicode[STIX]{x1D713}\circ \operatorname{Trace}_{k/\mathbb{F}_{q}},\quad \unicode[STIX]{x1D712}_{k/\mathbb{F}_{q}}:=\unicode[STIX]{x1D712}\circ \text{Norm}_{k/\mathbb{F}_{q}}.\end{eqnarray}$$

Then for $t\in k$ , we have

$$\begin{eqnarray}\displaystyle \operatorname{Trace}(\text{Frob}_{t,k}|{\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D)) & = & \displaystyle -\!\mathop{\sum }_{x\in k}\unicode[STIX]{x1D713}_{k/\mathbb{F}_{q}}(x^{D}+tx),\nonumber\\ \displaystyle \operatorname{Trace}(\text{Frob}_{t,k}|{\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D)) & = & \displaystyle -\!\mathop{\sum }_{x\in k^{\times }}\unicode[STIX]{x1D712}_{k/\mathbb{F}_{q}}(x)\unicode[STIX]{x1D713}_{k/\mathbb{F}_{q}}(x^{D}+tx).\nonumber\end{eqnarray}$$

When $D$ is odd, ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D)$ is symplectically self-dual toward $\overline{\mathbb{Q}_{\ell }}(-1)$ . When $D$ is even, ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D)$ has no autoduality.

When $D$ is odd, $p$ is odd, and $\unicode[STIX]{x1D712}$ is the quadratic character $\unicode[STIX]{x1D712}_{2}$ , then ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},D)$ is orthogonally self-dual toward $\overline{\mathbb{Q}_{\ell }}(-1)$ . No other ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D)$ with non-trivial $\unicode[STIX]{x1D712}$ is autodual.

Given a lisse $\overline{\mathbb{Q}_{\ell }}$ -sheaf ${\mathcal{F}}$ on $\mathbb{A}^{1}/\mathbb{F}_{q}$ of rank $r$ , we choose a geometric point $\overline{\unicode[STIX]{x1D702}}$ of $\mathbb{A}^{1}$ and view ${\mathcal{F}}$ as a representation of the fundamental group

$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D70B}_{1}^{\text{arith}}(\mathbb{A}^{1}/\mathbb{F}_{q},\overline{\unicode[STIX]{x1D702}}):=\unicode[STIX]{x1D70B}_{1}(\mathbb{A}^{1}/\mathbb{F}_{q},\overline{\unicode[STIX]{x1D702}}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D70C}_{{\mathcal{F}}}:\unicode[STIX]{x1D70B}_{1}^{\text{arith}}(\mathbb{A}^{1}/\mathbb{F}_{q},\overline{\unicode[STIX]{x1D702}})\rightarrow \operatorname{GL}({\mathcal{F}}_{\overline{\unicode[STIX]{x1D702}}})\cong \operatorname{GL}(r,\overline{\mathbb{Q}_{\ell }}). & \displaystyle \nonumber\end{eqnarray}$$

The Zariski closure of the image of $\unicode[STIX]{x1D70B}_{1}^{\text{arith}}$ is the arithmetic monodromy group $G_{\text{arith},{\mathcal{F}}}$ of ${\mathcal{F}}$ . The Zariski closure of the image of its normal subgroup

$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{1}^{\text{geom}}(\mathbb{A}^{1}/\mathbb{F}_{q},\overline{\unicode[STIX]{x1D702}}):=\unicode[STIX]{x1D70B}_{1}(\mathbb{A}^{1}/\overline{\mathbb{F}_{q}},\overline{\unicode[STIX]{x1D702}})\end{eqnarray}$$

is the geometric monodromy group $G_{\text{geom},{\mathcal{F}}}$ of ${\mathcal{F}}$ .

For ${\mathcal{F}}$ a lisse $\overline{\mathbb{Q}_{\ell }}$ -sheaf on $\mathbb{A}^{1}/\mathbb{F}_{q}$ of rank $r$ which is geometrically irreducible, if $p>2r+1$ , then ${\mathcal{F}}$ is Lie-irreducible, i.e., the identity component $G_{\text{geom}}^{0}$ of $G_{\text{geom}}$ acts irreducibly, cf. [Reference Katz15, Proposition 5], based on [Reference Feit and Thompson8]. Below we will be interested in situations where $p\leqslant 2r+1$ .

Geometrically, i.e., on $\mathbb{A}^{1}/\overline{\mathbb{F}_{q}}$ , the (restrictions to $\mathbb{G}_{m}$ of the) local systems ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},D)$ and ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D)$ are Kummer pullbacks, by the $D$ th power map, of certain Kloosterman, respectively hypergeometric, sheaves, cf. [Reference Katz17, 9.2.3 and 9.2.2]. The precise statement is this.

We have the following more precise information about determinants.

Proof. As shown in Corollary 2.2 part (1), the determinants in question are geometrically trivial, so each is of the form $\unicode[STIX]{x1D6FC}^{\text{deg}}$ for some scalar $\unicode[STIX]{x1D6FC}\in \overline{\mathbb{Q}_{\ell }}^{\times }$ . This scalar is then the common value of $\det (\text{Frob}_{\mathbb{F}_{q},t}|{\mathcal{F}})$ at points $t\in \mathbb{F}_{q}$ . Taking $t=0$ , the scalar is

$$\begin{eqnarray}\det (\text{Frob}_{\mathbb{F}_{q}}|H_{c}^{1}(\mathbb{A}^{1}/\overline{\mathbb{F}_{q}},{\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})})),\end{eqnarray}$$

for $\unicode[STIX]{x1D712}$ either $\unicode[STIX]{x1D7D9}$ or non-trivial.

Here is a simple trick which “unites” these cases. When $\unicode[STIX]{x1D712}$ is the trivial character $\unicode[STIX]{x1D7D9}$ , replace it by $j_{!}$ of itself; in other words, consider instead the cohomology group $H_{c}^{1}(\mathbb{G}_{m}/\overline{\mathbb{F}_{q}},{\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})})$ . When $\unicode[STIX]{x1D712}$ is non-trivial, this is the same group as before; when $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D7D9}$ , the cohomology group grows in dimension by one, adding an eigenvalue $1$ . So in both cases the determinant does not change.

To compute the determinant, we use the Hasse–Davenport method, cf. [Reference Katz15, p. 53] and [Reference Katz20, 2.2–2.3]. In terms of the elementary symmetric functions $S_{i}$ , the Newton symmetric functions $N_{i}$ are $\mathbb{Z}$ -polynomials in the $S_{j}$ . Then

$$\begin{eqnarray}\displaystyle & & \displaystyle \det (-\text{Frob}_{\mathbb{F}_{q}}|H_{c}^{1}(\mathbb{G}_{m}/\overline{\mathbb{F}_{q}},{\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})}))\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{S_{1},\ldots ,S_{D}\in \mathbb{F}_{q},S_{D}\neq 0}\unicode[STIX]{x1D712}(S_{D})\unicode[STIX]{x1D713}(N_{D}(S_{1},\ldots ,S_{D})).\nonumber\end{eqnarray}$$

The polynomial $N_{D}(S_{1},\ldots ,S_{D})$ is of the form

$$\begin{eqnarray}N_{D}=(-1)^{D+1}DS_{D}+(-1)^{D}\mathop{\sum }_{i=1}^{D-1}\text{i}S_{i}S_{D-i}+R,\end{eqnarray}$$

where $R$ is a polynomial in the $S_{i}$ which is isobaric of degree $D$ and in which every monomial has usual degree at least $3$ .

[To see this, begin with the identity

$$\begin{eqnarray}\log (1/(1-S_{1}T+S_{2}T^{2}+\cdots \,))=\mathop{\sum }_{i\geqslant 1}N_{i}T^{i}/i,\end{eqnarray}$$

apply $Td/dT$ to get

$$\begin{eqnarray}\frac{\mathop{\sum }_{i\geqslant 1}(-1)^{i+1}\text{i}S_{i}T^{i}}{1-S_{1}T+S_{2}T^{2}+\cdots }=\mathop{\sum }_{i\geqslant 1}N_{i}T^{i},\end{eqnarray}$$

and expand the denominator by the geometric series.]

When $D=2d+1$ is odd, this expression is of the form

$$\begin{eqnarray}N_{D}=DS_{D}-D\mathop{\sum }_{i=1}^{d}S_{i}S_{2d+1-i}+R.\end{eqnarray}$$

When $D=2d$ is even, it is of the form

$$\begin{eqnarray}N_{D}=-DS_{D}+dS_{d}^{2}+D\mathop{\sum }_{i=1}^{d-1}S_{i}S_{2d-i}+R.\end{eqnarray}$$

Exactly as in [Reference Katz20, 2.2–2.3], using this expression for $N_{D}$ we see that when $D=2d+1$ is odd, then $\det (-\text{Frob})$ is equal to

$$\begin{eqnarray}g(\unicode[STIX]{x1D713}_{D},\unicode[STIX]{x1D712})q^{d},\end{eqnarray}$$

and that when $D=2d$ is even, then $\det (-\text{Frob})$ is equal to

$$\begin{eqnarray}g(\unicode[STIX]{x1D713}_{-D},\unicode[STIX]{x1D712})g(\unicode[STIX]{x1D713}_{d},\unicode[STIX]{x1D712}_{2})q^{d-1}.\end{eqnarray}$$

Thus when $D=2d+1$ is odd, $\det (\operatorname{Frob})$ is given by

$$\begin{eqnarray}-g(\unicode[STIX]{x1D713}_{D},\unicode[STIX]{x1D712})q^{d},\end{eqnarray}$$

and when $D=2d$ , $\det (\operatorname{Frob})$ is given by

$$\begin{eqnarray}(-g(\unicode[STIX]{x1D713}_{-D},\unicode[STIX]{x1D712}))(-g(\unicode[STIX]{x1D713}_{d},\unicode[STIX]{x1D712}_{2}))q^{d-1}.\end{eqnarray}$$

With our $j_{!}$ convention, $-g(\unicode[STIX]{x1D713}_{D},\unicode[STIX]{x1D7D9})=1$ .◻

3 Review of the situation in large characteristic

Suppose we fix an integer $D\geqslant 3$ . In large (compared to $D$ ) characteristic $p$ , the local systems ${\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712},D)$ have very large geometric monodromy groups $G_{\text{arith}}$ . Here is a precise statement.

Proof. Case (1) is proven in [Reference Katz15, Theorem 19]. For case (2), we argue as follows. Because $p>2D$ , ${\mathcal{F}}$ on $\mathbb{A}^{1}/\mathbb{F}_{q}$ is geometrically Lie-irreducible. Because its determinant is geometrically trivial, its $G_{\text{geom}}$ is connected, cf. [Reference Katz15, Proposition 5]. Because ${\mathcal{F}}$ is pure of weight one, its $G_{\text{geom}}$ is a semisimple group. Therefore

$$\begin{eqnarray}G_{\text{geom}}=G_{\text{geom}}^{0}=(G_{\text{geom}}^{0})^{\text{der}}.\end{eqnarray}$$

Its highest $\infty$ -slope is $D/(D-1)$ , which occurs with multiplicity $D-1$ . Applying [Reference Katz17, 7.2.7], which lists the possible $(G_{\text{geom}}^{0})^{\text{der}}$ , we see that $G_{\text{geom}}$ is one of the listed groups, except that we must show that if $D$ is even, we cannot have $\operatorname{Sp}$ . In fact, when $D$ is even, no ${\mathcal{F}}$ is self dual. Indeed, the dual of a Fourier transform is given geometrically by

$$\begin{eqnarray}(\text{FT}({\mathcal{A}}))^{\vee }\cong \text{FT}([x\mapsto -x]^{\star }(({\mathcal{A}})^{\vee })).\end{eqnarray}$$

So by Fourier inversion, $\text{FT}({\mathcal{A}})$ is geometrically self dual if and only if there is a geometric isomorphism

$$\begin{eqnarray}{\mathcal{A}}\cong [x\mapsto -x]^{\star }(({\mathcal{A}})^{\vee }).\end{eqnarray}$$

Here our ${\mathcal{A}}$ is ${\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})}$ , so the requirement is

$$\begin{eqnarray}{\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D})}\cong {\mathcal{L}}_{\unicode[STIX]{x1D712}^{-1}(-x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(-(-x)^{D})},\end{eqnarray}$$

which is equivalent to having

$$\begin{eqnarray}{\mathcal{L}}_{\unicode[STIX]{x1D712}(x)}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(x^{D}+(-x)^{D})}\cong {\mathcal{L}}_{\unicode[STIX]{x1D712}^{-1}(-x)}.\end{eqnarray}$$

This is nonsense, because $D$ is even and $p>D>2$ , so the left side has $\text{Swan}_{\infty }=D$ , while the right side is tame at $\infty$ .◻

4 Kubert’s finiteness theorems: statements

In the Spring of 1986, Kubert lectured in my graduate course, proving that various Kloosterman and hypergeometric sheaves had finite geometric monodromy.

5 Proofs of Kubert’s theorems

As explained in 2.1 above, each local system ${\mathcal{F}}$ on $\mathbb{A}^{1}$ in question is geometrically, when restricted to $\mathbb{G}_{m}$ , the pullback by the $D$ th power map of an explicit hypergeometric sheaf. So to prove the theorems, it suffices to show in each case that the relevant hypergeometric sheaf has finite $G_{\text{geom}}$ . The proofs of Theorem 4.1 and part (1) of 4.2 are given in [Reference Katz22, 13.3]. To give the remaining proofs, we will make use of Kubert’s $V$ function [Reference Katz22, § 13]

$$\begin{eqnarray}V:(\mathbb{Q}/\mathbb{Z})_{\text{prime to }p}\rightarrow [0,1),\end{eqnarray}$$

giving the suitably normalized $p$ -adic ords of gauss sums. As explained in [Reference Katz22, § 13], this function has the following properties:

1. (1) $V(x)=0$ if and only if $x=0$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to }p}$ ;

2. (2) for $x$ non-zero in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , $V(x)+V(-x)=1$ ;

3. (3) $V(1/2)=1/2$ ;

4. (4) for any $x$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , $V(x)=V(px)$ ;

5. (5) for any $x$ and $y$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , $V(x)+V(y)\geqslant V(x+y)$ ;

6. (6) for any $x$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , and any integer $N\geqslant 1$ prime to $p$ , $\sum _{i~\text{mod}~N}V(x+i/N)=V(Nx)+(N-1)/2$ .

The application of this $V$ function to showing that a hypergeometric sheaf ${\mathcal{H}}$ has finite geometric monodromy (or equivalently [Reference Katz17, 8.14.4] that after a constant $\unicode[STIX]{x1D6FC}^{\text{deg}}$ twist to produce an ${\mathcal{F}}$ whose determinant is arithmetically of finite order, the resulting ${\mathcal{F}}$ has finite $G_{\text{arith}}$ ) is given by the following proposition, quoted from [Reference Katz22, 13.2].

Proposition 5.1. Given the hypergeometric sheaf ${\mathcal{H}}:={\mathcal{H}}(\unicode[STIX]{x1D713};\unicode[STIX]{x1D712}_{i}~^{\prime }s;\unicode[STIX]{x1D70C}_{j}~^{\prime }s)$ on $\mathbb{G}_{m}/k$ and its twist ${\mathcal{F}}$ , pick any multiplicative character $\operatorname{Teich}_{k}$ of $k^{\times }$ which is faithful, i.e., has order $\#k-1$ . Define a list of $n+m$ elements $(a_{1},\ldots ,a_{n},b_{1},\ldots ,b_{m})$ of $(1/(\#k-1))\mathbb{Z}/\mathbb{Z}$ by

$$\begin{eqnarray}\unicode[STIX]{x1D712}_{i}=\operatorname{Teich}_{k}^{-a_{i}(\#k-1)},\qquad \unicode[STIX]{x1D70C}_{j}=\operatorname{Teich}_{k}^{-b_{j}(\#k-1)}.\end{eqnarray}$$

Then ${\mathcal{F}}$ has finite $G_{\text{arith}}$ if and only if the following conditions hold. For every $N\in (\mathbb{Z}/(\#k-1)\mathbb{Z})^{\times }$ , and for every $x\in (\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , we have the inequality

$$\begin{eqnarray}\mathop{\sum }_{i}V(Na_{i}+x)+\mathop{\sum }_{j}V(-Nb_{j}-x)\geqslant (n-1)/2+(1/n)\mathop{\sum }_{i,j}V(Na_{i}-Nb_{j}).\end{eqnarray}$$

When ${\mathcal{H}}$ is a Kloosterman sheaf of the form

$$\begin{eqnarray}\text{Kl}(\unicode[STIX]{x1D713};~\text{all non-trivial }\unicode[STIX]{x1D712}~\text{of order dividing }D)\end{eqnarray}$$

for some prime to $p$ integer $D$ , the criterion is that for all $x\in (\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , we have

$$\begin{eqnarray}V(Dx)+1/2\geqslant V(x).\end{eqnarray}$$

When ${\mathcal{H}}$ is a hypergeometric sheaf of the form

$$\begin{eqnarray}{\mathcal{H}}(\unicode[STIX]{x1D713};~\text{all }\unicode[STIX]{x1D712}~\text{of order dividing }D;~\text{a single }\unicode[STIX]{x1D70C}~\text{of order }M)\end{eqnarray}$$

for $D$ and $M$ prime to $p$ integers such that $V(aD/M)=1/2$ for all integers $a$ prime to $M$ , the criterion is that

$$\begin{eqnarray}V(Dx)+V(a/M-x)\geqslant 1/2\end{eqnarray}$$

for every integer $a$ prime to $M$ .

To prove part (2) of Theorem 4.2, we take $D=(q+1)/2$ and $M=q+1$ . Here $D/M=1/2$ , so the condition $V(D/M)=1/2$ is met. We have

$$\begin{eqnarray}V(((q+1)/2)x)+V(a/(q+1)-x)\geqslant V(((q-1)/2)x+a/(q+1)),\end{eqnarray}$$

by property (5) of $V$ . Using property (4), we have

$$\begin{eqnarray}V(a/(q+1)-x)=V(aq/(q+1)-qx)=V(-a/(q+1)-qx),\end{eqnarray}$$

the second equality because $q$ is $-1~\text{mod}~q+1$ . So we also have the inequality

$$\begin{eqnarray}\displaystyle & & \displaystyle V(((q+1)/2)x)+V(a/(q+1)-x)\nonumber\\ \displaystyle & & \displaystyle \quad =V(((q+1)/2)x)+V(-a/(q+1)-qx)\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant V(((1-q)/2)x-a/(q+1)).\nonumber\end{eqnarray}$$

Adding these inequalities, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle 2(V(((q+1)/2)x)+V(a/(q+1)-x))\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant V(((q-1)/2)x+a/(q+1))+V(((1-q)/2)x-a/(q+1)).\nonumber\end{eqnarray}$$

For $x$ such that $((q-1)/2)x+a/(q+1)\neq 0$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , the two arguments are non-zero negatives of each other, and by property (2) of $V$ , they sum to $1$ and we are done.

In the remaining case,

$$\begin{eqnarray}((q+1)/2)x=x+((q-1)/2)x=x-a/(q+1),\end{eqnarray}$$

and hence

$$\begin{eqnarray}V(((q+1)/2)x)+V(a/(q+1)-x)=V(x-a/(q+1))+V(a/(q+1)-x),\end{eqnarray}$$

which is ${\geqslant}\!1$ unless $x=a/(q+1)$ . In that case $((q+1)/2)x=0$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ , but this is impossible, since $a/(q+1)$ has full order $q+1$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ .

To prove part (1) of 4.3, we take $D=(q^{n}+1)/(q+1)$ with $n\geqslant 3$ odd. We must show

$$\begin{eqnarray}V(((q^{n}+1)/(q+1))x)+1/2\geqslant V(x).\end{eqnarray}$$

This trivially holds for $x=0$ . For $x\neq 0$ , use $V(x)=1-V(-x)$ to write the criterion as

$$\begin{eqnarray}V(((q^{n}+1)/(q+1))x)+V(-x)\geqslant 1/2\quad \text{for all }x\neq 0.\end{eqnarray}$$

This sum is

$$\begin{eqnarray}{\geqslant}\!V(((q^{n}+1)/(q+1)-1)x).\end{eqnarray}$$

Replacing $V(((q^{n}+1)/(q+1))x)$ by $V(q((q^{n}+1)/(q+1))x)$ , and replacing $V(-x)$ by $V(-q^{n}x)$ , this same sum is

$$\begin{eqnarray}{\geqslant}\!V((q(q^{n}+1)/(q+1)-q^{n})x).\end{eqnarray}$$

The two quantities $((q^{n}+1)/(q+1)-1)x$ and $(q(q^{n}+1)/(q+1)-q^{n})x$ are negative of each other, i.e., they sum to

$$\begin{eqnarray}((q+1)(q^{n}+1)/(q+1)-1-q^{n})x=0.\end{eqnarray}$$

So we are done, unless $x$ is such that $((q^{n}+1)/(q+1))x=x$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ . In that case,

$$\begin{eqnarray}V(((q^{n}+1)/(q+1))x)+V(-x)=V(x)+V(-x)=1,\end{eqnarray}$$

as $x\neq 0$ .

To prove part (2) of 4.3, we take $D=(q^{n}+1)/(q+1)$ and we take $M=D(q+1)/m$ , for $m$ some proper divisor $m$ of $q+1$ . Then $D/M=m/(q+1)$ . For any $a$ prime to $M$ , we have $V(aD/M)=V(am/(q+1))=1/2$ , thanks to the Stickelberger identity [Reference Berndt, Evans and Williams1, 11.6.1]. We must show that

$$\begin{eqnarray}V(Dx)+V(a/M-x)\geqslant 1/2\end{eqnarray}$$

for all $a$ prime to $M$ . This sum is

$$\begin{eqnarray}{\geqslant}V((D-1)x+a/M).\end{eqnarray}$$

Replacing $V(Dx)$ by $V(qDx)$ and $V(a/M-x)$ by $V(q^{n}a/M-q^{n}x)$ , this same sum is

$$\begin{eqnarray}{\geqslant}V((qD-q^{n})x+q^{n}a/M).\end{eqnarray}$$

The two quantities $(D-1)x+a/M$ and $(qD-q^{n})x+q^{n}a/M$ are negatives of each other; they sum to

$$\begin{eqnarray}\displaystyle ((q+1)D-1-q^{n})x+(q^{n}+1)a/M & = & \displaystyle (q^{n}+1)a/M\nonumber\\ \displaystyle & = & \displaystyle (q^{n}+1)am/(D(q+1))\nonumber\\ \displaystyle & = & \displaystyle am=0\quad \text{in }(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}.\nonumber\end{eqnarray}$$

So we are done unless $x$ is such that $Dx=x-a/M$ in $(\mathbb{Q}/\mathbb{Z})_{\text{prime to}~p}$ . In that case,

$$\begin{eqnarray}V(Dx)=V(x-a/M),\end{eqnarray}$$

and thus

$$\begin{eqnarray}V(Dx)+V(a/M-x)=V(x-a/M)+V(a/M-x).\end{eqnarray}$$

So we are done unless $x=a/M$ . But $Dx-a/M$ , so $Dx=0$ , i.e., $D(a/M)=0$ . But

$$\begin{eqnarray}D(a/M)=D(am/(D(q+1)))=am/(q+1)\end{eqnarray}$$

is non-zero.

This completes the proofs of Kubert’s theorems.

6 Some numerology for $\operatorname{SL}(2,\mathbb{F}_{q})$

Suppose $q\geqslant 5$ is odd. The group $\operatorname{PSL}(2,q):=\operatorname{PSL}(2,\mathbb{F}_{q})$ is simple, and, with the exception of $q=9$ , its Schur cover group is the double cover $\operatorname{SL}(2,q):=\operatorname{SL}(2,\mathbb{F}_{q})$ . The group $\operatorname{SL}(2,q)$ has two irreducible representations of degree $(q-1)/2$ , and two of degree $(q+1)/2$ . The characters of these four representations all take values in the (ring of integers of the) field $\mathbb{Q}(\sqrt{\unicode[STIX]{x1D716}q})$ , with $\unicode[STIX]{x1D716}=(-1)^{(q-1)/2}$ . When $q$ is an odd power of $p$ , the characters of the two representations of each of the degrees $(q\pm 1)/2$ are algebraically conjugate, by the Galois group of $\mathbb{Q}(\sqrt{\unicode[STIX]{x1D716}q})/\mathbb{Q}$ .

If $q$ is $1~\text{mod}~4$ , the two representations of even degree $(q-1)/2$ are both symplectically self-dual, and the two of odd degree $(q+1)/2$ are both orthogonally self-dual. Those of odd degree $(q+1)/2$ factor through $\operatorname{PSL}(2,q)$ , but not those of even degree $(q-1)/2$ .

If $q$ is $3~\text{mod}~4$ , none of these four representations is self-dual. Those of odd degree $(q-1)/2$ factor through $\operatorname{PSL}(2,q)$ , but not those of even degree $(q+1)/2$ .

7 Some numerology for $\operatorname{SU}(\text{odd}~n,\mathbb{F}_{q})$

In this $\operatorname{SU}$ discussion, we assume throughout that $n\geqslant 3$ is odd. We assume further that either $q>2$ or $n>3$ , i.e., we rule out the case of $\operatorname{SU}(3,\mathbb{F}_{2})$ . The group $\operatorname{PSU}(n,q):=\operatorname{PSU}(n,\mathbb{F}_{q})$ is simple.

The group $\operatorname{SU}(n,q):=\operatorname{SU}(n,\mathbb{F}_{q})$ has one representation of degree $(q^{n}+1)/(q+1)-1$ , and it has $q$ representations of degree $(q^{n}+1)/(q+1)$ , cf. [Reference Hiss and Malle14, Theorem 16]. The representation of degree

$$\begin{eqnarray}(q^{n}+1)/(q+1)-1\end{eqnarray}$$

is symplectically self-dual. When $q$ is odd, precisely one of the $q$ representations of degree $(q^{n}+1)/(q+1)$ is self-dual, and its autoduality is orthogonal. If $q$ is even, none of the $q$ representations of degree $(q^{n}+1)/(q+1)$ is self-dual.

The representation of degree $(q^{n}+1)/(q+1)-1$ factors through $\operatorname{PSU}(n,q)$ . When $q$ is odd, the unique self-dual representation of degree $(q^{n}+1)/(q+1)$ also factors through $\operatorname{PSU}(n,q)$ .

The order of the center of $\operatorname{SU}(n,q)$ is $\text{gcd}(n,q+1)$ . Of the $q+1$ representations of degree either $(q^{n}+1)/(q+1)-1$ or $(q^{n}+1)/(q+1)$ , precisely $(q+1)/\text{gcd}(n,q+1)$ of them factor through $\operatorname{PSU}(n,q)$ . [Notice that if $n=q+1$ , then $q$ must be even (as $n$ is odd), and none of the $q$ representations of degree $(q^{n}+1)/(q+1)$ factors through $\operatorname{PSU}(n,q)$ .]

8 The conjectures: preparations

In this section, we refine our determinant calculations for the local systems which will figure in the conjectures.

We begin with the $\operatorname{SL}(2)$ case.

Lemma 8.1. Suppose $q$ is odd. Denote by $\unicode[STIX]{x1D713}_{-2}$ the additive character $\unicode[STIX]{x1D713}_{-2}:x\mapsto \unicode[STIX]{x1D713}(-2x)$ , and define

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}:=\unicode[STIX]{x1D6FD}(\mathbb{F}_{q},\unicode[STIX]{x1D713},(q+1)/2):=-g(\unicode[STIX]{x1D713}_{-2},\unicode[STIX]{x1D712}_{2}).\end{eqnarray}$$

1. (1) Suppose $q$ is $1~\text{mod}~4$ . Then the twisted local system

   $$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2):={\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2)\otimes \unicode[STIX]{x1D6FD}^{-\text{deg}}\end{eqnarray}$$
   has
   $$\begin{eqnarray}G_{\text{geom}}\subset G_{\text{arith}}\subset \operatorname{Sp}((q-1)/2,\overline{\mathbb{Q}_{\ell }}),\end{eqnarray}$$
   and the twisted local system
   $$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2):={\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2)\otimes \unicode[STIX]{x1D6FD}^{-\text{deg}}\end{eqnarray}$$
   has
   $$\begin{eqnarray}G_{\text{geom}}\subset G_{\text{arith}}\subset \operatorname{SO}((q+1)/2,\overline{\mathbb{Q}_{\ell }}).\end{eqnarray}$$

2. (2) Suppose $q$ is $3~\text{mod}~4$ . Then the twisted local system

   $$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2):={\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2)\otimes \unicode[STIX]{x1D6FD}^{-\text{deg}}\end{eqnarray}$$
   has
   $$\begin{eqnarray}G_{\text{geom}}\subset G_{\text{arith}}\subset \operatorname{SL}((q-1)/2,\overline{\mathbb{Q}_{\ell }}),\end{eqnarray}$$
   and the twisted local system
   $$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2):={\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2)\otimes \unicode[STIX]{x1D6FD}^{-\text{deg}}\end{eqnarray}$$
   has
   $$\begin{eqnarray}G_{\text{geom}}\subset G_{\text{arith}}\subset \operatorname{SL}((q+1)/2,\overline{\mathbb{Q}_{\ell }}).\end{eqnarray}$$

Proof. For $p$ the characteristic of $\mathbb{F}_{q}$ , all traces a priori lie in $\mathbb{Q}(\unicode[STIX]{x1D701}_{p})$ . The Galois group of $\mathbb{Q}(\unicode[STIX]{x1D701}_{p})/\mathbb{Q}$ is $\mathbb{F}_{p}^{\times }$ , and the Galois group of $\mathbb{Q}(\unicode[STIX]{x1D701}_{p})/\mathbb{Q}(\sqrt{\unicode[STIX]{x1D716}_{p}p})$ is the subgroup of squares in $\mathbb{F}_{p}^{\times }$ . For $\unicode[STIX]{x1D712}$ either $\unicode[STIX]{x1D7D9}$ or $\unicode[STIX]{x1D712}_{2}$ , and $k/\mathbb{F}_{q}$ a finite extension, the trace at $t\in k$ is

$$\begin{eqnarray}(1/g)\mathop{\sum }_{x\in k}\unicode[STIX]{x1D712}(x)\unicode[STIX]{x1D713}(x^{(q+1)/2}+tx),\end{eqnarray}$$

with $g$ the gauss sum $g:=g(\unicode[STIX]{x1D713}_{-2},\unicode[STIX]{x1D712}_{2})$ . Each of these sums is invariant under the effect of $a^{2}$ , for any $a\in \mathbb{F}_{p}^{\times }$ , the “trick” being that $a^{2}=(a^{2})^{(q+1)/2}$ . Indeed, the effect of $a^{2}$ is to map this sum to

$$\begin{eqnarray}\displaystyle & & \displaystyle (1/g)\mathop{\sum }_{x\in k}\unicode[STIX]{x1D712}(x)\unicode[STIX]{x1D713}(a^{2}x^{(q+1)/2}+ta^{2}x)\nonumber\\ \displaystyle & & \displaystyle \quad =(1/g)\mathop{\sum }_{x\in k}\unicode[STIX]{x1D712}(x)\unicode[STIX]{x1D713}((a^{2}x)^{(q+1)/2}+ta^{2}x)\nonumber\\ \displaystyle & & \displaystyle \quad =(1/g)\mathop{\sum }_{x\in k}\unicode[STIX]{x1D712}(a^{2}x)\unicode[STIX]{x1D713}((a^{2}x)^{(q+1)/2}+ta^{2}x)\nonumber\\ \displaystyle & & \displaystyle \quad =(1/g)\mathop{\sum }_{x\in k}\unicode[STIX]{x1D712}(x)\unicode[STIX]{x1D713}(x^{(q+1)/2}+tx).\nonumber\end{eqnarray}$$

When $q$ is an even power of $p$ , then any $a\in \mathbb{F}_{p}^{\times }$ becomes a square $b^{2}$ for some $b\in \mathbb{F}_{q}^{\times }$ , and then each sum is invariant under $a$ , by the substitution $x\mapsto b^{2}x$ (now because $b^{2}=(b^{2})^{(q+1)/2}$ ).◻

We now turn to the $\operatorname{SU}(\text{odd }n)$ case. We assume that $n\geqslant 3$ is odd, and that either $q>2$ or $n>3$ .

Proof. Parts (1) and (3) result from the $D$ odd case of the determinant lemma. Here

$$\begin{eqnarray}D=(q^{n}+1)/(q+1)=1+q(q-1)\mathop{\sum }_{i=0}^{(n-3)/2}q^{2i}\end{eqnarray}$$

is odd, and $D$ is $1~\text{mod}~p$ . In both cases, the trace function of ${\mathcal{G}}$ has real (in fact, integer, because $a^{D}=a$ for $a\in \mathbb{F}_{p}^{\times }$ and such an $a$ is a square, indeed a $q+1$ ’st power, in $\mathbb{F}_{q^{2}}$ ) values. As ${\mathcal{G}}$ is pure of weight zero, it is self-dual, and we argue as in the proof of Lemma 8.1.

To prove (2), for $\unicode[STIX]{x1D713}$ a non-trivial additive character of $\mathbb{F}_{q}$ and $\unicode[STIX]{x1D712}$ a non-trivial multiplicative character of $\mathbb{F}_{q^{2}}^{\times }$ of order $m$ dividing $q+1$ , we have the Stickelberger determination [Reference Berndt, Evans and Williams1, Theorem 11.6.1] of $g(\unicode[STIX]{x1D713},\unicode[STIX]{x1D712})$ over $\mathbb{F}_{q^{2}}$ ; it is equal to $q$ if $q$ is even, and to $(-1)^{(q+1)/m}q$ if $q$ is odd. [In the cited reference, the $\unicode[STIX]{x1D713}$ is a $\unicode[STIX]{x1D6F9}$ that comes from $\mathbb{F}_{p}$ . So our $\unicode[STIX]{x1D713}$ is of the form $\unicode[STIX]{x1D6F9}_{\unicode[STIX]{x1D706}}$ for some $\unicode[STIX]{x1D706}\in \mathbb{F}_{q}^{\times }$ . But any such $\unicode[STIX]{x1D706}$ is a $q+1$ ’st power in $\mathbb{F}_{q^{2}}$ (surjectivity of the norm), so $g(\unicode[STIX]{x1D713},\unicode[STIX]{x1D712})=g(\unicode[STIX]{x1D6F9},\unicode[STIX]{x1D712})$ .]◻

9 The conjectures

10 Comments on the conjectures

Suppose we use a non-trivial additive character $\unicode[STIX]{x1D713}$ of $\mathbb{F}_{q}$ which comes (by composition with the trace) from a character of the prime field $\mathbb{F}_{p}$ . Then both the local systems

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2)\quad \text{and}\quad {\mathcal{F}}(\mathbb{F}_{q},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2)\end{eqnarray}$$

on $\mathbb{A}^{1}/\mathbb{F}_{q}$ come, by extension of scalars, from the local systems

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2)\quad \text{and}\quad {\mathcal{F}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2)\end{eqnarray}$$

on $\mathbb{A}^{1}/\mathbb{F}_{p}$ . Moreover, if we use the gauss sum $-g(\unicode[STIX]{x1D713}_{-2},\unicode[STIX]{x1D712}_{2})$ over $\mathbb{F}_{p}$ as the twisting factor, we get descents

$$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q+1)/2)\quad \text{and}\quad {\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q+1)/2)\end{eqnarray}$$

to $\mathbb{A}^{1}/\mathbb{F}_{p}$ of the corresponding ${\mathcal{G}}$ on $\mathbb{A}^{1}/\mathbb{F}_{q}$ . The $G_{\text{geom}}$ groups for the descents do not change, but their $G_{\text{arith}}$ groups can grow. A natural guess for what they are is the following. The Galois group of $\mathbb{F}_{q}/\mathbb{F}_{p}$ acts coefficientwise on the group $\operatorname{SL}(2,q)$ and fixes the isomorphism classes of both representations of degree $(q-1)/2$ and of both representations of degree $(q+1)/2$ (because the Galois action preserves each of the two conjugacy classes of unipotent elements, which are distinguished by whether the upper right entry is a square or not). So each of these representations extends to the semidirect product of $\operatorname{SL}(2,q)$ with $\operatorname{Gal}(\mathbb{F}_{q}/\mathbb{F}_{p})$ . The natural guess is that $G_{\text{arith}}$ for the descended ${\mathcal{G}}$ is the image of this semidirect product in the corresponding representation.

Similarly, when $\unicode[STIX]{x1D713}$ starts life over $\mathbb{F}_{p}$ , for each odd $n\geqslant 3$ the local system

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{q^{2}},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q^{n}+1)/(q+1))\end{eqnarray}$$

and, if $q$ is odd, the local system

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{q^{2}},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q^{n}+1)/(q+1))\end{eqnarray}$$

descend to $\mathbb{A}^{1}/\mathbb{F}_{p}$ ; just replace $\mathbb{F}_{q^{2}}$ by $\mathbb{F}_{p}$ in the name. In the $q$ odd case, using $-g(\unicode[STIX]{x1D713}_{(-1)^{(n-1)/2}},\unicode[STIX]{x1D712}_{2})$ over $\mathbb{F}_{p}$ as the twisting factor, we get a descent ${\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q^{n}+1)/(q+1))$ to $\mathbb{A}^{1}/\mathbb{F}_{p}$ of the corresponding ${\mathcal{G}}$ . When $q$ is $3~\text{mod}~4$ , this same twisting factor (or any quadratic gauss sum over $\mathbb{F}_{p}$ ) gives a descent ${\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q^{n}+1)/(q+1))$ . We do not know the “right”Footnote ⁵ twisting factor to use when $q$ is not $3~\text{mod}~4$ . Just as in the paragraph above, the $G_{\text{geom}}$ groups for the descents do not change, but their $G_{\text{arith}}$ groups can grow. The Galois group of $\mathbb{F}_{q^{2}}/\mathbb{F}_{p}$ acts coefficientwise on the group $\operatorname{SU}(n,q)$ and fixes the isomorphism class of its unique irreducible representation of degree

$$\begin{eqnarray}(q^{n}+1)/(q+1)-1\end{eqnarray}$$

(precisely by the uniqueness). When $q$ is odd, the Galois action also fixes the isomorphism class of the unique orthogonal irreducible representation of degree $(q^{n}+1)/(q+1)$ (again by uniqueness). So each of these representations extends to the semidirect product of $\operatorname{SU}(n,q)$ with $\operatorname{Gal}(\mathbb{F}_{q}/\mathbb{F}_{p})$ . The natural guess is that $G_{\text{arith}}$ for the descended ${\mathcal{G}}$ is the image of this semidirect product in the corresponding representation.

Once we think in terms of these descents, the following question arises. Suppose we are in a given characteristic $p$ , and are told that one of the conjectures applies to ${\mathcal{F}}(\mathbb{F}_{p},\unicode[STIX]{x1D7D9},D)$ (in which case it also applies to ${\mathcal{F}}(\mathbb{F}_{p},\unicode[STIX]{x1D712}_{2},D)$ when $p$ is odd). Can we be sure which conjecture?

Can there be more than one way of writing $D$ as $(q^{n}+1)/(q+1)$ for $q$ some power of $p$ and $n$ some odd integer ${\geqslant}3$ ? The answer is no. Begin with the identity

$$\begin{eqnarray}(q^{n}+1)/(q+1)=1-q+q^{2}\cdots +q^{n-1}=1+q(q-1)[1+q^{2}+\cdots +q^{n-3}].\end{eqnarray}$$

When $q$ is $p^{r}$ , the base $p$ expansion of $q-1$ is a sequence of $r$ digits, each $p-1$ . So the base $p$ digit expression of $(q^{n}+1)/(q+1)-1$ is $(n-1)/2$ successive strings, each consisting of $r$ digits $p-1$ followed by $r$ digits $0$ . For example, in base $p$ we have

$$\begin{eqnarray}(p^{15}+1)/(p^{3}+1)-1=XXX000XXX000\quad \text{with }X=p-1.\end{eqnarray}$$

So both $r$ and $n$ are determined by the base $p$ expression of $D-1$ .

When $p$ is odd, we must also distinguish $(q^{n}+1)/(q+1)$ , $q$ some power of $p$ and $n\geqslant 3$ odd, from $(p^{a}+1)/2$ , $a\geqslant 1$ . If $p\geqslant 5$ , two such expressions can never be equal, because the first is $1~\text{mod}~p$ and the second is $1/2~\text{mod}~p$ . To do the general case, subtract $1$ from each and compare the base $p$ expressions. That of $(p^{a}-1)/2$ consists of a sequence of $a$ digits, each $(p-1)/2$ . We must also note that if $D=(p^{a}+1)/2$ , then $p^{a}$ is determined, it is $2D-1$ .

A slight variant on this question is this. Still in a fixed characteristic $p$ , can there be more than one local system of a given rank $R$ to which one of the conjectures applies? If $R$ is to be either $(q^{n}+1)/(q+1)$ or $(q_{1}^{m}+1)/(q_{1}+1)-1$ , we can tell which of the two, because the first is $1~\text{mod}~p$ and the second is $0~\text{mod}~p$ . Then looking at the base $p$ expansion of $R$ tells the rest. Similarly, looking at the base $p$ expansions of $(p^{a}+1)/2=(p^{a}-1)/2+1$ and of $(p^{a}-1)/2$ allows us to separate these cases from each other and from the $\operatorname{SU}$ cases.

On the other hand, for a given rank, different characteristics can give rise to cases of the conjectures. For example, any time we have twin primes or twin prime powers (e.g. $9,11$ or $23,25$ or $27,29$ or $79,81$ or $81,83$ ), say $q_{2}=q_{1}+2$ , then ${\mathcal{F}}(\mathbb{F}_{q_{1}},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(q_{1}+1)/2)$ in characteristic $p_{1}$ and ${\mathcal{F}}(\mathbb{F}_{q_{2}},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(q_{2}+1)/2)$ in characteristic $p_{2}$ are local systems of the same rank $(q_{1}+1)/2=(q_{2}-1)/2$ .

More interesting examples involve ranks having $\operatorname{SL}$ conjectures in some characteristics and $\operatorname{SU}$ conjectures in others. For example, with rank $6$ we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{3},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(3^{3}+1)/(3+1))\end{eqnarray}$$

in characteristic $3$ , and the two $\operatorname{SL}$ cases for the twin primes $11,13$ .

For rank $7$ , we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{3},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(3^{3}+1)/(3+1)),\end{eqnarray}$$

which has $G_{\text{geom}}$ the image of $\operatorname{SU}(3,3)$ in its unique orthogonal irreducible representation of degree $7$ , cf. [Reference Katz20, 4.15], and we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{13},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(13+1)/2)\end{eqnarray}$$

which has $G_{\text{geom}}=\operatorname{PSL}(2,13)$ , cf. [Reference Katz20, 4.13]. In both of these rank $7$ cases, $G_{\text{geom}}$ is a subgroup of the exceptional group $G_{2}$ .

For rank $10$ , we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{2},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(2^{5}+1)/(2+1))\end{eqnarray}$$

in characteristic $2$ , and we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{19},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(19+1)/2)\end{eqnarray}$$

in characteristic $19$ .

For rank $11$ , we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{2},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{3},(2^{5}+1)/(2+1))\end{eqnarray}$$

in characteristic $2$ , with $\unicode[STIX]{x1D712}_{3}$ either of the two characters of $\mathbb{F}_{4}^{\times }$ of order $3$ , and we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{23},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(23+1)/2)\end{eqnarray}$$

in characteristic $23$ .

For rank $12$ we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{2},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(4^{3}+1)/(4+1))\end{eqnarray}$$

in characteristic $2$ , and the two $\operatorname{SL}$ cases for the twin prime powers $23,25$ .

For rank $13$ we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{2},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{5},(4^{3}+1)/(4+1))\end{eqnarray}$$

in characteristic $2$ , with $\unicode[STIX]{x1D712}_{5}$ any of the four characters of order $5$ of $\mathbb{F}_{4^{2}}^{\times }$ , and the two $\operatorname{SL}$ cases for the twin prime powers $25,27$ .

For rank $20$ we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{5},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(5^{3}+1)/(5+1))\end{eqnarray}$$

in characteristic $5$ , and we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{41},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(41+1)/2)\end{eqnarray}$$

in characteristic $41$ .

For rank $21$ , we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{5},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(5^{3}+1)/(5+1))\end{eqnarray}$$

in characteristic $5$ , and the two $\operatorname{SL}$ cases for the twin primes $41,43$ .

For rank $993$ , we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{2},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(2^{15}+1)/(2^{5}+1))\end{eqnarray}$$

in characteristic $2$ , and we have

$$\begin{eqnarray}{\mathcal{F}}(\mathbb{F}_{1987},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(1987+1)/2)\end{eqnarray}$$

in characteristic $1987$ .

11 Verification, for $\operatorname{SL}(2,p)$

Proof. Let ${\mathcal{G}}$ be one of the local systems in question, i.e., either

$$\begin{eqnarray}{\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(p+1)/2)\quad \text{or}\quad {\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(p+1)/2).\end{eqnarray}$$

Its $G_{\text{geom}}$ has order divisible by $p$ (the wild inertia group $P_{\infty }$ acts non-trivially on ${\mathcal{G}}$ ), its determinant is trivial, and it is a primitive irreducible subgroup of $\operatorname{SL}(n,\overline{\mathbb{Q}_{\ell }})$ with $n$ one of $(p\pm 1)/2$ . [It is primitive because if the representation were induced, ${\mathcal{G}}$ would be, geometrically, the direct image of a local system on a finite étale connected covering of $\mathbb{A}^{1}/\overline{\mathbb{F}_{p}}$ of some degree $d>1$ dividing $n$ . But $n<p$ , hence $d<p$ , and $\mathbb{A}^{1}/\overline{\mathbb{F}_{p}}$ has no finite étale connected covers of degree $1<d<p$ .] The larger group $G_{\text{arith}}$ has the same properties.

We have already seen that the trace functions of both these local systems take values in $\mathbb{Q}(\sqrt{\unicode[STIX]{x1D716}p})$ for $\unicode[STIX]{x1D716}=(-1)^{(p-1)/2}$ . Because $p\geqslant 5$ , the only roots of unity in this field are $\pm 1$ . So the only scalars which can possibly lie in $G_{\text{geom}}$ or $G_{\text{arith}}$ are among $\pm 1$ . As these groups are irreducible subgroups of the ambient $\operatorname{SL}(n,\overline{\mathbb{Q}_{\ell }})$ , we conclude that

$$\begin{eqnarray}\displaystyle Z(G_{\text{arith}}) & = & \displaystyle Z(G_{\text{geom}})=\{1\}\quad \text{if }n\text{ is odd},\nonumber\\ \displaystyle Z(G_{\text{arith}}) & \subset & \displaystyle Z(G_{\text{geom}})\subset \pm 1\quad \text{if }n\text{ is even}.\nonumber\end{eqnarray}$$

Consider first the case of ${\mathcal{G}}:={\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},(p+1)/2)$ . Here the representation has dimension $n=(p-1)/2$ , and so $p=2n+1$ . By [Reference Brauer2, (2B), (2C)], both $G_{\text{geom}}/Z(G_{\text{geom}})$ and $G_{\text{arith}}/Z(G_{\text{arith}})$ are isomorphic to $\operatorname{PSL}(2,p)$ . If $n=(p-1)/2$ is odd, then the centers are trivial, and we are done. If $n=(p-1)/2$ is even, then neither center can be trivial, since $\operatorname{PSL}(2,p)$ has no irreducible representation of degree $n=(p-1)/2$ . Thus both $G_{\text{geom}}$ and $G_{\text{arith}}$ contain $\pm 1$ , and their quotients by $\pm 1$ are $\operatorname{PSL}(2,p)$ . Neither $G_{\text{geom}}$ or $G_{\text{arith}}$ can be isomorphic to the product $\pm 1\times \operatorname{PSL}(2,p)$ , again because $\operatorname{PSL}(2,p)$ has no irreducible representation of degree $n=(p-1)/2$ . So both $G_{\text{geom}}$ and $G_{\text{arith}}$ are isomorphic to the Schur double cover of $\operatorname{PSL}(2,p)$ , which is $\operatorname{SL}(2,p)$ , and again we are done.

In the case of ${\mathcal{G}}:={\mathcal{G}}(\mathbb{F}_{p},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},(p+1)/2)$ , the representation dimension is $n=(p+1)/2$ . In this case $p=2n-1>n+1$ (because $p\geqslant 5$ ), and Brauer, cf. [Reference Brauer2, (2B)] and [Reference Brauer3], tells us that, by a theorem of Feit [Reference Feit7], only the first power of $p$ divides $\#G_{\text{arith}}/Z(G_{\text{arith}})$ or divides $\#G_{\text{geom}}/Z(G_{\text{geom}})$ . As the centers have order dividing $2$ , only the first power of $p$ divides either $\#G_{\text{arith}}$ or divides $\#G_{\text{geom}}$ . Furthermore, the $p$ -Sylow subgroups of $G_{\text{geom}}$ are not normal subgroups. Otherwise the quotient of $G_{\text{geom}}$ by the $p$ -Sylow subgroup would be a prime to $p$ quotient of $\unicode[STIX]{x1D70B}_{1}(\mathbb{A}^{1}/\overline{\mathbb{F}_{p}})$ , so trivial, hence $G_{\text{geom}}$ would be a $p$ -group, in fact of order $p$ , which is nonsense, as it has an irreducible representation of degree $1<n<p$ . Because $p$ only divides $\#G_{\text{arith}}$ to the first power, its $p$ -Sylows are conjugate to $p$ -Sylows of $G_{\text{geom}}$ , so they are certainly not normal subgroups of $G_{\text{arith}}$ .

This information allows us to apply a theorem of Tuan [Reference Tuan28, p. 111, first four paragraphs], which tells us that for $p>7$ , both $G_{\text{geom}}/Z(G_{\text{geom}})$ and $G_{\text{arith}}/Z(G_{\text{arith}})$ are isomorphic to $\operatorname{PSL}(2,p)$ . For $p=5$ (respectively $p=7$ ) we could also have the alternating group $A_{6}$ (respectively $A_{7}$ ) in addition to $\operatorname{PSL}(2,p)$ .

In the case $p=5$ , we have shown in [Reference Katz20, 3.9] that $G_{\text{geom}}$ is $\operatorname{PSL}(2,5)=A_{5}$ . The group $A_{5}$ is its own normalizer in $\operatorname{SO}(3,\overline{\mathbb{Q}_{\ell }})$ , hence we have $G_{\text{arith}}=A_{5}$ as well.

In the case $p=7$ , we invoke an extraordinary isomorphismFootnote ⁶ , where we exploit the fact that for the rank three local system

$$\begin{eqnarray}{\mathcal{G}}_{3}:={\mathcal{G}}(\mathbb{F}_{7},\unicode[STIX]{x1D713},\unicode[STIX]{x1D7D9},4),\end{eqnarray}$$

we already know that $G_{\text{geom}}=G_{\text{arith}}=\operatorname{PSL}(2,7)$ . We wish to prove that for

$$\begin{eqnarray}{\mathcal{G}}_{4}:={\mathcal{G}}(\mathbb{F}_{7},\unicode[STIX]{x1D713},\unicode[STIX]{x1D712}_{2},4),\end{eqnarray}$$

both $G_{\text{geom}}/Z(G_{\text{geom}})$ and $G_{\text{arith}}/Z(G_{\text{arith}})$ are isomorphic to $\operatorname{PSL}(2,7)$ . The extraordinary isomorphism is

$$\begin{eqnarray}\operatorname{Sym}^{2}({\mathcal{G}}_{3})\cong \unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4}).\end{eqnarray}$$

Granting this, which we will prove below, we argue as follows. The images of both $\unicode[STIX]{x1D70B}_{1}^{\text{geom}}(\mathbb{A}^{1}/\mathbb{F}_{7})$ and of $\unicode[STIX]{x1D70B}_{1}^{\text{arith}}(\mathbb{A}^{1}/\mathbb{F}_{7})$ acting on ${\mathcal{G}}_{3}$ are $\operatorname{PSL}(2,7)$ , so this is also their image acting on $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ (the homomorphism

$$\begin{eqnarray}\operatorname{Sym}^{2}:\operatorname{SL}(3,\overline{\mathbb{Q}_{\ell }})\rightarrow \operatorname{SL}(6,\overline{\mathbb{Q}_{\ell }})\end{eqnarray}$$

is injective). Therefore their images acting on $\unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4})$ are also $\operatorname{PSL}(2)$ . The homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}^{2}:\operatorname{SL}(4,\overline{\mathbb{Q}_{\ell }})\rightarrow \operatorname{SO}(6,\overline{\mathbb{Q}_{\ell }})\end{eqnarray}$$

(which is the spin double cover of $\operatorname{SO}(6)$ ) has kernel $\pm 1$ , so the images of both $\unicode[STIX]{x1D70B}_{1}^{\text{geom}}(\mathbb{A}^{1}/\mathbb{F}_{7})$ and of $\unicode[STIX]{x1D70B}_{1}^{\text{arith}}(\mathbb{A}^{1}/\mathbb{F}_{7})$ acting on ${\mathcal{G}}_{4}$ have $G_{\text{geom}}/Z(G_{\text{geom}})=G_{\text{arith}}/Z(G_{\text{arith}})=\operatorname{PSL}(2,7)$ .

Once we know that both $G_{\text{geom}}/Z(G_{\text{geom}})$ and $G_{\text{arith}}/Z(G_{\text{arith}})$ are isomorphic to $\operatorname{PSL}(2,p)$ , we argue as follows.

If $p$ is $1~\text{mod}~4$ , then $(p+1)/2$ is odd, and the ambient group $\operatorname{SO}((p+1)/2,\overline{\mathbb{Q}_{\ell }})$ contains no scalars other than $1$ . So in this case the centers $Z(G_{\text{geom}})$ and $Z(G_{\text{arith}})$ are trivial, and we conclude that $G_{\text{geom}}=G_{\text{arith}}=\operatorname{PSL}(2,p)$ .

If $p$ is $3~\text{mod}~4$ , then $(p+1)/2$ is even, and $\operatorname{PSL}(2,p)$ has no irreducible representation of dimension $(p+1)/2$ . Therefore neither of the groups $G_{\text{geom}}$ or $G_{\text{arith}}$ can be either $\operatorname{PSL}(2,p)$ or the product $\pm 1\times \operatorname{PSL}(2,p)$ . Therefore each of these groups is the Schur double cover of $\operatorname{PSL}(2,p)$ , which is $\operatorname{SL}(2,p)$ .

It remains to show that when we choose a non-square $\unicode[STIX]{x1D706}\in \mathbb{F}_{p}^{\times }$ , and replace $\unicode[STIX]{x1D713}$ by $\unicode[STIX]{x1D713}_{\unicode[STIX]{x1D706}}:x\mapsto \unicode[STIX]{x1D713}(\unicode[STIX]{x1D706}x)$ , then in both cases we replace $G_{\text{geom}}$ by the image of $\operatorname{SL}(2,p)$ in its other representation of the same dimension. In both cases the trace functions of the two representations of the given dimension (either $(p-1)/2$ or $(p+1)/2$ ) are known to be Galois conjugates of each other, by $\operatorname{Gal}(\mathbb{Q}(\sqrt{\unicode[STIX]{x1D716}p})/\mathbb{Q})$ . The replacement of $\unicode[STIX]{x1D713}$ by $\unicode[STIX]{x1D713}_{\unicode[STIX]{x1D706}}$ performs the conjugation by the non-trivial element in this Galois group.◻

12 Existence of the extraordinary isomorphism

We first show the existence of a geometric isomorphism between $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ and $\unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4}).$ For this, we use the following.

Proof. Because ${\mathcal{A}}$ and ${\mathcal{B}}$ and $\overline{{\mathcal{A}}}\otimes {\mathcal{B}}$ are pure, they are geometrically semisimple, hence the $H_{c}^{2}$ in question is the group $\operatorname{Hom}_{\text{geom}}({\mathcal{A}},{\mathcal{B}})$ . Because ${\mathcal{A}}$ is geometrically irreducible and of the same rank as ${\mathcal{B}}$ , any non-zero hom is automatically an isomorphism. So if this $H_{c}^{2}$ is non-zero, ${\mathcal{B}}$ is also geometrically irreducible. The one-dimensionality then follows from Schur’s lemma.

If the $H_{c}^{2}$ vanishes, then the Euler–Poincaré formula gives

$$\begin{eqnarray}\displaystyle \operatorname{dim}H_{c}^{1} & = & \displaystyle -\unicode[STIX]{x1D712}_{c}(\mathbb{A}^{1}/\overline{\mathbb{F}_{q}},\overline{{\mathcal{A}}}\otimes {\mathcal{B}})\nonumber\\ \displaystyle & = & \displaystyle \text{Swan}(\overline{{\mathcal{A}}}\otimes {\mathcal{B}})-\operatorname{rank}(\overline{{\mathcal{A}}}\otimes {\mathcal{B}})\leqslant \unicode[STIX]{x1D706}r^{2}-r^{2}=(\unicode[STIX]{x1D706}-1)r^{2}.\nonumber\end{eqnarray}$$

The sum of traces

$$\begin{eqnarray}\mathop{\sum }_{t\in k}\operatorname{Trace}(\text{Frob}_{t,k}|\overline{{\mathcal{A}}})\operatorname{Trace}(\text{Frob}_{t,k}|{\mathcal{B}})\end{eqnarray}$$

is minus the trace of $\operatorname{Frob}_{k}$ on the $H_{c}^{1}$ , so by Deligne’s Weil II estimate, its absolute value is bounded by $\operatorname{dim}(H_{c}^{1})\sqrt{\#k}$ .◻

We now apply this lemma, with ${\mathcal{A}}=\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ and with ${\mathcal{B}}=\unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4})$ . Both are of rank $6$ , pure of weight zero, and with $\infty$ -slopes ${\leqslant}4/3$ . The quantity $(\unicode[STIX]{x1D706}-1)r^{2}$ is thus $12$ .

To see that $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ is geometrically irreducible, we argue as follows. We already know that $G_{\text{geom}}$ for ${\mathcal{G}}_{3}$ is $\operatorname{PSL}(2,7)$ in one of its three-dimensional irreducible representations, call it $\unicode[STIX]{x1D70C}_{3}$ . The character table of $\operatorname{PSL}(2,7)$ shows that $\operatorname{Sym}^{2}(\unicode[STIX]{x1D70C}_{3})$ is the unique irreducible six-dimensional representation of $\operatorname{PSL}(2,7)$ .

Calculation in Magma shows that the sum of traces over the field $k=GF(7^{3})$ is approximately

$$\begin{eqnarray}18.4662642527365302439092364832\sqrt{7^{3}}.\end{eqnarray}$$

This shows that the $H_{c}^{2}$ cannot vanish, and hence we have the asserted geometric isomorphism. Because both sides are geometrically irreducible, there exists a lisse rank one ${\mathcal{L}}$ on $\mathbb{A}^{1}/\mathbb{F}_{7}$ for which there exists an arithmetic isomorphism of $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ with ${\mathcal{L}}\otimes \unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4})$ . The Swan conductor at $\infty$ of ${\mathcal{L}}$ is ${\leqslant}\!4/3$ , hence either $0$ or $1$ . Thus ${\mathcal{L}}$ is of the form $\unicode[STIX]{x1D6FC}^{\text{degree}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D713}(at)}$ for some scalar $\unicode[STIX]{x1D6FC}\in \overline{\mathbb{Q}_{\ell }}^{\times }$ , and some $a\in \mathbb{F}_{7}$ . At both the points $t=0$ and $t=1$ , $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ and $\unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4})$ have equal $\mathbb{F}_{7}$ traces, namely $2$ and $-1$ respectively. So we have two equalities

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D713}(0)=1,\qquad \unicode[STIX]{x1D6FC}\unicode[STIX]{x1D713}(a)=1.\end{eqnarray}$$

Thus $\unicode[STIX]{x1D6FC}=1$ and $a=0$ , i.e., ${\mathcal{L}}$ is arithmetically trivial. Thus there exists an arithmetic isomorphism of $\operatorname{Sym}^{2}({\mathcal{G}}_{3})$ with $\unicode[STIX]{x1D6EC}^{2}({\mathcal{G}}_{4})$ .

13 $\operatorname{PGL}(2,q)$ d’apres Gross, and hypergeometric sheaves

We write $\operatorname{PGL}(2,\mathbb{F}_{q}):=\operatorname{PGL}(2,q)$ , $q$ any prime power, though later we will specialize to the case $q$ odd. In [Reference Gross10], Gross, using results of Lusztig [Reference Lusztig25], explains how to view the Deligne–Lusztig curve as a $\operatorname{PGL}(2,q)$ torsor ${\mathcal{T}}$ on $\mathbb{G}_{m}/\mathbb{F}_{q}$ . On ${\mathcal{T}}$ , the inertia and wild inertia groups at $0$ and $\infty$ are given explicitly in terms of the Borel $B$ , its unipotent radical $U$ , the split torus $T_{\text{spl}}$ , of order $q-1$ , and the non-split torus $T_{\text{nspl}}$ , of order $q+1$ . We have

$$\begin{eqnarray}\displaystyle & \displaystyle I_{\infty }=B{\vartriangleright}P_{\infty }=U,\qquad I_{\infty }/P_{\infty }=T_{\text{spl}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I_{0}=T_{\text{nspl}},\qquad P_{0}=\{1\}. & \displaystyle \nonumber\end{eqnarray}$$

Concretely, the complete Deligne–Lusztig curve in this case is $\mathbb{P}^{1}/\mathbb{F}_{q}$ , on which $\operatorname{PGL}(2,q)$ acts in the usual way by fractional linear transformation. This action is free on $\mathbb{P}^{1}\setminus \mathbb{P}^{1}(\mathbb{F}_{q^{2}})$ . The quotient of $\mathbb{P}^{1}/\mathbb{F}_{q}$ by $\operatorname{PGL}(2,q)$ is $\mathbb{P}^{1}/\mathbb{F}_{q}$ . The map to the quotient is given explicitly, in terms of a coordinate $x$ upstairs, by

$$\begin{eqnarray}x\mapsto \frac{((x^{q^{2}}-x)/(x^{q}-x))^{q+1}}{(x^{q}-x)^{q(q-1)}}=\frac{((x^{q}-x)^{q-1}+1)^{q+1}}{(x^{q}-x)^{q(q-1)}}.\end{eqnarray}$$

It maps $\mathbb{P}^{1}(\mathbb{F}_{q})$ to $\infty$ , and it maps $\mathbb{P}^{1}(\mathbb{F}_{q^{2}})\setminus \mathbb{P}^{1}(\mathbb{F}_{q})$ to $0$ . Using the fact that $\operatorname{PGL}(2,q)$ is generated by the transformations $x\mapsto ax,a\in \mathbb{F}_{q}^{\times }$ , $x\mapsto 1/x$ , and by the translations $x\mapsto x+b,b\in \mathbb{F}_{q}$ , one checks that the map is $\operatorname{PGL}(2,q)$ -equivariant. For $t$ in an overfield of $\mathbb{F}_{q}$ , the fibre over $t$ consists of the roots of the polynomial

$$\begin{eqnarray}f(x):=((x^{q}-x)^{q-1}+1)^{q+1}-t(x^{q}-x)^{q(q-1)}.\end{eqnarray}$$

Its derivative is easily computed to be

$$\begin{eqnarray}f^{\prime }(x)=[(x^{q}-x)^{q-1}+1]^{q}(x^{q}-x)^{q-2},\end{eqnarray}$$

whose only zeroes lie in $\mathbb{F}_{q^{2}}$ , all points of which map to either $\infty$ (for $x\in \mathbb{F}_{q}$ ) or to $0$ (for $x\in \mathbb{F}_{q^{2}}\setminus \mathbb{F}_{q}$ ). Thus the map makes $\mathbb{P}^{1}\setminus \mathbb{P}^{1}(\mathbb{F}_{q^{2}})$ into a finite étale covering of $\mathbb{G}_{m}/\mathbb{F}_{q}$ of degree $q(q-1)(q+1)=\#\operatorname{PGL}(2,q)$ . As the map is $\operatorname{PGL}(2,q)$ -equivariant, it must be the quotient map.

Now take an irreducible $\overline{\mathbb{Q}_{\ell }}$ -representation

$$\begin{eqnarray}\unicode[STIX]{x1D70C}:\operatorname{PGL}(2,q)\rightarrow \operatorname{GL}(\operatorname{dim}(\unicode[STIX]{x1D70C}),\overline{\mathbb{Q}_{\ell }}),\end{eqnarray}$$

of dimension ${>}\!1$ , and denote by ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ the local system on $\mathbb{G}_{m}/\mathbb{F}_{q}$ obtained by “pushing out” the torsor ${\mathcal{T}}$ by $\unicode[STIX]{x1D70C}$ . Thanks to [Reference Gross10, Corollary p. 2537], we have a great deal of information. First of all, $\text{Swan}_{\infty }({\mathcal{W}}_{\unicode[STIX]{x1D70C}})=1$ . This fact implies that ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ is, geometrically (i.e., on $\mathbb{G}_{m}/\overline{\mathbb{F}_{q}}$ ), a multiplicative translate of a hypergeometric sheaf, cf. [Reference Katz17, 8.5.3.1].

In what follows, we fix a choice of additive character $\unicode[STIX]{x1D713}$ of $\mathbb{F}_{q}$ , and write simply

$$\begin{eqnarray}{\mathcal{H}}(\text{character data})\end{eqnarray}$$

for the hypergeometric sheaf

$$\begin{eqnarray}\operatorname{Hyp}(!,\unicode[STIX]{x1D713}\text{ the same character data})\end{eqnarray}$$

in the notation of [Reference Katz17, 8.2.2].

For the Steinberg representation, $\text{St}$ of dimension $q$ , we have $\operatorname{dim}(({\mathcal{W}}_{\text{St}})^{I_{\infty }})=1$ and $\operatorname{dim}(({\mathcal{W}}_{\text{St}})^{I_{0}})=0$ . For all other irreducibles $\unicode[STIX]{x1D70C}$ of dimension ${>}\!1$ , we have $\operatorname{dim}(({\mathcal{W}}_{\unicode[STIX]{x1D70C}})^{I_{\infty }})=0$ and $\operatorname{dim}(({\mathcal{W}}_{\unicode[STIX]{x1D70C}})^{I_{0}})=1$ .

The irreducible $\unicode[STIX]{x1D70C}$ of dimension ${>}\!1$ have dimension either $q-1$ , $q$ , or $q+1$ . Looking at the character table of $\operatorname{PGL}(2,q)$ , we see that the trace of a non-trivial unipotent element in one of these representations is

$$\begin{eqnarray}\displaystyle -1\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q-1,\nonumber\\ \displaystyle 0\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q,\nonumber\\ \displaystyle 1\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q+1.\nonumber\end{eqnarray}$$

This means that, writing Reg for the regular representation of $P_{\infty }$ , the action of $\unicode[STIX]{x1D70C}|P_{\infty }$ is given by

$$\begin{eqnarray}\displaystyle \text{Reg}-\unicode[STIX]{x1D7D9}\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q-1,\nonumber\\ \displaystyle \text{Reg}\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q,\nonumber\\ \displaystyle \text{Reg}+\unicode[STIX]{x1D7D9}\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q+1.\nonumber\end{eqnarray}$$

This in turn means that the $I_{\infty }$ -representation is of the form

$$\begin{eqnarray}\displaystyle \text{Wild}_{q-1}\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q-1,\nonumber\\ \displaystyle \text{Wild}_{q-1}\oplus (1\text{ dim tame})\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q,\nonumber\\ \displaystyle \text{Wild}_{q-1}\oplus (2\text{ dim tame})\quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C}) & = & \displaystyle q+1,\nonumber\end{eqnarray}$$

where we write $\text{Wild}_{q-1}$ for a totally wild $I_{\infty }$ -representation of dimension $q-1$ and $\operatorname{Swan}1$ .

The $I_{0}$ -representation, being tame, consists of various characters of order dividing $q+1$ . No character can occur more than once, otherwise the local monodromy at $0$ will not be of finite order, cf. [Reference Katz17, 8.4.5(5)]. Thus our hypergeometric must be a multiplicative translate of

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{H}}(\text{all but two char.}^{\prime }\text{s of order dividing }q+1;\emptyset ),\nonumber\\ \displaystyle & & \displaystyle \quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C})=q-1,\nonumber\\ \displaystyle & & \displaystyle {\mathcal{H}}(\text{all but one char. of order dividing }q+1;\text{one char. of order dividing }q-1),\nonumber\\ \displaystyle & & \displaystyle \quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C})=q,\nonumber\\ \displaystyle & & \displaystyle {\mathcal{H}}(\text{all char.}^{\prime }\text{s of order dividing }q+1;\text{two char.}^{\prime }\text{s of order dividing }q-1),\nonumber\\ \displaystyle & & \displaystyle \quad \text{if }\operatorname{dim}(\unicode[STIX]{x1D70C})=q+1.\nonumber\end{eqnarray}$$

These facts, together with what we already know about ${\mathcal{W}}_{\text{St}}$ , show that ${\mathcal{W}}_{\text{St}}$ is, geometrically, a multiplicative translate of

$$\begin{eqnarray}{\mathcal{H}}(\text{all non-triv. char.}^{\prime }\text{s of order dividing }q+1;\unicode[STIX]{x1D7D9}).\end{eqnarray}$$

We now bring to bear the fact that every irreducible $\unicode[STIX]{x1D70C}$ is orthogonally self-dual. Then from [Reference Katz17, 8.8.1 and 8.8.2] we see that for any other (i.e., other than Steinberg) irreducible $\unicode[STIX]{x1D70C}$ of dimension ${>}\!1$ , its ${\mathcal{H}}$ must have:

1. (1) In dimension $q-1$ , the two omitted characters are $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ with $\unicode[STIX]{x1D712}\neq \overline{\unicode[STIX]{x1D712}}$ , and $\unicode[STIX]{x1D712}$ of order dividing $q+1$ .

2. (2) In dimension $q$ , with $q$ odd, the omitted character is the quadratic character $\unicode[STIX]{x1D712}_{2}$ , and the bottom character is $\unicode[STIX]{x1D712}_{2}$ . [If $q$ is even, $\text{St}$ is the only irreducible of dimension $q$ .]

3. (3) In dimension $q+1$ , the two “downstairs” characters must be $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ with $\unicode[STIX]{x1D712}\neq \overline{\unicode[STIX]{x1D712}}$ and $\unicode[STIX]{x1D712}$ of order dividing $q-1$ .

When $q$ is odd, there are precisely $(q-1)/2$ unordered pairs $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ as in (1) above, and there are precisely $(q-3)/2$ unordered pairs $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ as in (3) above. When $q$ is even, there are $q/2$ unordered pairs $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ as in (1) above, and there are precisely $(q-2)/2$ unordered pairs $\unicode[STIX]{x1D712},\overline{\unicode[STIX]{x1D712}}$ as in (3) above. These are precisely the number of irreducible $\unicode[STIX]{x1D70C}$ of dimensions $q-1$ and $q+1$ respectively. So each ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ with $\unicode[STIX]{x1D70C}$ an irreducible of dimension ${>}1$ is, geometrically, a multiplicative translate of a hypergeometric of specified form, and every hypergeometric of that specified form is, geometrically, a multiplicative translate of a ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ .

14 Descent of hypergeometric sheaves

Let us denote by

$$\begin{eqnarray}{\mathcal{H}}_{\unicode[STIX]{x1D70C}\text{-}\text{data}}\end{eqnarray}$$

the hypergeometric sheaf

$$\begin{eqnarray}\operatorname{Hyp}(!,\unicode[STIX]{x1D713},\text{character data determined by }\unicode[STIX]{x1D70C})\end{eqnarray}$$

which occurs geometrically as a multiplicative translate of ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ . Each such has a canonical descent to $\mathbb{G}_{m}/\mathbb{F}_{q}$ , the point being that in all cases both the “upstairs” characters and the “downstairs” characters are $\operatorname{Gal}(\overline{\mathbb{F}_{q}}/\mathbb{F}_{q})$ -stable sets of characters. Let us call any such hypergeometric “descendable to $\mathbb{F}_{q}$ ”. In the description [Reference Katz17, 8.2] of a hypergeometric as the multiplicative convolution of a Kloosterman sheaf $\text{Kl}_{\text{up}}$ with the “upstairs” characters with the $[x\mapsto 1/x]$ pullback of a Kloosterman sheaf $\text{Kl}_{\text{down}}$ with the inverses of the “downstairs” characters (and the additive character $\overline{\unicode[STIX]{x1D713}}$ ), both factors have canonical descents, cf. [Reference Katz16, 8.8], and the convolution of these descents is the desired descent

$$\begin{eqnarray}{\mathcal{H}}_{\unicode[STIX]{x1D70C}\text{-}\text{data, desc}}\end{eqnarray}$$

of our ${\mathcal{H}}_{\unicode[STIX]{x1D70C}\text{-}\text{data}}$ .

Lemma 14.1. Given a hypergeometric ${\mathcal{H}}$ on $\mathbb{G}_{m}/\overline{\mathbb{F}_{q}}$ which is descendable to $\mathbb{F}_{q}$ and a character $\unicode[STIX]{x1D712}$ of  $\mathbb{F}_{q}^{\times }$ , the formation of the canonical descent commutes with the operation of tensoring with ${\mathcal{L}}_{\unicode[STIX]{x1D712}}$ :

$$\begin{eqnarray}({\mathcal{H}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D712}})_{\text{desc}}\cong {\mathcal{H}}_{\text{desc}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D712}}\quad \text{on }\mathbb{G}_{m}/\mathbb{F}_{q}.\end{eqnarray}$$

In particular, if ${\mathcal{H}}$ is geometrically isomorphic to ${\mathcal{H}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D712}}$ , then

$$\begin{eqnarray}{\mathcal{H}}_{\text{desc}}\cong {\mathcal{H}}_{\text{desc}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D712}}\quad \text{on }\mathbb{G}_{m}/\mathbb{F}_{q}.\end{eqnarray}$$

Proof. The second assertion is a special case of the first. The first is clear from the explicit description of the canonical descent of a descendable ${\mathcal{H}}$ , which separately breaks the upstairs and downstairs characters into orbits under $\operatorname{Gal}(\overline{\mathbb{F}_{q}}/\mathbb{F}_{q})$ , and then reduces to the case of Kloosterman sheaves formed on single orbit, i.e., those of the form

$$\begin{eqnarray}\text{Kl}(\unicode[STIX]{x1D6EC},\unicode[STIX]{x1D6EC}^{q},\ldots ,\unicode[STIX]{x1D6EC}^{q^{r-1}})\end{eqnarray}$$

for some $r\geqslant 1$ and some character $\unicode[STIX]{x1D6EC}$ of $\mathbb{F}_{q^{r}}$ which has $r$ distinct Galois conjugates under the action of $\operatorname{Gal}(\mathbb{F}_{q^{r}}/\mathbb{F}_{q})$ .◻

At the same time, we have the local system ${\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ on $\mathbb{G}_{m}/\mathbb{F}_{q}$ with which we began. It is geometrically isomorphic to a multiplicative translate of ${\mathcal{H}}_{\unicode[STIX]{x1D70C}\text{-}\text{data},\text{desc}}$ . Recall from [Reference Katz17, 8.5.4] that a hypergeometric sheaf is isomorphic to no non-trivial translate of itself. That the translation is by a point of $\mathbb{G}_{m}(\mathbb{F}_{q})$ results from the following rationality lemma.

Proof. The uniqueness is obvious, from condition (1). Now consider the local system ${\mathcal{C}}$ on $G\times _{k}G$ , coordinates $(t,g)$ , given by

$$\begin{eqnarray}{\mathcal{C}}(t,g)=({\mathcal{A}}(g))^{\vee }\otimes {\mathcal{B}}(tg).\end{eqnarray}$$

In fancier terms, we have the multiplication map $m:G\times _{k}G\rightarrow G,(t,g)\mapsto tg$ , and our ${\mathcal{C}}$ is

$$\begin{eqnarray}{\mathcal{C}}:=pr_{2}^{\star }({\mathcal{A}}^{\vee })\otimes m^{\star }({\mathcal{B}}).\end{eqnarray}$$

For $d$ the relative dimension of $G$ over $k$ , the sheaf $R^{2d}(pr_{1})_{!}({\mathcal{C}})$ on $G$ is supported at $\unicode[STIX]{x1D6FE}$ . Therefore $\unicode[STIX]{x1D6FE}$ is rational over the perfection of $k$ . As $k$ is perfect, $\unicode[STIX]{x1D6FE}$ lies in $G(k)$ .◻

Thus there exists a unique $a_{\unicode[STIX]{x1D70C}}\in k^{\times }$ such that there exists a geometric isomorphism

$$\begin{eqnarray}{\mathcal{W}}_{\unicode[STIX]{x1D70C}}\cong [x\mapsto a_{\unicode[STIX]{x1D70C}}x]_{\star }{\mathcal{H}}_{\unicode[STIX]{x1D70C}\text{-}\text{data},\text{desc}}:={\mathcal{H}}_{a_{\unicode[STIX]{x1D70C}},\unicode[STIX]{x1D70C}\text{-}\text{data},\text{desc}}.\end{eqnarray}$$

As both are geometrically isomorphic, there exists a unique $\unicode[STIX]{x1D6FC}_{\unicode[STIX]{x1D70C}}\in \overline{\mathbb{Q}_{\ell }}^{\times }$ such that on $\mathbb{G}_{m}/\mathbb{F}_{q}$ we have an arithmetic isomorphism

$$\begin{eqnarray}{\mathcal{W}}_{\unicode[STIX]{x1D70C}}\cong {\mathcal{H}}_{a_{\unicode[STIX]{x1D70C}},\unicode[STIX]{x1D70C}\text{-}\text{data},\text{desc}}\otimes \unicode[STIX]{x1D6FC}_{\unicode[STIX]{x1D70C}}^{\text{deg}}.\end{eqnarray}$$

Theorem 14.3. For each irreducible representation $\unicode[STIX]{x1D70C}$ of $\operatorname{PGL}(2,q)$ of dimension ${>}\!1$ , the local system

$$\begin{eqnarray}{\mathcal{H}}_{\unicode[STIX]{x1D70C}}:={\mathcal{H}}_{a_{\unicode[STIX]{x1D70C}},\unicode[STIX]{x1D70C}\text{-}\text{data},\text{desc}}\otimes \unicode[STIX]{x1D6FC}_{\unicode[STIX]{x1D70C}}^{\text{deg}}\end{eqnarray}$$

on $\mathbb{G}_{m}/\mathbb{F}_{q}$ has $G_{\text{geom}}=G_{\text{arith}}=\text{the image of }\,\text{PGL}(2,q)$ in the corresponding representation $\unicode[STIX]{x1D70C}$ .

15 Pulling back from $\operatorname{PGL}(2,q)$ to $\operatorname{PSL}(2,q)$

In this section, we assume that $q$ is odd. Then $\operatorname{PSL}(2,q)$ is a subgroup of index two in $\operatorname{PGL}(2,q)$ . If we think of the Galois theory diagram of the geometric covering ${\mathcal{T}}\rightarrow \mathbb{G}_{m}/\mathbb{F}_{q}$ as a finite étale covering with group $\operatorname{PGL}(2,q)$ , then we have a diagram

$$\begin{eqnarray}{\mathcal{T}}\rightarrow S\rightarrow \mathbb{G}_{m}/\mathbb{F}_{q}\end{eqnarray}$$

in which ${\mathcal{T}}\rightarrow S$ is a $\operatorname{PSL}(2,q)$ torsor, and $S\rightarrow \mathbb{G}_{m}/\mathbb{F}_{q}$ is a finite étale covering of degree two. As Gross explains in [Reference Gross10, bottom of p. 2537], this last covering is the squaring map

$$\begin{eqnarray}[2]:x\mapsto x^{2}\end{eqnarray}$$

of $\mathbb{G}_{m}/\mathbb{F}_{q}$ as a covering of itself. What this means concretely is that for $\unicode[STIX]{x1D70C}$ an irreducible representation of $\operatorname{PGL}(2,q)$ of dimension ${>}\!1$ , its restriction to $\operatorname{PSL}(2,q)$ is given by the local system on $\mathbb{G}_{m}/\mathbb{F}_{q}$ which is the pullback $[2]^{\star }{\mathcal{W}}_{\unicode[STIX]{x1D70C}}$ .

Let us recall the following irreducibility lemma, which we will apply both with $k=\overline{\mathbb{F}_{q}}$ and with $k=\mathbb{F}_{q}$ .

Proof. We use Frobenius reciprocity to compute

$$\begin{eqnarray}\displaystyle \langle [2]^{\star }{\mathcal{H}},[2]^{\star }{\mathcal{H}}\rangle & = & \displaystyle \langle {\mathcal{H}},[2]_{\star }[2]^{\star }{\mathcal{H}}\rangle =\langle {\mathcal{H}},{\mathcal{H}}\otimes [2]_{\star }\overline{\mathbb{Q}_{\ell }}\rangle \nonumber\\ \displaystyle & = & \displaystyle \langle {\mathcal{H}},{\mathcal{H}}\otimes (\overline{\mathbb{Q}_{\ell }}\oplus {\mathcal{L}}_{\unicode[STIX]{x1D712}_{2}})\rangle =\langle {\mathcal{H}},{\mathcal{H}}\rangle +\langle {\mathcal{H}},{\mathcal{H}}\otimes {\mathcal{L}}_{\unicode[STIX]{x1D712}_{2}}\rangle .\nonumber\end{eqnarray}$$

This makes (1) obvious. If we are in situation (2), then $\langle [2]^{\star }{\mathcal{H}},[2]^{\star }{\mathcal{H}}\rangle =2$ , hence $[2]^{\star }{\mathcal{H}}$ is ${\mathcal{A}}\oplus {\mathcal{B}}$ with ${\mathcal{A}}$ not isomorphic to ${\mathcal{B}}$ (otherwise $[2]^{\star }{\mathcal{H}}$ is ${\mathcal{A}}\oplus {\mathcal{A}}$ , in which case $\langle [2]^{\star }{\mathcal{H}},[2]^{\star }{\mathcal{H}}\rangle$ would be $4$ ). Let ${\mathcal{A}}$ have $\operatorname{rank}({\mathcal{A}})\leqslant \operatorname{rank}({\mathcal{B}})$ . Then

$$\begin{eqnarray}\langle {\mathcal{H}},[2]_{\star }{\mathcal{A}}\rangle =\langle [2]^{\star }{\mathcal{H}},{\mathcal{A}}\rangle =1,\end{eqnarray}$$

so ${\mathcal{H}}$ occurs in $[2]_{\star }{\mathcal{A}}$ . But as $\operatorname{rank}({\mathcal{A}})\leqslant \operatorname{rank}({\mathcal{B}})$ , we have $\operatorname{rank}({\mathcal{H}})\geqslant \operatorname{rank}([2]_{\star }{\mathcal{A}})$ . Therefore we have ${\mathcal{H}}\cong [2]_{\star }{\mathcal{A}}$ . In particular, ${\mathcal{H}}$ has even rank and ${\mathcal{A}}$ has rank half that of ${\mathcal{H}}$ . [The same argument then applies to ${\mathcal{B}}$ , and so ${\mathcal{H}}\cong [2]_{\star }{\mathcal{B}}$ .] Then

$$\begin{eqnarray}[2]^{\star }{\mathcal{H}}\cong [2]^{\star }[2]_{\star }{\mathcal{A}}\cong {\mathcal{A}}\oplus [x\mapsto -x]^{\star }{\mathcal{A}}.\Box\end{eqnarray}$$