Proof. In terms of the auxiliary sheaf

$$ \begin{align*}{\mathcal G} :=({\mathcal F})^{\otimes a} \otimes ({\mathcal F}^\vee)^{\otimes b},\end{align*} $$

which is $\iota $ -pure of weight zero, and hence geometrically semisimple (cf. [Reference DeligneDe2, 3.4.1(iii)]), we have

$$ \begin{align*}M_{a,b}=\dim H^{2d}_c(X_{\overline{k}},{\mathcal G}).\end{align*} $$

Our asserted formula for this dimension is

$$ \begin{align*}\limsup_{\mathrm{finite\ extensions\ }L/k}\Big{|}\frac{1}{\#X(L)}\sum_{x \in X(L)}\mathrm{Trace}(\mathrm{Frob}_{x,L}|{\mathcal G})\Big{|}.\end{align*} $$

By the Lefschetz trace formula, this is

$$ \begin{align*}\limsup_{\mathrm{finite\ extensions\ }L/k}\Big{|}\frac{1}{\#X(L)}\sum_{i=0}^{2d}(-1)^i\mathrm{Trace}(\mathrm{Frob}_{L}|H^i_c(X_{\overline{k}},{\mathcal G}))\Big{|}.\end{align*} $$

By Deligne’s fundamental estimate [Reference DeligneDe2, 3.4], $H^i_c$ is $\iota $ -mixed of weight $\le i$ , while $H^{2d}_c$ is $\iota $ -pure of weight $2d$ . But $\#X(L)=(\# L)^d + O((\# L)^{d-1/2})$ , and hence, the $H^i_c$ summands with $i < 2d$ contribute $0$ to the $\limsup $ . So we must prove that $ \dim H^{2d}_c(X_{\overline {k}},{\mathcal G})$ is

$$ \begin{align*}\limsup_{\mathrm{finite\ extensions\ }L/k}\Big{|}\frac{1}{\#X(L)}\mathrm{Trace}(\mathrm{Frob}_L|H^{2d}_c(X_{\overline{k}},{\mathcal G}))\Big{|}.\end{align*} $$

If this $H^{2d}_c$ vanishes, we are done.

If $H^{2d}_c$ is nonzero, the eigenvalues of $\mathrm {Frob}_k$ on this $H^{2d}_c$ are each of the form $(\#k)^d\alpha _i$ , for $i=1, \ldots , \dim H^{2d}_c$ , and each of these $\alpha _i$ has complex absolute value $|\alpha _i|=1$ . Thus, for $L/k$ a finite extension, we have

$$ \begin{align*}\frac{1}{\#X(L)}\mathrm{Trace}(\mathrm{Frob}_L|H^{2d}_c(X_{\overline{k}},{\mathcal G}))= \frac{(\# L)^d}{\#X(L)}\sum_{i=1}^{\dim H^{2d}_c}(\alpha_i)^{\deg(L/k)}.\end{align*} $$

For any $L/k$ , this last expression visibly has absolute value at most

$$ \begin{align*}(1/\#X(L))(\# L)^d{\dim H^{2d}_c}.\end{align*} $$

As $L/k$ grows, the tuple $(\alpha _1^{\deg (L/k)},\ldots , \alpha _{\dim H^{2d}_c}^{\deg (L/k)})$ will, infinitely often, come arbitrarily close to $(1,\ldots ,1)$ , while the ratio $\#X(L)/(\# L)^d$ has limit $1$ as L grows.

Proof. Each geometrically irreducible component of $X_{\overline {k}}$ is defined over some finite extension of k, so at the expense of replacing k by a finite extension of itself, we reduce to the case where each geometrically irreducible component Z is defined over k (i.e., is a geometrically irreducible k-scheme of dimension $e_Z \le d$ ). The result then follows from the Lang-Weil estimate, that for each such component Z, $\#Z(L) =(\# L)^{e_Z} + O( (\# L)^{e_Z -1/2}).$

Proof. Recall that $\Sigma _d$ denotes the Fermat form $x^d+y^d-z^d-w^d$ for any $d \in {\mathbb Z}_{\geq 1}$ . By Theorem 2.1, $M_{2,2}({\mathcal F}_\chi )$ is the limsup over L of the sums

$$ \begin{align*}& \frac{1}{(\# L)^2}\sum_{(x,y,z,w) \in {\mathbb A}^4(L),~\Sigma_a=\Sigma_b=0}\psi_L(f(x)+f(y)-f(z)-f(w))\chi_L(xy)\bar\chi_L(zw)\\= & \frac{1}{(\# L)^2}\sum_{(x,y,z,w) \in {\mathbb A}^4(L),~\Sigma_a=\Sigma_b=0}\psi_L(\sum_{i\in{\mathcal E}}A_i\Sigma_i(x,y,z,w))\chi_L(xy)\bar\chi_L(zw).\end{align*} $$

The key observation is that the affine variety

$$ \begin{align*}\Sigma_{a,b} := \{\Sigma_a=\Sigma_b=0\}\end{align*} $$

in ${\mathbb A}^4$ is homogeneous, the affine cone over the projective variety $\Sigma _{a,b,\mathrm {proj}} \subset {\mathbb P}^3$ defined by these same equations. We may omit the origin $(0,0,0,0) \in {\mathbb A}^4$ without changing the large L limit. Then we choose, for each point in $\Sigma _{a,b,\mathrm {proj}}(L)$ , a representative $(x_0,y_0,z_0,w_0) \in \Sigma _{a,b}(L)$ . Then every point $(x,y,z,w)$ in $\Sigma _{a,b}(L) \smallsetminus \{ 0\}$ is uniquely of the form $(rx_0,ry_0,rz_0,rw_0)$ with $r \in L^\times $ and $(x_0,y_0,z_0,w_0) \in \Sigma _{a,b}(L)$ a chosen representative. Moreover,

$$ \begin{align*}\chi_L(xy)\bar\chi_L(zw)=\chi_L(x_0y_0)\bar\chi_L(z_0w_0).\end{align*} $$

Thus, we are looking at the limsup over L of the sums

$$ \begin{align*}\frac{1}{(\# L)^2}\sum_{\small {\begin{array}{c}(x_0,y_0,z_0,w_0)\\ \mathrm{\ chosen\ rep.\ over\ }L\end{array}}}\chi_L(x_0y_0)\bar\chi_L(z_0w_0)~\sum_{r \in L^\times}\psi_L\big(\sum_{i\in{\mathcal E}}A_i\Sigma_i(x_0,y_0,z_0,w_0)r^i\big).\end{align*} $$

The innermost sum is $O(\# L)^{1/2}$ so long as the polynomial

$$ \begin{align*}\sum_{i\in{\mathcal E}}A_i\Sigma_i(x_0,y_0,z_0,w_0)r^i\end{align*} $$

in r is not Artin-Schreier trivial. The number of L-valued points on $\Sigma _{a,b,\mathrm {proj}}$ is $O(\# L)$ , so the Artin-Schreier nontrivial cases contribute $O((\# L)^{3/2})/(\# L)^2$ to the sum, and hence contribute $0$ to the large L limit.

Because $f(x)$ is Artin-Schreier reduced, the only way the polynomial

$$ \begin{align*}\sum_{i\in{\mathcal E}}A_i\Sigma_i(x_0,y_0,z_0,w_0)r^i\end{align*} $$

in r can be Artin-Schreier trivial is for every $\Sigma _i(x_0,y_0,z_0,w_0)$ with $i\in {\mathcal E}$ to vanish, in which case the inner sum is $\# L-1$ . Thus, our large L limiting sum is

$$ \begin{align*}\frac{1}{(\# L)^2}\sum_{\small {\begin{array}{c}(x,y,z,w) \in {\mathbb P}^3(L),\\ \Sigma_a=\Sigma_b=0,~\Sigma_i=0,~\forall i \in {\mathcal E}\end{array}}}(\# L-1)\chi_L(xy)\bar\chi_L(zw).\end{align*} $$

We break the domain of summation into finitely many closed points and the one-dimensional geometrically irreducible components Z of the projective variety $\Sigma _{\mathrm {proj}}(S)$ defined by

$$ \begin{align*}\Sigma_a=\Sigma_b=0,~\Sigma_i=0,~\forall i \in {\mathcal E},\end{align*} $$

each of which is defined over some finite extension of k. At the expense of enlarging k, we may assume each Z is defined over k. Then

$$ \begin{align*}\#Z(L)=\# L+O(\sqrt{\# L}).\end{align*} $$

So our $\limsup $ is the $\limsup $ of the sum

$$ \begin{align*}\sum_{1\tiny{\mbox{-dim irred. compt's }}Z}\frac{1}{(\# L)^2}\sum_{(x,y,z,w) \in Z(L)}(\# L-1)\chi_L(xy)\bar\chi_L(zw).\end{align*} $$

When $\chi (xy)\bar \chi (zw)$ is geometrically trivial on (the dense open set where $xyzw\neq 0$ of) Z, this sum over Z contributes $1$ to the $\limsup $ , while if $\chi (xy)\bar \chi (zw)$ is geometrically nontrivial on (the dense open set where $xyzw\neq 0$ of) Z, it contributes $0$ to the limsup. Thus, we have

So is the number of geometrically irreducible components of dimension one in $\Sigma _{\mathrm {proj}}(S)$ , while $M_{2,2}({\mathcal F}_\chi )$ is the number of those components on which $\chi (xy)\bar \chi (zw)$ is geometrically trivial.

Now assume that all integers in S are odd. Then $\Sigma _{\mathrm {proj}}(S)$ contains the line $x+y=0=z+w$ . For any character $\chi $ of $k^\times $ , the sum of $\chi _L(xy)\bar \chi _L(zw)$ over this line is $\# L-1$ if and $0$ otherwise. Thus, if , this line is an irreducible component on which $\chi (xy)\bar \chi (zw)$ is geometrically nontrivial, and hence the asserted inequality

if .

Proof. The proof is essentially identical to that of the previous Theorem 2.3, which is the case $n=2$ . Let us denote

$$ \begin{align*}B:=\{b_1,\ldots , b_n\}.\end{align*} $$

The role of $\Sigma _{a,b}$ there is played by $\Sigma _B:=\cap _i\Sigma _{b_i}$ here. The affine variety $\Sigma _B$ is homogeneous, the affine cone over the projective variety $\Sigma _{B,\mathrm {proj}}$ defined by the same equations. Because $n \ge 2$ , the projective variety $\Sigma _{B,\mathrm {proj}}$ has dimension at most one (i.e., all its geometrically irreducible components have dimension $\le 1$ ), so over any finite extension, $L/k$ has $O(\# L) L$ -valued points. From here on, the proof is identical.

Proof. That ${\mathcal F}(S,\chi )$ has its $M_{2,2}$ given by the same recipe, purely in terms of the data $(S,\chi )$ , as did ${\mathcal F}(f,a,b,\chi )$ , is the special case $f(x)=x^A$ , $n=r$ , and $b_i=B_{r+1-i}$ , of Theorem 2.4.

Proof. By Theorem 2.1 and the argument of Theorem 2.3, $M_{2,2}$ for ${\mathcal F}_\chi $ is the $\limsup $ over L of $\frac {1}{(\# L)(\# L-1)}$ times

$$ \begin{align*}\sum_{s\in L, t\in L^\times}\frac{1}{(\# L)^2}\sum_{x,y,z,w \in L} \psi_L(s\Sigma_a +t\Sigma_b +f(x)+f(y)-f(z) -f(w)) \chi_L(xy)\bar\chi_L(zw).\end{align*} $$

If the summation were over all $(s,t)\in L^2$ , this would be

$$ \begin{align*}\frac{1}{(\# L)(\# L-1)}\sum_{x,y,z,w \in L,~\Sigma_a=\Sigma_b=0}\psi_L(f(x)+f(y)-f(z)-f(w))\chi_L(xy)\bar\chi_L(zw),\end{align*} $$

and just as in the proof of Theorem 2.3, we would get

However, the summation is only over $(s,t)\in L\times L^\times $ . So we must subtract, for each $L/k$ , the expression

$$ \begin{align*}& \frac{1}{(\# L)(\# L-1)}\sum_{s\in L}\frac{1}{(\# L)^2}\sum_{x,y,z,w \in L}\psi_L(s\Sigma_a +f(x)+f(y)-f(z) -f(w))\chi_L(xy)\bar\chi_L(zw)\\= & \frac{1}{(\# L)^2(\# L-1)}\sum_{x,y,z,w \in L,~\Sigma_a=0}\psi_L(f(x)+f(y)-f(z) -f(w))\chi_L(xy)\bar\chi_L(zw).\end{align*} $$

So long as $f(x)$ contains monomials of degree $e_i \neq a$ , the ray calculation used in the proof of Theorem 2.3 shows that this limit (not just $\limsup $ ) vanishes. The assertion about is proven exactly as in Theorem 2.3.

Suppose now that $f(x)$ is a constant multiple of $x^a$ and . Then the term we are subtracting is equal to

$$ \begin{align*}\frac{1}{(\# L)^2(\# L-1)}\sum_{x,y,z,w \in L,~\Sigma_a=0}\psi_L(0)=\#\Sigma_a(L)/((\# L)^2(\# L-1)),\end{align*} $$

which tends to $1$ as L grows, simply because $\Sigma _{a}$ is the affine cone over the smooth surface $\Sigma _{a,\mathrm {proj}}$ .

Suppose finally that $f(x)$ is a constant multiple of $x^a$ and . Then the sum we are subtracting is

$$ \begin{align*}\frac{1}{(\# L)^2(\# L-1)}\sum_{x,y,z,w \in L,~xyzw \neq 0,~\Sigma_a=0}\chi_L(xy/zw).\end{align*} $$

This sum will be $O(1/\sqrt {\# L})$ , and thus have large L limit zero, if the Kummer sheaf ${\mathcal L}_{\chi (xy/zw)}$ is geometrically nontrivial on the dense open set U of $\Sigma _{a,\mathrm {proj}}$ where $xyzw$ is invertible. Thus, U is the open set in the affine surface $x^a+y^a=z^a+1$ where $xyz$ is invertible, and our sheaf is ${\mathcal L}_{\chi (xy/z)}$ on U. We will show that this sheaf has a geometrically nontrivial pullback.

Choose an element $\alpha \in {\mathbb F}_{p^2} \smallsetminus {\mathbb F}_p$ , and $\beta $ with $\beta ^a=\alpha $ . It suffices to show the pullback of ${\mathcal L}_{\chi (xy/z)}$ to the closed subscheme $y=\beta $ of U is geometrically nontrivial. This pullback is ${\mathcal L}_{\chi (\beta x/z)}$ , on the open set of the curve

$$ \begin{align*}{\mathcal C}: x^a+\alpha=z^a +1\end{align*} $$

where $xz$ is invertible. But the function $\beta x/z$ on ${\mathcal C}$ has a simple zero at each point $(0,\gamma )$ with $\gamma $ one of the a distinct roots of the polynomial $T^a=\alpha -1$ . Hence, ${\mathcal L}_{\chi (\beta x/z)}$ is geometrically nontrivial on ${\mathcal C}$ .

So in this case when $f(x)$ is a constant multiple of $x^a$ and , we have

$$ \begin{align*}M_{2,2}({\mathcal F}_\chi) \leq \limsup_{\# L \to \infty}\frac{\#\Sigma(S)(L)}{(\# L)^2}.\end{align*} $$

[Of course, in this case, the set $S =\{a,b\}$ .] The argument in the proof of Theorem 2.3 shows that if $a,b$ are both odd, but , then

$$ \begin{align*}M_{2,2}({\mathcal F}_\chi) < \limsup_{\# L \to \infty}\frac{\#\Sigma(S)(L)}{(\# L)^2}.\\[-45pt] \end{align*} $$

Proof. Consider the set ${\mathcal E}_f$ of exponents which occur in f:

$$ \begin{align*}{\mathcal E}_f:=\{i \in {\mathbb Z},~A_i \neq 0\}\end{align*} $$

and $S_f:=\{a,b\} \cup {\mathcal E}_f$ .

Consider also the set ${\mathcal E}_{f,+}$ of exponents with occur in $f(x)+g(x^2)$ :

$$ \begin{align*}{\mathcal E}_{f,+}:=\{i \in {\mathbb Z},~A_i \neq 0\}\cup \{2j,~B_j \neq 0\}\end{align*} $$

and $S_{f,+}:=\{a,b\} \cup {\mathcal E}_{f,+}$ . Then from Theorems 2.3 and 2.6, we know that

$$ \begin{align*}\kern-10pt M_{2,2}({\mathcal F})=\limsup_{\# L \to \infty}\frac{\#\Sigma(S_f)(L)}{(\# L)^2}.\end{align*} $$

$$ \begin{align*}M_{2,2}({\mathcal G})=\limsup_{\# L \to \infty}\frac{\#\Sigma(S_{f,+})(L)}{(\# L)^2}.\end{align*} $$

As $S_f \subset S_{f,+}$ , we trivially have $M_{2,2}({\mathcal G}) \le M_{2,2}({\mathcal F})$ . Because $S_f$ consists entirely of odd integers, among the two-dimensional geometrically irreducible components of $\Sigma (S_f)$ is the locus $x+y=0=z+w$ .

It suffices to show that this locus $x+y=0=z+w$ does not lie in $\Sigma (S_{f,+})$ . Indeed, $S_{f,+}$ contains some nonzero even integer $2j $ , and hence, $\Sigma (S_{f,+})$ lies inside the hypersurface of equation $x^{2j}+y^{2j}=z^{2j}+w^{2j}$ . So it suffices to show that the locus $x+y=0=z+w$ does not lie in this hypersurface. The intersection of this hypersurface with the locus $x+y=0=z+w$ is the locus in $(x,z)$ space defined by $x^{2j} +(-x)^{2j}=z^{2j} +(-z)^{2j}$ . As we are in odd characteristic, this intersection is the locus $x^{2j}=z^{2j}$ , which is the union of $2j$ lines.

Proof. (i) Assume that $\varphi (\mathsf {Res}(f,g)) \neq 0$ . Let $f(x)$ be of degree d and with leading term $ax^d$ , and let $g(x)$ be of degree e and with leading term $bx^e$ . Suppose that $\varphi (a)=\varphi (b)=0$ , so that $\varphi (f) \in S[x]$ has degree $<d$ and $\varphi (g) \in S[x]$ has degree $<e$ . In this case, $\varphi (\mathsf {Res}(f,g))=0$ , a contradiction. So we may assume that $\varphi (a) \neq 0$ , so that $\varphi (f) \in S[x]$ has degree d. Now, if $\varphi (g)$ has degree $e' \leq e$ , then

(3.1.1) $$ \begin{align} \varphi(\mathsf{Res}(f,g)) = \pm \varphi(a)^{e-e'}\mathsf{Res}(\varphi(f),\varphi(g)), \end{align} $$

and hence, $\mathsf {Res}(\varphi (f),\varphi (g)) \neq 0$ .

(ii) follows from (3.1.1) (with $e'=e$ ).

(iii) follows from (i), (3.1.1), and the assumption that S is an integral domain.

Proof. The statement is obvious when $p = 2$ , so we will assume $p> 2$ . Replacing q by $q^e$ , we may assume that $\gcd (m_1, \ldots ,m_n)=e=1$ . Suppose $2|m_i$ , $2 \nmid m_j$ , and $\zeta \in \mu _{total}({\mathcal Q})$ . Since $\zeta ^{q^{m_j}-1}=-1$ and $m_j$ is odd, we see that the $2$ -part $2^f$ of the order of $\zeta $ is $2(q^{m_j}-1)_2 = 2(q-1)_2$ , twice the $2$ -part of $q-1$ . As $p> 2$ , $2^f$ divides $(q^2-1)_2$ , which in turn divides $q^{m_i}-1$ because $2|m_i$ , and this contradicts the equality $\zeta ^{q^{m_i}-1} = -1$ .

Assume now that $2 \nmid m_i$ for all i, so that $2 \nmid (q^{m_i}-1)/(q-1)$ , and choose a primitive ${(2q-2)}^{\mathrm {th}}$ root of unity $\theta \in \overline {{\mathbb F}_p}$ . Then $-1=\theta ^{q-1}=\theta ^{q^{m_i}-1}$ , and hence, $\zeta \in \mu _{total}({\mathcal Q})$ if and only if $(\zeta \theta )^{q^{m_i}-1}=1$ for all i. There are exactly

$$ \begin{align*}\gcd\big(q^{m_1}-1, \ldots,q^{m_n}-1\big)=q^{\gcd(m_1, \ldots,m_n)}-1 = q-1\end{align*} $$

possibilities for such $\zeta \theta $ .

Proof. For (i), consider any point $P=(x,y,z,w) \in \Sigma (L)$ . Then $x+y=z+w$ . Certainly, $\Sigma $ contains the two planes

$$ \begin{align*}(x=z,~y=w) \mbox{ and } (x=w,~y=z)\end{align*} $$

which contribute $2(\# L)^2-\# L$ points to $\Sigma (L)$ . So we have to count the points $P \in \Sigma (L)$ for which $z \neq x,y$ . For these points, we can use the parametrization

(3.3.1) $$ \begin{align} x = (A+1)z-Ay = (A+1)u +y,~z=y+u,~w=Az-(A-1)y = Au+y, \end{align} $$

for P in terms of $A,u,y$ , where $u:=z-y \neq 0$ and $A := (x-z)/(z-y) \neq 0$ . The condition $\Sigma _b(P)=0$ now reads

(3.3.2) $$ \begin{align} \big( (A+1)u +y \big)^b +y^b-(y+u)^b-(Au+y)^b=0. \end{align} $$

First, we look at such points P with $y=0$ . Since $u \neq 0$ , (3.3.2) implies $(A+1)^b-A^b-1=0$ . The leading term of this polynomial equation in A is $bA^{b-1}$ . Since $p \nmid b$ , there are at most $b-1$ such A’s, which contributes at most $(b-1)\# L$ points to $\Sigma (L)$ . This dies in the large L-limit.

So we may now assume $y \neq 0$ , and replace $(A,y,u)$ by $(A,y,v)$ , where $v:= u/y \neq 0$ . Since $y \neq 0$ , now (3.3.2) becomes

(3.3.3) $$ \begin{align} \big( (A+1)v+1\big)^b-(Av+1)^b-(v+1)^b+1 = 0. \end{align} $$

Note that the coefficient for $v^j$ in the left-hand side of (3.3.3) is

$$ \begin{align*}\binom{b}{j}\big( (A+1)^j-A^j-1\big)\end{align*} $$

when $2 \leq j \leq b$ , and $0$ if $j = 0,1$ . So the condition $P \in \Sigma (L)$ now reads

(3.3.4) $$ \begin{align} F_b(A,v)=F_c(A,v)=0. \end{align} $$

Furthermore, if $A \in {\mathcal M}_p(b)$ , then (3.3.3) is vacuously true. Hence, if

$$ \begin{align*}A \in {\mathcal M}_p(b) \cap {\mathcal M}_p(c),\end{align*} $$

then (3.3.4) is vacuously true, and each A contributes $(\# L-1)^2$ points to $\Sigma (L)$ with $y,v \neq 0$ , which do not belong to the two planes $(x=z,~y=w)$ and $(x=w,~y=z)$ . This yields the lower bound in (i).

Now we look at $A \notin {\mathcal M}_p(b) \cap {\mathcal M}_p(c)$ , and assume that $R(A) \not \equiv 0$ as a function of A. Applying Lemma 3.1 to the specialization homomorphism $A \mapsto \gamma $ at any point $\gamma $ where $R(\gamma ) \neq 0$ , we see that (3.3.4) has no solution v when $A = \gamma $ . Thus, (3.3.4) can have solutions in v only at $A=\gamma $ with $R(\gamma )=0$ . This implies that the number of A for which (3.3.4) has a common solution in v is bounded independently of L (in fact by $2bc$ , an upper bound for the degree of $R(A)$ ). If $A \notin {\mathcal M}_p(b)$ for instance, then $F_b(A,v)$ is a nonzero polynomial in v, and hence has at most b zeros once A is fixed. Thus, each such A contributes at most $\max (b,c)(\# L-1)$ points to $\Sigma (L)$ (with y running), and again this dies in the large L-limit. This proves the equality in (ii).

Suppose $b=2 < p$ . Then $F_2(A,v)=2$ , and hence, (3.3.4) has no solutions. Furthermore, ${{\mathcal M}_p(2)=\varnothing }$ , proving (iii).

Suppose $b=3 < p$ . Then $F_3(A,v)=3((A+1)v+2)$ . Hence, (3.3.4) is equivalent to $(A+1)v=-2$ and $(-1)^c-(-v-1)^c-(v+1)^c+1=0$ . If $2|c$ , this shows that $(v+1)^c=1$ . Thus, there are at most c pairs $(A,v)$ that satisfy (3.3.4), contributing at most $c(\# L-1)$ points to $\Sigma (L)$ , and this dies in the large L-limit. Suppose $2 \nmid c$ . This argument then shows that there are exactly $\# L-2$ pairs $(A,v)$ that satisfy (3.3.4) and $A,v \neq 0$ (namely, one for each $v \neq 0,-2$ ). This gives $(\# L-1)(\# L-2)$ more points to $\Sigma (L)$ , proving (iv).

Next, suppose that $b= q+1$ with $q := p^f \geq p$ . Then (3.3.3) becomes

(3.3.5) $$ \begin{align} (A^q+A)v^{q+1}=0, \end{align} $$

which shows that

(3.3.6) $$ \begin{align} A \in {\mathcal M}_p(q+1) \mbox{ if and only if }A^{q-1}=-1, \end{align} $$

that is, $A \in \mu _{total}(\{q\})$ . Now, if $A \notin {\mathcal M}_p(b)$ , then (3.3.5) has no solution since $v \neq 0$ , and hence, (3.3.3), respectively (3.3.4), has no solution. If $A \in {\mathcal M}_p(b) \smallsetminus {\mathcal M}_p(c)$ , then we have at most $b-2=q-1$ possibilities for A, for each of which $F_c(A,v)=0$ yields at most c possibilities for v. This contributes at most $(b-2)c(\# L-1)$ points to $\Sigma (L)$ , and this dies in the large L-limit. Hence, we have to count only the A’s in ${\mathcal M}_p(b) \cap {\mathcal M}_p(c)$ , and hence, (v) holds.

For (vi), note that the coefficient for $v^{j-2}$ in $F_b(A,v)$ is

$$ \begin{align*}\frac{1}{A}\binom{b}{j}\big( (A+1)^j-A^j-1\big) \equiv j\binom{b}{j} = b\binom{b-1}{j-1} \ \ \ \ (\mathrm{mod} \ A)\end{align*} $$

when $2 \leq j \leq b$ . Hence,

$$ \begin{align*}F_b(0,v) = b\sum^b_{j=2}\binom{b-1}{j-1}v^{j-2} = b\frac{(v+1)^{b-1}-1}{v}.\end{align*} $$

Thus, the only roots of $F_b(0,v)$ are the elements of $\mu _{(b-1)_{p'}} \smallsetminus \{1\}$ (subtracted by $1$ ). Similarly, the set of roots of $F_c(0,v)$ is $\mu _{(c-1)_{p'}} \smallsetminus \{1\}$ (translated by $-1$ ). So the assumption $\gcd \big ( (b-1)_{p'},(c-1)_{p'}\big ) = 1$ implies that $F_b(0,v)$ and $F_c(0,v)$ have no common root. Furthermore, the specialization $A \mapsto 0$ preserves the degree $b-2$ of $F_b(A,v)$ (as $p \nmid b$ ). It follows from Lemma 3.1 that $R(0) \neq 0$ , and so $R(A) \not \equiv 0$ .

Note that (vi) implies (iii) and (v), since $(b-1)_{p'} = 1$ when $b=p^f+1$ with $f \geq 0$ .

Assume now that $\gcd (b-1,c-1)=1$ . If $p \nmid b(b-1)$ , then the $j=2$ condition in the definition of ${\mathcal M}_p(b)$ is the vanishing of

$$ \begin{align*}\binom{b}{2}\big( (A+1)^2-A^2-1 \big) = b(b-1)A,\end{align*} $$

and hence, ${\mathcal M}_p(b) = \varnothing $ , implying $\lim _{\# L \to \infty }\#\Sigma (L)/(\# L)^2 = 2$ by (vi). If $p \nmid c(c-1)$ , the same $j=2$ condition shows that ${\mathcal M}_p(c) = \varnothing $ , with the same conclusion that $\lim _{\# L \to \infty }\#\Sigma (L)/(\# L)^2 = 2$ by (vi).

For (viii), note that $(b-1)_p=p^{fm}$ implies that $p \nmid \binom {b}{j}$ for $j = p^{fm}+1$ . Now

$$ \begin{align*}\binom{b}{j}\big( (A+1)^j-A^j-1 \big) = \binom{b}{j}\big( A^{q^m}+A \big),\end{align*} $$

where $q:= p^f$ . Thus, ${\mathcal M}_p(b)$ is contained in $\{A \mid A^{q^m-1}=-1\}$ . Similarly, ${\mathcal M}_p(c)$ is contained in $\{A \mid A^{q^n-1}=-1\}$ . By Lemma 3.2, the set $\{A \mid A^{q^m-1}=A^{q^n-1}=-1\}$ is empty, and so we are done by (vi).

Proof. The hypothesis implies that $n-m = \sum _{i \geq 0}(n_i-m_i)p^i$ is the base p expansion of $n-m$ . Now for any $j \geq 0$ , we have

$$ \begin{align*}\lfloor \frac{m}{p^j} \rfloor + \lfloor \frac{n-m}{p^j} \rfloor = \sum_{i \geq j}m_ip^{i-j} + \sum_{i \geq j}(n_i-m_i)p^{i-j} = \sum_{i \geq j}n_ip^{i-j} = \lfloor \frac{n}{p^j} \rfloor.\end{align*} $$

Since $\sum _{j \geq 0}\lfloor \frac {n}{p^j} \rfloor $ is the exponent of the highest power of p that divides $n!$ , and similarly for $m!$ and $(n-m)!$ , the above equalities imply the claim.

Proof. We will follow the proof of Proposition 3.3 and count the points

$$ \begin{align*}P=(x,y,z,w) \in \Sigma(L)\end{align*} $$

that lie outside of the two planes $(x=z,y=w)$ and $(x=w,y=z)$ , for which we can use the parametrization (3.3.1). For these points, the condition $P \in \Sigma $ reads

$$ \begin{align*}F_a(A,v)=F_{b_1}(A,v)= \ldots = F_{b_r}(A,v)=0;\end{align*} $$

cf. (3.3.4). Since $\Sigma _a(P)=0$ and $a=p^n+1$ , we have $A^{p^n-1}=-1$ ; see (3.3.5). Now the proof of Proposition 3.3(v) can be repeated verbatim to show that

$$ \begin{align*}M=2+ \#\big({\mathcal M}_p(a) \cap \bigcap^r_{i=1}{\mathcal M}_p({b_i})\big),\end{align*} $$

where ${\mathcal M}_p(a)$ and ${\mathcal M}_p({b_i})$ are defined in (3.2.1).

We will assume that $M> 2$ if $p> 2$ , $M> 3$ if $p=2$ , and aim to show that we are in (c) with $M=p^e+1 \geq 4$ or in (d) with $M=2^e+1 \geq 5$ . Note that when $p=2$ , $1 \in {\mathcal M}_p(b)$ for any integer $b \geq 3$ . Hence, our assumption implies that

(3.5.1) $$ \begin{align} \mbox{For all }i,~{\mathcal M}_p({b_i} )\neq \varnothing \mbox{ if }p>2, \mbox{ and }{\mathcal M}_p({b_i}) \supset \{1\} \mbox{ if }p=2. \end{align} $$

Consider the base p expansion

$$ \begin{align*}c=\sum_{i\geq 0}c_ip^i\end{align*} $$

of $c:=b_1$ . We already noted that ${\mathcal M}_p(c) = \varnothing $ if $2 < p \nmid c(c-1)$ , contrary to (3.5.1). However, if $p=2$ , then $2 \nmid c$ and so $p|(c-1)$ . Henceforth, we may assume that $p|(c-1)$ , whence $c_0=1$ .

Consider any digit $c_i \geq 1$ of c, with $i \geq 1$ . By Lemma 3.4, $p \nmid \binom {c}{p^i+1}$ . Taking $j:=p^i+1$ in the definition (3.2.1) of ${\mathcal M}_p(c)$ , we get

$$ \begin{align*}0=(A+1)^{p^i+1}-A^{p^i+1}-1 = (A^{p^i}+1)(A+1)-A^{p^i+1}-1 = A^{p^i}+A\end{align*} $$

for $A \in {\mathcal M}_p(c)$ . As $A \neq 0$ , we get

(3.5.2) $$ \begin{align} A^{p^i-1}=-1, \end{align} $$

in particular,

(3.5.3) $$ \begin{align} A^{p^i}=-A,~A^{2p^i+1}=A^3. \end{align} $$

Assume in addition that $c_i \geq 2$ (and so $p>2$ as $c_i \leq p-1$ .) Then by Lemma 3.4, we have $p \nmid \binom {c}{2p^i+1}$ . Taking $j:=2p^i+1$ in the definition (3.2.1) of ${\mathcal M}_p(c)$ , we get

$$ \begin{align*}0 & =(A+1)^{2p^i+1}-A^{2p^i+1}-1\\ & = (A^{p^i}+1)^2(A+1)-A^{2p^i+1}-1\\ & = (-A+1)^2(A+1)-A^3-1\\ & =-A(A+1),\end{align*} $$

and so $A=-1$ . But this is impossible by (3.5.2) (since $p^i \geq 3$ is odd in the case under consideration), and so ${\mathcal M}_p(c)=\varnothing $ , again contradicting (3.5.1).

We have shown that any positive digit $c_i$ of c must be equal to $1$ . Suppose now that $c_i=1=c_j$ for some $i> j \geq 1$ . Then (3.5.2) holds for both $A^{p^i}$ and $A^{p^j}$ , and so

$$ \begin{align*}A^{p^i}=A^{p^j}=-A,~A^{p^i+p^j+1}=A^3.\end{align*} $$

Furthermore, by Lemma 3.4, we have $p \nmid \binom {c}{p^i+p^j+1}$ . Taking $j:=p^i+p^j+1$ in the definition (3.2.1) of ${\mathcal M}_p(c)$ , we now get

$$ \begin{align*}0 & =(A+1)^{p^i+p^j+1}-A^{p^i+p^j+1}-1\\ & = (A^{p^i}+1)(A^{p^j}+1)(A+1)-A^{p^i+p^j+1}-1\\ & = (-A+1)^2(A+1)-A^3-1\\ & =-A(A+1),\end{align*} $$

and so $A=-1$ . If $p> 2$ , then this is again impossible by (3.5.2), and so ${\mathcal M}_p(c)=\varnothing $ , contrary to (3.5.1). If $p=2$ , then ${\mathcal M}_p(c) \subseteq \{1\}$ , contradicting (3.5.1).

We have shown that $b_1=c$ has only two positive digits, $c_0$ and $c_{m_1}$ , and both are equal to $1$ . Thus, $b_1=p^{m_1}+1$ . Applying the same argument to any $b_i$ , we see that $b_i=p^{m_i}+1$ . Hence,

$$ \begin{align*}{\mathcal M}_p({b_i} )= \{ A \in \overline{{\mathbb F}_p} \mid A^{p^{m_i}-1}=-1\}.\end{align*} $$

Let $e:=\gcd (n,m_1, \ldots ,m_r)$ . If $p> 2$ , then it follows from Lemma 3.2 that

$$ \begin{align*}\#\big({\mathcal M}_p(a) \cap \bigcap^r_{i=1}{\mathcal M}_p({b_i})\big)\end{align*} $$

equals $p^e+1$ if all $n/e$ and $m_i/e$ are odd, and $0$ otherwise, and thus, we arrive at (c). Similarly, if $p=2$ , then using Lemma 3.2, we arrive at (d).

Proof. Arguing as in the proof of Proposition 3.5, we have

$$ \begin{align*}\limsup_{\# L \to \infty}\#\Sigma(L)/(\# L)^2 = 2+ \# \cap^r_{i=1} {\mathcal M}_p(q_i+1).\end{align*} $$

According to (3.3.6), $\cap ^r_{i=1} {\mathcal M}_p(q_i+1)$ is precisely $\mu _{total}({\mathcal Q})$ for ${\mathcal Q}:= \{q_1, \ldots ,q_r\}$ . The statement now follows from Lemma 3.2.

Proof. The idea is to make use of the limsup formulas of Theorems 2.3 and 2.6 to compute $M_{2,2}$ .

Consider first the case when $B=2A$ . Then the two equations

$$ \begin{align*}\Sigma_A=0,~\Sigma_{2A}=0,\end{align*} $$

which we view as the equations $\Sigma _1=\Sigma _2=0$ applied to the variables $x^A,y^A,z^A, w^A$ , show that we have an equality of sets

$$ \begin{align*}\{x^A,y^A\}=\{z^A,w^A\}.\end{align*} $$

If any of $x,y,z,w$ vanishes, this equality of sets has $O(\# L)$ solutions, so we may assume that each of $x,y,z,w$ is nonzero. Then we are in one of $2A^2$ cases, as follows. For each ordered pair $\zeta ,\eta $ of $A^{\mathrm {th}}$ roots of unity in $\mu _A(\overline {k})$ , either

$$ \begin{align*}[z,w] =[\zeta x,\eta y]\ \ \mathrm{or\ }\ [w,z] =[\zeta x,\eta y].\end{align*} $$

In this first case of $[z,w] =[\zeta x,\eta y]$ , we use the various $\Sigma _{C_i}$ equations, that $x^{C_i}+y^{C_i}=z^{C_i}+w^{C_i}$ , to get

$$ \begin{align*}x^{C_i}+y^{C_i}=\zeta^{C_i}x^{C_i} +\eta^{C_i}y^{C_i}\ \ \mathrm{(i.e.,\ we\ have\ }(\zeta^{C_i}-1)x^{C_i}_i+(\eta^{C_i}-1)y^{C_i} =0).\end{align*} $$

This equation has $O(\# L)$ solutions unless we have $\zeta ^{C_i}=\eta ^{C_i}=1$ . But $\gcd (\mathrm {the\ }C_i,A,2A)=1$ , and hence, $\gcd (\mathrm {the\ }C_i,A)=1$ . So in order to have more than $O(\# L)$ solutions, we must have

$$ \begin{align*}\zeta^{C_i}=\eta^{C_i}=1 \ \mathrm{for\ each\ } C_i.\end{align*} $$

As both $\zeta ,\eta $ are $A^{\mathrm {th}}$ roots of unity, and $\gcd (\mathrm {the\ }C_i,A)=1$ , these equalities force $\zeta =1=\eta $ , Thus, in this first case, we have the solution $[z,w] =[x,y]$ , with its $(\# L)^2$ points, and $A^2-1$ other solutions, each with $\# L$ points. The treatment of the second case, $[w,z] =[\zeta x,\eta y]$ , is identical.

Consider now the case when $B=3A$ . Then the two equations

$$ \begin{align*}\Sigma_A=0,~\Sigma_{3A}=0,\end{align*} $$

which we view as the equations $\Sigma _1=\Sigma _3=0$ applied to the variables $x^A,y^A,z^A, w^A$ , show that either we have an equality of sets

$$ \begin{align*}\{x^A,y^A\}=\{z^A,w^A\}\end{align*} $$

or we have the relations

$$ \begin{align*}x^A+y^A=0=z^A+w^A.\end{align*} $$

Exactly as in the $B=2A$ discussion above, we use the fact that

$$ \begin{align*}\gcd(\mathrm{the\ }C_i,A)=1\end{align*} $$

to show that from the equality of sets $\{x^A,y^A\}=\{z^A,w^A\}$ we get, up to $O(\# L)$ , the $(\# L)^2$ solutions $\{x,y\}=\{z,w\}.$

It remains to deal with with the equation $x^A+y^A=0=z^A+w^A.$ Fix an $A^{\mathrm {th}}$ root $\tau $ of $-1$ . Then this breaks into the $A^2$ cases $ y =\tau \zeta x, z=\tau \eta w$ , for each pair $\zeta ,\eta $ of $A^{\mathrm {th}}$ roots of unity. We then use the $\Sigma _{C_i}$ equations to obtain the relations

$$ \begin{align*}x^{C_i}(1 +(\tau\zeta)^{C_i})=0, \ \ z^{C_i}(1 +(\tau\eta)^{C_i})=0.\end{align*} $$

In order to get more than $O(\# L)$ solutions, we must have

$$ \begin{align*}1 +(\tau\zeta)^{C_i} =0,\ \ 1 +(\tau\eta)^{C_i}=0 \ \mathrm{for\ each\ } C_i.\end{align*} $$

Suppose first that A is odd. Then we take $\tau :=-1$ , and our equations become

$$ \begin{align*}\zeta^{C_i}=-(-1)^{C_i}=\eta^{C_i} \ \mathrm{for\ each\ } C_i.\end{align*} $$

If all $C_i$ are odd, these are the equations

$$ \begin{align*}\zeta^{C_i}=1=\eta^{C_i} \ \mathrm{for\ each\ } C_i.\end{align*} $$

In order to get more than $O(\# L)$ solutions, we must have $\zeta =1=\eta $ .

Suppose next that A is odd but some $C_i$ is even, say $C_1$ is even. (This can only happen if we are in odd characteristic, as f is Artin-Schreier reduced, and $p \nmid ab$ .) Then we have the equation

$$ \begin{align*}\zeta^{C_1}=-1 =\eta^{C_1}.\end{align*} $$

But $\zeta $ and $\eta $ are roots of unity of odd order, so no powers of either can be $-1$ . So in this case, we have only $x=y=z=w=0$ .

Finally, consider the case when A is even. Then $\gcd (\mathrm {the\ }C_i,A)=1$ , so there is some odd $C_i$ , say $C_1$ is odd. Then the two equations

$$ \begin{align*}1 +(\tau\zeta)^{C_1} =0,\ \ 1 +(\tau\eta)^{C_1}=0,\end{align*} $$

rewritten as

$$ \begin{align*}(\tau\zeta)^{C_1}=-1= (\tau\eta)^{C_1}\end{align*} $$

and raised to the A power, give

$$ \begin{align*}(\tau\zeta)^{AC_1}=(-1)^A=1, (\tau\eta)^{AC_1}=(-1)^A=1.\end{align*} $$

But $\zeta ^A=\eta ^A=1$ , so we get $\tau ^{AC_1}=1=\eta ^{AC_1}$ . But $\tau ^A=-1$ and $C_1$ is odd, so we get $-1=1$ , which is nonsense. Thus, in this case as well, the only solution is $x=y=z=w=0$ .

Proof. The local system ${\mathcal F}$ is pure of weight zero, so geometrically semisimple, and of rank

$$ \begin{align*}\deg(f)-1 \ge 2,\end{align*} $$

so has $M_{2,2}({\mathcal F}) \ge 2$ . Thus, it suffices to show that $M_{2,2}({\mathcal F}) \le 2$ under the stated hypotheses. Now apply Theorem 2.3 and Proposition 3.3(vii), (viii).

Proof. The first statement summarizes Theorems 10.2.6, 10.3.13 and 11.2.3 of [Reference Katz and TiepKT6]. The second statement follows from the explicit determination of $G_{\mathrm {geom}}$ and [Reference Guralnick and TiepGT2, Theorem 1.5], if we assume in addition that $A> 9$ in the cases of 4.1(ii), (iii) with $q=2$ . Assume we are in the cases of 4.1(ii), (iii) with $q=2$ and $A =2^n+1 \leq 9$ . Now if $B_r =1$ (so we are in 4.1(ii) with $r \geq 2$ ), then $M_{2,2}=3$ by Corollary 3.6. Thus, we are left with the cases where $p=2$ and, moreover, $(A,B_1, \ldots ,B_r)= (5,3)$ , $(9,5)$ , $(9,5,3)$ . The third case has $M_{2,2}=3$ by Theorem 3.7. The two remaining local systems of rank $8$ and $4$ , with $r=1$ and $(A,B_1)=(9,5)$ , $(5,3)$ , are dealt with in the next result, which also resolves some open cases left in [Reference Katz and TiepKT6, Chapter 8].

Proof. (a) First, we note that both ${\mathcal H}_{95}$ and ${\mathcal H}_{53}$ satisfy $\mathrm {(\mathbf {S+})}$ by [Reference Katz and TiepKT3, Theorem 3.13]. Furthermore, each of ${\mathcal F}_{531}$ , ${\mathcal F}_{9531}$ , ${\mathcal F}_{951}$ , ${\mathcal F}_{931}$ has $M_{2,2}= 3$ by Corollary 3.6. We also use the facts that if $\varphi $ denotes the character of the underlying representation for the arithmetic monodromy group $G_{\mathrm {arith},{\mathbb F}_2}$ of any of the listed sheaves over ${\mathbb F}_2$ , then $\varphi $ is irreducible of symplectic type; in particular, ${\mathbf Z}(G_{\mathrm {arith},{\mathbb F}_2}) \leq C_2$ . (Indeed, $\varphi $ is visibly real-valued, and its restriction to $G_{\mathrm {geom}}$ of ${\mathcal H}_{95}$ , respectively ${\mathcal H}_{53}$ , is symplectically self-dual by [Reference KatzKa2, 8.8.1-2].) Furthermore, the restriction of $\varphi $ to the arithmetic monodromy group $G_{\mathrm {arith},{\mathbb F}_4}$ of any of the listed sheaves over ${\mathbb F}_4$ is rational-valued by [Reference Katz and TiepKT6, Theorem 7.1.2].

(b) Let $\tilde {\mathcal F}$ denote any of the systems ${\mathcal F}_{9531}$ , ${\mathcal F}_{951}$ , ${\mathcal F}_{953}$ , and let $\tilde {G}$ denote its geometric monodromy group. By the above, $\tilde {G}$ is a finite irreducible subgroup of $\mathrm {Sp}_8({\mathbb C})$ with $M_{2,2}=3$ . Now we can apply [Reference Guralnick and TiepGT2, Theorem 1.5] to $\tilde {G}$ , and note that case (B) cannot occur because the dimension $D=8$ , whereas case (D) cannot occur because $\varphi $ is of symplectic type. It follows that we are in case (C) of [Reference Guralnick and TiepGT2, Theorem 1.5]:

(4.4.1) $$ \begin{align} 2^{1+6}_- \cong E \lhd \tilde{G} \leq {\mathbf N}_{\mathrm{Sp}_8({\mathbb C})}(E) = E \cdot \mathrm{O}^-_6(2), \end{align} $$

and $\tilde {G}/E \leq \mathrm {O}^-_6(2)$ acts transitively on $27$ (nonzero) singular vectors and on $36$ nonsingular vectors of the natural module ${\mathbb F}_2^6$ for $\mathrm {O}^-_6(2)$ . In particular, $27$ divides $|\tilde {G}/E|$ . In fact, the observations in (a) imply that (4.4.1) also holds for $G_{\mathrm {arith},{\mathbb F}_2,\tilde {\mathcal F}}$ , the arithmetic monodromy group of $\tilde {\mathcal F}$ over ${\mathbb F}_2$ .

Next, observe that a pullback of $\tilde {\mathcal F}$ yields ${\mathcal F}_{95}$ , which is a Kummer pullback of ${\mathcal H}_{95}$ . In particular, if G denotes the geometric monodromy group of ${\mathcal F}_{95}$ and H denotes that of ${\mathcal H}_{95}$ , then $G \lhd H$ , $H/G \hookrightarrow C_9$ , and $G \hookrightarrow \tilde {G}$ . Clearly, $5$ divides $|H|$ , so it also divides $|\tilde {G}|$ and $|\tilde {G}/E|$ . Thus, $27 \cdot 5$ divides $|\tilde {G}/E|$ . Using the list of maximal subgroups of $\mathrm {O}^-_6(2)$ [Reference Conway, Curtis, Norton, Parker and WilsonAtlas], we deduce that $\tilde {G}/E$ is either $\Omega ^-_6(2)$ or $\mathrm {O}^-_6(2)$ . In the latter case, [Reference Katz and TiepKT6, Proposition 8.2.4] implies, however, that $|\varphi (g)| = \sqrt {2}$ for some $g \in \tilde {G}$ , which is impossible by (a). Hence, we conclude that $\tilde {G} = E \cdot \Omega ^-_6(2)$ , and the same holds for $G_{\mathrm {arith},{\mathbb F}_4,\tilde {\mathcal F}}$ . However, the Frobenius at $(1,0,\ldots ,0)$ over ${\mathbb F}_2$ (where $1$ is the coefficient for $x^5$ ) has trace $-2/\sqrt {2}$ and hence does not belong to $G_{\mathrm {arith},{\mathbb F}_4,\tilde {\mathcal F}}$ . Together with (4.4.1), this implies that $G_{\mathrm {arith},{\mathbb F}_2,\tilde {\mathcal F}}=E \cdot \mathrm {O}^-_6(2)$ .

(c) To identify H, the $G_{\mathrm {geom}}$ for ${\mathcal H}_{95}$ , we recall that H satisfies $\mathrm {(\mathbf {S+})}$ by (a). First, suppose that H is an extraspecial normalizer. Together with (a), this implies that

(4.4.2) $$ \begin{align} 2^{1+6}_- \cong E_1 \lhd H \leq {\mathbf N}_{\mathrm{Sp}_8({\mathbb C})}(E_1) = E_1 \cdot \mathrm{O}^-_6(2). \end{align} $$

We already mentioned that each of $C_9$ and $C_5$ injects in H, hence also in $H/E_1 \leq \mathrm {O}^-_6(2)$ . Again using the list of maximal subgroups of $\Omega ^-_6(2)$ [Reference Conway, Curtis, Norton, Parker and WilsonAtlas], we deduce that $H/E_1$ is either $\Omega ^-_6(2)$ or $\mathrm {O}^-_6(2)$ . In the latter case, [Reference Katz and TiepKT6, Proposition 8.2.4] implies, however, that $|\varphi (g)| = \sqrt {2}$ for some $g \in H$ , which is impossible by (a). Hence, we conclude that $H = E_1 \cdot \Omega ^-_6(2)$ (in fact, the same holds for $G_{\mathrm {arith},{\mathbb F}_4,{\mathcal H}_{95}}$ because it normalizes ${\mathbf O}_2(H)=E_1$ and hence also satisfies (4.4.2)). In particular, H is perfect. Since $H/G \hookrightarrow C_9$ , we also have $G=H$ . Knowing now that

$$ \begin{align*}G \leq G_{\mathrm{arith},{\mathbb F}_4,{\mathcal F}_{95}} \leq G_{\mathrm{arith},{\mathbb F}_4,{\mathcal F}_{951}} = G,\end{align*} $$

we conclude that $G_{\mathrm {arith},{\mathbb F}_4,{\mathcal F}_{95}}=G$ . Next, again using the Frobenius at $s=1$ of ${\mathcal F}_{95}$ with trace $-\sqrt {2}$ , we see that this Frobenius is in $G_{\mathrm {arith},{\mathbb F}_2,{\mathcal F}_{951}}$ but not in its subgroup G of index $2$ . This shows that $G_{\mathrm {arith},{\mathbb F}_2,{\mathcal F}_{95}} = G_{\mathrm {arith},{\mathbb F}_2,{\mathcal F}_{951}}=E \cdot \mathrm {O}^-_6(2)$ . As $G_{\mathrm {arith},{\mathbb F}_2,{\mathcal F}_{95}}$ is a subgroup of $G_{\mathrm {arith},{\mathbb F}_2,{\mathcal H}_{95}}$ , which normalizes ${\mathbf O}_2(H)=E_1$ and hence satisfies (4.4.2), we deduce that $G_{\mathrm {arith},{\mathbb F}_2,{\mathcal H}_{95}}=2^{1+6}_- \cdot \mathrm {O}^-_6(2)$ .

Assume now that H is almost quasisimple, with R the unique non-abelian composition factor. Then $G^{(\infty )} = H^{(\infty )}$ is a cover of R with center

$$ \begin{align*}{\mathbf Z}(H^{(\infty)}) \leq {\mathbf Z}(H) \leq C_2={\mathbf Z}(E_1)={\mathbf Z}(E);\end{align*} $$

cf. (4.4.1), (4.4.2). However, $E \cap G^{(\infty )}$ is a normal $2$ -subgroup of $G^{(\infty )}$ , so

$$ \begin{align*}E \cap G^{(\infty)} \leq {\mathbf Z}(E) \cap G^{(\infty)} = {\mathbf Z}(G^{(\infty)}).\end{align*} $$

We also know from $G \leq \tilde {G} = E\cdot S$ that

$$ \begin{align*}G^{(\infty)}/(E \cap G^{(\infty)}) \cong G^{(\infty)}E/E \leq \tilde{G}/E = S \cong \mathrm{SU}_4(2).\end{align*} $$

It follows that R is a simple subquotient of $\mathrm {SU}_4(2)$ . Using [Reference Conway, Curtis, Norton, Parker and WilsonAtlas], we readily see that $R=\mathsf {A}_5$ , $\mathsf {A}_6$ , or $\mathrm {SU}_4(2)$ ; in particular, $\mathrm {Out}(R)$ is a $2$ -group. Recalling that

$$ \begin{align*}R \lhd H/{\mathbf Z}(H) \leq \mathrm{Aut}(R), ~{\mathbf Z}(H) \leq C_2, ~C_9 \hookrightarrow H,\end{align*} $$

we have that $C_9 \hookrightarrow R$ . This rules out the possibilities $\mathsf {A}_5$ and $\mathsf {A}_6$ , and so $R = \mathrm {SU}_4(2)$ . But H acts irreducibly on ${\mathcal H}_{95}$ of dimension $8$ , so we must have that $H \cong \mathrm {Sp}_4(3) \cdot 2$ . This is, however, impossible because $H={\mathbf O}^2(H)$ .

(d) In dimension $8$ , it remains to determine $G_1$ , the $G_{\mathrm {geom}}$ for ${\mathcal F}_{931}$ , which also has $M_{2,2}=3$ . As in the case of $\tilde {G}$ , this equality implies by [Reference Guralnick and TiepGT2, Theorem 1.5] that

(4.4.3) $$ \begin{align} 2^{1+6}_- \cong E_2 \lhd G_1 \leq {\mathbf N}_{\mathrm{Sp}_8({\mathbb C})}(E_2) = E_2 \cdot \mathrm{O}^-_6(2). \end{align} $$

Moreover, $G_1/E_2 \leq \mathrm {O}^-_6(2)$ still acts acts transitively on $27$ (nonzero) singular vectors and on $36$ non-singular vectors of the natural module $W={\mathbb F}_2^6$ for $\mathrm {O}^-_6(2)$ ; in particular, $27$ divides $|G_1/E_2|$ . Using the list of maximal subgroups of $\mathrm {O}^-_6(2)$ [Reference Conway, Curtis, Norton, Parker and WilsonAtlas], we deduce that $G_1/E_2$ is either $\mathrm {O}^-_6(2)$ , $\Omega ^-_6(2)$ , a subgroup of $M:=\mathrm {O}^-_2(2) \wr \mathsf {S}_3$ , or a subgroup of $N:=\mathrm {GU}_3(2) \cdot 2 \cong 3^{1+2}_+ \rtimes 2\mathsf {S}_4$ . The first case is impossible since $|G_1| \leq |\tilde {G}| = |E_2| \cdot |\Omega ^-_6(2)|$ . To rule out the second possibility, we make use of [Reference Katz and TiepKT6, Corollary 7.1.5], which shows that

(4.4.4) $$ \begin{align} \varphi(x) \equiv -1 \ \ \ \ (\mathrm{mod} \ 3) \end{align} $$

for any odd-order element $x \in G_1$ . Indeed, in this case, we have $G_1 = \tilde {G}$ since $|G_1|=|\tilde {G}|$ and $G_1 \leq \tilde {G}$ ; in particular, $G_1$ contains an element $g_1$ of order $5$ which has rational trace. The latter condition implies that $\varphi (g_1) \in \{-2,3\}$ , violating (4.4.4). In the third possibility, we can realize M as the stabilizer of the decomposition

$$ \begin{align*}W = \langle e_1,e_2 \rangle_{{\mathbb F}_2} \oplus \langle e_3,e_4 \rangle_{{\mathbb F}_2} \oplus \langle e_5,e_6 \rangle_{{\mathbb F}_2},\end{align*} $$

where the quadratic form Q on W takes value

$$ \begin{align*}x_1^2+x_1x_2+x_2^2+x_3^2+x_3x_4+x_4^2+x_5^2+x_5x_6+x_6^2\end{align*} $$

at the vector $\sum ^6_{i=1}x_ie_i$ . But then the vectors $u:=e_1+e_2$ and $v:=\sum ^6_{i=1}e_i$ have $Q(u)=Q(v)=1$ and belong to different M-orbits, showing that M is not transitive on the non-singular vectors of W. This leaves only the fourth possibility: $G_1/E_2 \leq N$ . In particular, $G_1$ is solvable.

Now we use the embedding $G_1 \hookrightarrow \tilde {G}=G_{\mathrm {geom},{\mathcal F}_{9531}} = E \cdot \Omega ^-_6(2)$ . Then

$$ \begin{align*}3=M_{2,2}(G_1) \geq M_{2,2}(EG_1) \geq M_{2,2}(\mathrm{Sp}_8({\mathbb C})) = 3,\end{align*} $$

showing that $M_{2,2}(EG_1)=3$ . Thus, $EG_1/E$ is a solvable subgroup of $\Omega ^-_6(2)$ which acts transitively on $27$ singular vectors and on $36$ non-singular vectors of ${\mathbb F}_2^6$ . Using the list of maximal subgroups of $\Omega ^-_6(2)$ as in the preceding paragraph, we see that $EG_1/E$ is contained in $N_1 \cong \mathrm {GU}_3(2)$ . Recalling E is a $2$ -group and $G_1={\mathbf O}^{2'}(G_1)$ (as the $G_{\mathrm {geom}}$ for a local system on ${\mathbb A}^2/\overline {{\mathbb F}_2}$ ), we then have

$$ \begin{align*}EG_1/E \leq {\mathbf O}^{2'}(N_1) \cong \mathrm{SU}_3(2) \cong 3^{1+2}_+ \rtimes Q_8.\end{align*} $$

Moreover, $27$ and $36$ both divide $|G_1/E_2| = |G_1|/|E|$ , so in fact, we have

(4.4.5) $$ \begin{align} 3^{1+2}_+ \rtimes C_4 \leq EG_1/E \leq 3^{1+2}_+ \rtimes Q_8. \end{align} $$

Suppose that $EG_1/E = 3^{1+2}_+ \rtimes C_4$ in (4.4.5). Note that we can turn the quadratic space $W = {\mathbb F}_2^6$ into the Hermitian space $W_1:={\mathbb F}_4^3$ for $\mathrm {SU}_3(2)$ in such a way that the set $N(W)$ of $36$ non-singular vectors of W is exactly the set $N(W_1)$ of $36$ non-singular vectors of $W_1$ . Since $EG_1/E$ acts transitively on $N(W)=N(W_1)$ , the stabilizer of any $w \in N(W_1)$ has order $3$ , which implies that a fixed involution j in $EG_1/E$ does not fix any $w \in N(W_1)$ . The Sylow $2$ -subgroups of $\mathrm {SU}_3(2)$ are isomorphic to $Q_8$ , so any involution in $\mathrm {SU}_3(2)$ is conjugate to j and hence does not fix any $w \in N(W_1)$ . But this is a contradiction, since the stabilizer of any $w \in N(W_1)$ in $\mathrm {SU}_3(2)$ is $\mathrm {SU}_2(2) \cong \mathsf {S}_3$ , which clearly contains an involution. We have therefore shown that