Conjecture 3.7

Let U be a scheme of finite type over $\mathbb{Z}$ and let $\mathcal{A} \to U$ be a family of g-dimensional, PPAVs with full monodromy. For a prime power q, let

\begin{equation*} H'(q,t) = \frac{\#\{u \in U(\mathbb{F}_q) : q+1-t=\#\mathcal{A}_u(\mathbb{F}_q) \}}{\#U(\mathbb{F}_q)}, \end{equation*}

seen as a measure on $\mathbb{Z}$. Let $\nu^{\prime}$ be as in (10). As $q \to \infty$ along prime powers, we have $d(H'(q, \cdot), \nu'(q, \cdot)) \to 0$, where $H'(q, \cdot)$ and $\nu^{\prime}(q, \cdot)$ are considered as probability measures on $\mathbb{Z}$.

Proof. The proof is virtually identical to that of [Reference Lee39, theorem 1]: if one simply replaces every occurrence of det with $\operatorname{mult}$ in the proof of [Reference Lee39, theorem 1] everything goes through without difficulty. More precisely, let

\begin{equation*} e(x) = \begin{cases} 1 \,\text{if } x = \zeta, \\ 0 \,\text{otherwise}. \end{cases} \end{equation*}

Throughout the proof, several instances of $\det(d_\alpha)=\alpha^n$ are replaced by $\operatorname{mult}(d_\alpha)=\alpha$, where $d_\alpha = \begin{pmatrix} \operatorname{Id}_n & 0 \\ 0 & \alpha \operatorname{Id}_n \end{pmatrix}$. In particular, the sums $\sum_{\alpha \in \mathbb{F}_q^\times} e(\alpha^n)$ are replaced by $\sum_{\alpha \in \mathbb{F}_q^\times} e(\alpha)$. In the proof of [Reference Lee39, theorem 1], the sum $\sum_{\alpha \in \mathbb{F}_q^\times} e(\alpha^n)$ evaluates to the number S of nth roots of ζ in $\mathbb{F}_q^\times$; in our case, the sum $\sum_{\alpha \in \mathbb{F}_q^\times} e(\alpha)$ simply evaluates to 1 for all $\zeta \in \mathbb{F}_q^\times$.

Proof. The numerator of (20) is given by (18) (with n = g, $q=\ell$, $\zeta=m$ and $\eta=t$). Note that $\ell^{g^2-1} \prod_{j=1}^g \left(\ell^{2j}-1 \right)$ is exactly $\frac{\#\operatorname{GSp}_{2g}(\mathbb{F}_\ell)}{\ell\varphi(\ell)}$. Thus, the ratio in (20) is given by

\begin{equation*} 1 + \frac{E}{\frac{1}{\ell(\ell-1)} \#\operatorname{GSp}_{2g}(\mathbb{F}_\ell)}. \end{equation*}

Since

\begin{align*} \frac{1}{\ell(\ell-1)} \#\operatorname{GSp}_{2g}(\mathbb{F}_\ell) = \frac{1}{\ell(\ell-1)} (\ell-1) \#\operatorname{Sp}_{2g}(\mathbb{F}_\ell) = \ell^{g^2-1} \prod_{j=1}^g(\ell^{2j}-1) \gg \ell^{2g^2+g-1}, \end{align*}

we obtain that (20) is

\begin{equation*} 1+O\left( \ell^{\frac{3}{2}g^2 + \frac{1}{2}g -\frac{3}{4} - (2g^2+g-1)} \right) = 1+O\left( \ell^{-\frac{1}{2}g^2 - \frac{1}{2}g +\frac{1}{4}} \right), \end{equation*}

which is $1+O(\ell^{-2})$ for all $g \geq 2$.

Proof. We view X as a subvariety of the affine space $\mathbb{A}_{\mathbb{F}_\ell}^{(2g)^2}$, considered as the space of matrices of size $2g \times 2g$. The variety X is the intersection of $\operatorname{GSp}_{2g, \mathbb{F}_\ell}^{m} \cong \operatorname{Sp}_{2g, \mathbb{F}_\ell}$ with the hyperplane H defined by the condition $\operatorname{Tr}(M) = t$. The hyperplane section $\operatorname{GSp}_{2g, \mathbb{F}_\ell}^{m} \cap H$ is smooth at a point $x \in X(\overline{\mathbb{F}_\ell})$ unless the (tangent space to the) hyperplane H contains the tangent space of $\operatorname{GSp}_{2g, \overline{\mathbb{F}_\ell}}^{m}$ at the point x. Take any point $x \in X(\overline{\mathbb{F}_\ell})$. Since x has multiplier m, left multiplication by $x \in \operatorname{GSp}_{2g}(\overline{\mathbb{F}_\ell})$ gives an isomorphism L_x between $\operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}$ and $\operatorname{GSp}_{2g, \overline{\mathbb{F}_\ell}}^{m}$. The differential of L_x gives an isomorphism between the tangent space at $\operatorname{Id}$ and the tangent space at x. If we identify both tangent spaces to subspaces of the tangent space to $\mathbb{A}_{\overline{\mathbb{F}_\ell}}^{(2g)^2}$ (that is, to matrices of size $2g \times 2g$), the differential in question is simply multiplication by x itself. Thus, we may view the tangent space at x as the image via x of the tangent space at $\operatorname{Id}$, which is the Lie algebra of $\operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}$. This can be written down explicitly: choose the anti-symmetric bilinear form represented by the matrix

\begin{equation*} \Omega := \begin{pmatrix} 0 & \operatorname{Id}_g \\ -\operatorname{Id}_g & 0 \end{pmatrix}. \end{equation*}

Differentiating the condition ${}^tM \Omega M = \Omega$, we find that the Lie algebra of $\operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}$ is given by those matrices M that satisfy ${}^tM\Omega + \Omega M = 0$. Writing M in block form, we obtain that $\operatorname{Lie} \operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}$ is the vector space of $\overline{\mathbb{F}_\ell}$-matrices

\begin{equation*} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \end{equation*}

with ${}^t B= B, {}^tC = C, {}^tD = -A$ (see [Reference Fulton and Harris22, § 16.1] for the identical calculation over the complex numbers). From the previous arguments, it follows that x can only be a singular point if

\begin{equation*} x \operatorname{Lie}(\operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}) \subseteq \{\operatorname{Tr} = 0\}, \end{equation*}

which is to say

\begin{equation*} \operatorname{Tr} (xL) = 0 \quad \forall L \in \operatorname{Lie}(\operatorname{Sp}_{2g, \overline{\mathbb{F}_\ell}}). \end{equation*}

Write $x = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ and $L=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ with $B, C$ symmetric and $D=-{}^tA$. This easily gives $\operatorname{Tr} (\beta C) = \operatorname{Tr}(\gamma B) = 0$ for all symmetric $B, C$ (which implies that $\beta, \gamma$ are anti-symmetric) and

\begin{equation*} \operatorname{Tr}(\alpha A - \delta \cdot {}^t A ) = \operatorname{Tr}(\alpha A - A \cdot {}^t \delta ) = \operatorname{Tr}(\alpha A - {}^t \delta \cdot A ) = 0 \end{equation*}

for all A (which implies $\alpha={}^t\delta$).

Thus, the locus of non-smooth points is contained in the linear space defined by the equations

\begin{equation*}{}^t\beta=-\beta, {}^t\gamma=-\gamma, {}^t\delta=\alpha.\end{equation*}

This linear space has dimension $g^2 + 2 \frac{g(g-1)}{2} = 2g^2-g$, and hence codimension at least $2g-1 \geq 3$ in X, each of whose irreducible components has dimension at least $\dim \operatorname{GSp}_{2g, \mathbb{F}_\ell}^{m}-1 = \dim \operatorname{Sp}_{2g, \mathbb{F}_\ell}-1 = 2g^2+g-1$ (at least one irreducible component has exactly this dimension). We now observe that by the Lang–Weil estimates [Reference Lang and Weil37, theorem 1] we have $\#X^{\operatorname{sing}}(\mathbb{F}_\ell) = O( \ell^{\dim X^{\operatorname{sing}}} ) = O(\ell^{\dim X - 3})$, with an implicit constant that depends only on X and not $\ell$. Taking into account the obvious decomposition $X^{\operatorname{smooth}}(\mathbb{F}_\ell) \bigsqcup X^{\operatorname{sing}}(\mathbb{F}_\ell) = X(\mathbb{F}_\ell)$ and the fact that $ \#X(\mathbb{F}_\ell) = \frac{\#\operatorname{GSp}_{2g, \mathbb{F}_\ell}(\mathbb{F}_\ell)}{\ell \varphi(\ell)} (1 + O(\ell^{-2})) $ by lemma 4.5, we obtain the desired estimate $\#X^{\operatorname{smooth}}(\mathbb{F}_\ell) = \frac{\#\operatorname{GSp}_{2g, \mathbb{F}_\ell}(\mathbb{F}_\ell)}{\ell \varphi(\ell)} (1 + O(\ell^{-2})).$

Proof. Let $X:=X_t^q$. We apply [Reference Oesterlé44, Property (U), p. 326] to

\begin{equation*} X(\mathbb{Z}_\ell) = \{M \in \operatorname{GSp}_{2g}(\mathbb{Z}_\ell) : \operatorname{Tr} M =t, \operatorname{mult} M = m q \} \end{equation*}

\begin{equation*} m=1, \quad N = (2g)^2, \quad n=n, \quad B = x_0 + \ell \mathbb{Z}_\ell^{(2g)^2} \end{equation*}

where $x_0 \bmod \ell$ is a matrix lying in $X^{\operatorname{sing}}(\mathbb{F}_\ell)$. We first assume that $X_{\mathbb{Z}_\ell}$ is irreducible. Considering X as a scheme over the spectrum of the DVR $\mathbb{Z}_\ell$, [1, lemma 0B2J] shows that $X_{\mathbb{F}_\ell}$ is equidimensional of some dimension d, and Oesterlé’s result gives

\begin{equation*} \#\{\text{closed balls}\ A\ \text{of radius}\ \ell^{-n} : A \cap X \neq \emptyset \,\text{and } A \subseteq B \} \leq C \ell^{\dim X (n-1)} \end{equation*}

for a constant C that depends only on the degree in dimension d [Reference Oesterlé44, § 0.6] of $X_{\mathbb{F}_\ell}$, which is clearly bounded independently of $\ell$. On the other hand, we have

\begin{align*} &\#\{\text{closed balls A of radius}\ \ell^{-n}: A \cap X \neq \emptyset \,\text{and } A \subseteq B \} \\= &\quad \# \left\{M \in \operatorname{GSp}_{2g}(\mathbb{Z}/\ell^n\mathbb{Z}) : \begin{array}{c} \exists \tilde{M} \in X(\mathbb{Z}_\ell) \\ \tilde{M} \equiv M\ (\mathrm{mod}\ {\ell^n}) \\ M \equiv x_0\ (\mathrm{mod}\ {\ell}) \end{array} \right\}. \end{align*}

Hence, summing over the points $x_0 \in X^{\operatorname{sing}}(\mathbb{F}_\ell)$, we obtain

(21)\begin{equation} \# \left\{M \in \operatorname{GSp}_{2g}(\mathbb{Z}/\ell^n\mathbb{Z}) : \begin{array}{c} \exists \tilde{M} \in X(\mathbb{Z}_\ell) \\ \tilde{M} \equiv M\ (\mathrm{mod}\ {\ell^n}) \\ M \bmod \ell \in X^{\operatorname{sing}}(\mathbb{F}_\ell) \end{array} \right\} \leq C \#X^{\operatorname{sing}}(\mathbb{F}_\ell) \ell^{(n-1) \dim X}. \end{equation}

If $X_{\mathbb{Z}_\ell}$ is not irreducible, we can repeat the above argument with each irreducible component X_i. If C_i is the constant that corresponds to the component X_i, applying the previous argument to X_i and summing over i we obtain

\begin{align*} &\# \left\{M \in \operatorname{GSp}_{2g}(\mathbb{Z}/\ell^n\mathbb{Z}) : \begin{array}{c} \exists \tilde{M} \in X(\mathbb{Z}_\ell) \\ \tilde{M} \equiv M\ (\mathrm{mod}\ {\ell^n}) \\ M \bmod \ell \in X^{\operatorname{sing}}(\mathbb{F}_\ell) \end{array} \right\} \\ & \leq \left(\sum_i C_i\right)\#X^{\operatorname{sing}}(\mathbb{F}_\ell) \ell^{(n-1) \dim X}. \end{align*}

Note that the number of irreducible components is bounded independently of $\ell$, and so is the constant $\left(\sum_{i} C_i\right)$ (because the degrees are bounded in terms of the equations of X, which are independent of $\ell$). The conclusion is that there exists a constant C such that (21) holds for all n and all but finitely many $\ell$.

Recall now the definition of $\nu_\ell(q, t)$ from Eq. (7): it is the limit over k of the ratio

(22)\begin{equation} \frac{\# \operatorname{Im}\left( X(\mathbb{Z}_\ell) \to \operatorname{GSp}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z}) \right) }{\#\operatorname{GSp}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z}) / (\ell^k\varphi(\ell^k))}. \end{equation}

Clearly, a matrix M counted in the numerator of this expression in particular reduces modulo $\ell$ to a point in $X(\mathbb{F}_\ell)$. For a fixed $x_0 \in X(\mathbb{F}_\ell)$, denote by $N(x_0, k)$ the quantity

\begin{equation*} N(x_0, k) = \#\left\{ M \in \operatorname{Im}\left( X(\mathbb{Z}_\ell) \to \operatorname{GSp}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z}) \right) : M \equiv x_0\ (\mathrm{mod}\ {\ell}) \right\}. \end{equation*}

When x ₀ is a smooth point of $X(\mathbb{F}_\ell)$, Hensel’s lemma shows that x ₀ has precisely $\ell^{(k-1)\dim X_{\mathbb{F}_\ell}}$ lifts to $X(\mathbb{Z}/\ell^k\mathbb{Z})$, and each of these further lifts to a point in $X(\mathbb{Z}_\ell)$ (note that a smooth point necessarily lies on a component of dimension equal to $\dim X_{\mathbb{F}_\ell}$: indeed, X is a hyperplane section of a smooth variety, so every smooth point lies on a component of maximal dimension). Therefore, we have $N(x_0, k)=\ell^{(k-1)\dim X_{\mathbb{F}_\ell}}$ for such x ₀. On the other hand, Eq. (21) and lemma 4.6 show that $\sum_{x_0 \in X^{\operatorname{sing}}(\mathbb{F}_\ell)} N(x_0, k) = O(\ell^{k \dim X_{\mathbb{F}_\ell}-3})$.

Thus, the numerator of (22) is given by

\begin{equation*} \begin{aligned} \sum_{x_0 \in X(\mathbb{F}_\ell)} N(x_0, k) & = \sum_{x_0 \in X^{\operatorname{smooth}}(\mathbb{F}_\ell)} N(x_0, k) + \sum_{x_0 \in X^{\operatorname{sing}}(\mathbb{F}_\ell)} N(x_0, k) \\ & = \#X^{\operatorname{smooth}}(\mathbb{F}_\ell) \ell^{(k-1)\dim X_{\mathbb{F}_\ell}} + O(\ell^{k \dim X_{\mathbb{F}_\ell}-3}) \\ & = \frac{\#\operatorname{GSp}_{2g}(\mathbb{F}_\ell)}{\ell\varphi(\ell)} (1+O(\ell^{-2})) \cdot \ell^{(k-1)\dim X_{\mathbb{F}_\ell}} + O(\ell^{k \dim X_{\mathbb{F}_\ell}-3}), \end{aligned} \end{equation*}

where in the last equality we have applied lemma 4.6. Using $\dim X_{\mathbb{F}_\ell} = \dim \operatorname{GSp}_{2g, \mathbb{F}_\ell} - 2$ and dividing by

\begin{equation*} \frac{\#\operatorname{GSp}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z})}{\ell^k\varphi(\ell^k)} = \frac{\#\operatorname{GSp}_{2g}(\mathbb{F}_\ell)}{\ell\varphi(\ell)} \ell^{(k-1) \dim X_{\mathbb{F}_\ell}}, \end{equation*}

we obtain that (22) is $1+O(\ell^{-2})$. The claim follows upon passing to the limit in k.

Proof. By lemma 4.7, we have $\nu_{\ell}(q, t)=1+O(\ell^{-2})$ as $\ell$ ranges over primes $\ell \geq 3$ that do not divide q. The factors $\nu_\infty(q,t)$, $\nu_2(q, t)$ and $\nu_p(q,t)$ are well defined, as already argued. It follows that the infinite product $\prod_{\ell \lt \infty}\nu_{\ell}(q, t)$ converges.

Proof. Since the infinite product defining $\nu(q,t)$ converges, it suffices to show that each factor in this product is non-zero. This is well known to be true for the infinite factor $\nu_\infty(q,t)$, whose support is the interval $[-2g\sqrt{q}, 2g\sqrt{q}]$. To show that $\nu_\ell(q,t)$ is non-zero (including for $\ell= p$) we proceed as follows. Let $X_t^q$ be as in definition 4.3 (for the ring $R=\mathbb{Q}_\ell$) and let for simplicity $X^q := \operatorname{GSp}_{2g, \mathbb{Q}_\ell}^{q}$. We rewrite the definition of $\nu_\ell(q,t)$ in the form of remark 3.8, namely,

\begin{equation*} \nu_{\ell}(q, t) = \lim_{k \to \infty} \frac{\#\operatorname{Im}\left( X^q_t(\mathbb{Q}_\ell) \cap \operatorname{Mat}_{2g}(\mathbb{Z}_\ell) \to \operatorname{Mat}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z}) \right)}{\#\operatorname{Im}\left( X^q(\mathbb{Q}_\ell) \cap \operatorname{Mat}_{2g}(\mathbb{Z}_\ell) \to \operatorname{Mat}_{2g}(\mathbb{Z}/\ell^k\mathbb{Z}) \right) / \ell^k}. \end{equation*}

Set $d := \dim \operatorname{GSp}_{2g,\mathbb{Q}_\ell}-2 = 2g^2+g-1$ and multiply both numerator and denominator by $\ell^{-kd}$. We see both X^q and $X^q_t$ as subschemes of $\mathbb{A}_{\mathbb{Q}_\ell}^{(2g)^2}$, so that their $\mathbb{Q}_\ell$-points are subsets of $\mathbb{Q}_\ell^{(2g)^2}$. Let $Y_t^q := X_t^q(\mathbb{Q}_\ell) \cap \mathbb{Z}_\ell^{(2g)^2}$ and $Y^q := X^q(\mathbb{Q}_\ell) \cap \mathbb{Z}_\ell^{(2g)^2}$. The sets $Y_t^q$ and Y^q are closed analytic subsets of $\mathbb{Z}_\ell^{(2g)^2}$. Note that X^q is smooth and irreducible of dimension d + 1, hence $X_t^q$—which is a subscheme of X^q defined by a single non-trivial equation—has dimension d: slicing with a hyperplane makes the dimension drop at most by 1; on the other hand, the dimension must drop (if $X_t^q$ had a component of dimension d + 1, by the irreducibility of X^q we would have $X_t^q \supseteq X^q$, which is not the case). More precisely, by the same argument, every irreducible component of $X_t^q$ has dimension d. We can thus write

(23)\begin{equation} \nu_\ell(q,t) = \lim_{k \to \infty} \frac{\ell^{-dk} \#\operatorname{Im}(Y_t^q \to (\mathbb{Z}/\ell^k\mathbb{Z})^{(2g)^2})}{\ell^{-(d+1)k} \#\operatorname{Im}(Y^q \to (\mathbb{Z}/\ell^k\mathbb{Z})^{(2g)^2})}. \end{equation}

Recall from [Reference Oesterlé44, § 3] the notion of measure in dimension d of a closed analytic subset Y of $\mathbb{Z}_\ell^{(2g)^2}$ of dimension $\leq d$ (denoted by $\mu_d(Y)$). By [Reference Oesterlé44, théorème 2], the numerator and denominator of (23) admit limit as $k \to \infty$, and these limits are given by $\mu_d(Y_t^q)$ and $\mu_{d+1}(Y^q)$, respectively. Hence, $\nu_\ell(q,t) = \frac{\mu_d(Y_t^q)}{\mu_{d+1}(Y^q)}$.

To conclude, it suffices to show that $\mu_{d+1}(Y^q)$ and $\mu_d(Y_t^q)$ are both strictly positive. Note that Y^q is open in $X^q(\mathbb{Q}_\ell)$ for the $\ell$-adic topology, since it is the intersection of $X^q(\mathbb{Q}_\ell)$ with the $\ell$-adically open set $(\mathbb{Z}_\ell)^{(2g)^2}$; a similar comment applies to $X_t^q$. We claim that to check the positivity of $\mu_{d+1}(Y^q)$ and $\mu_d(Y_t^q)$ it suffices to show that $Y^q, Y_t^q$ contain at least one smooth point of $X^q(\mathbb{Q}_\ell), X_t^q(\mathbb{Q}_\ell)$ respectively. To show this implication, we argue as follows (we discuss the case of $X^q_t$, but the case of X^q is completely analogous, and in fact easier since X^q is smooth). The $\ell$-adic analytic variety $X^q_t(\mathbb{Q}_\ell)$ is of pure dimension d, and its smooth points $(X^q_t)_{\operatorname{smooth}}$ form an $\ell$-adically open set, so the intersection $Y^q_t \cap (X^q_t)_{\operatorname{smooth}}$ is $\ell$-adically open (recall that $Y^q_t$ is $\ell$-adically open). In particular, if $Y^q_t$ contains at least one smooth point of $X^q_t(\mathbb{Q}_\ell)$, then it contains an open set of smooth points. The local dimension at each smooth point of $X^q_t(\mathbb{Q}_\ell)$ is d. By construction of the measure µ_d (see again [Reference Oesterlé44, § 3]), an open subset of $X^q_t(\mathbb{Q}_\ell)$ consisting of smooth points has positive measure: indeed, in the case of constant dimension d that we are considering here, µ_d is constructed locally by taking an analytic isometry between a ball in $(X^q_t)_{\operatorname{smooth}}$ and an open ball in $\mathbb{Q}_\ell^{d}$, and pulling back the Haar measure ν of $\mathbb{Q}_\ell^{d}$, normalized by $\nu(\mathbb{Z}_\ell^{d})=1$. It is then clear that any open set in $(X^q_t)_{\operatorname{smooth}}$ has positive measure with respect to µ_d, and we have shown that $Y^q_t$ contains an open set of smooth points of $X^q_t(\mathbb{Q}_\ell)$ as soon as it contains one. We are thus reduced to checking that $Y^q, Y_t^q$ contain at least one smooth point of $X^q(\mathbb{Q}_\ell), X_t^q(\mathbb{Q}_\ell)$ respectively.

For X^q, which is smooth, this amounts to constructing a symplectic matrix with coefficients in $\mathbb{Z}_\ell$ and given multiplier; this follows immediately from either proposition 6.3 and remark 6.5 or from remark 4.2 after observing that the identity matrix lies in $\operatorname{Sp}_{2g}(\mathbb{Z}_\ell)$. For $X^q_t$, we construct the relevant point explicitly.

We observe that $X^q_t$ arises as a fibre of the trace map:

\begin{equation*} \text{trace}: X^q \rightarrow \mathbb{A}^1 \end{equation*}

i.e., $X^q_t=\text{trace}^{-1}(t)$. A sufficient condition for a point $P \in X^q_t$ to be smooth is the existence of a curve $C \subseteq X^q$ containing P such that the restriction of the trace map

\begin{equation*} \text{trace}: C \rightarrow \mathbb{A}^1 \end{equation*}

has non-vanishing differential at P. To see this, notice that the dimension of the tangent space at P in $X^q_t$ is the dimension of the tangent space at P in X^q minus the dimension of the image of the differential of the trace map (restricted to X^q) at P. Let us fix the symplectic form

\begin{equation*} \Omega= \begin{pmatrix} 0 & \operatorname{Id}_g \\ -\operatorname{Id}_g & 0 \end{pmatrix}. \end{equation*}

We consider the curve M_a, parametrized by $a \in \mathbb{A}^1$, given by

\begin{equation*} M_a= \begin{bmatrix} a & z & a-q & z \\ {}^t z & q\operatorname{Id}_{g-1} & {}^t z & 0_{g-1} \\ 1 & z & 1 & z \\ {}^t z & 0_{g-1} & {}^t z & \operatorname{Id}_{g-1} \end{bmatrix} \end{equation*}

where z is the $1 \times (g-1)$ vector $(0,\ldots,0)$. One checks that $M_a \in X^q(\mathbb{Q}_\ell)$: up to a suitable change of basis, the symplectic form is represented by $\operatorname{diag}\left( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \ldots, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right)$, and in the same basis M_a becomes the matrix $\operatorname{diag}\left( \begin{pmatrix} a & a-q \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & q \end{pmatrix}, \ldots, \begin{pmatrix} 1 & 0 \\ 0 & q \end{pmatrix} \right)$, which is manifestly symplectic since every $2 \times 2$ block has determinant q. Moreover, $\text{trace}(M_a)=a+qg-q+g$; the composition

\begin{equation*} a \rightarrow M_a \rightarrow \text{trace}(M_a)=a+qg-q+g \end{equation*}

is just a translation of $\mathbb{A}^1$, which implies that the differential of the trace map at M_a is surjective. Therefore, the point $M_{t-qg+q-g} \in X^{q}_t$ is smooth and its entries are elements of $\mathbb{Z}_{\ell}$. This concludes the proof.

Proof. Write $P_{C}(t)=\sum_{i=0}^{2g} a_i t^i \in \mathbb{Z}[t]$ and $f_{C}(t)=\sum_{i=0}^{2g} b_i t^i$. By theorem 1.7, we have the equality $b_i=a_{2g-i}$, and since q is odd we also have $ b_i=a_{2g-i}=q^{g-i}a_i \equiv a_i\ (\mathrm{mod}\ 2) $.

Proof. As above, let $\infty$ be the unique point at infinity of C, and for $i=1,\ldots,2g+1$ let $Q_i=(\alpha_i,0)-\infty \in \operatorname{Div}_{C}(\overline{\mathbb{F}_q})$. Write P_i for the image of Q_i in the $\mathbb{F}_2$-vector space $\operatorname{Div}_{C}(\overline{\mathbb{F}_q})\otimes \mathbb{F}_2$, and let V be the $(2g+1)$-dimensional $\mathbb{F}_2$-vector subspace of $\operatorname{Div}_{C}(\overline{\mathbb{F}_q})\otimes \mathbb{F}_2$ spanned by the P_i. There is a natural action of $\operatorname{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ on V, which we consider as a representation $\rho:\operatorname{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q) \to \operatorname{GL}(V)$. By Galois theory, it is clear that $\operatorname{Frob}$ acts on the set $\{\alpha_i\}_{i=1}^{2g+1}$ with r orbits, one corresponding to each irreducible factor of f(x). The lengths of the orbits are given by the degrees d_i of the factors $f_i(x)$. This means that, in the natural basis of V given by the P_i, the action of Frobenius is given by a permutation matrix corresponding to a permutation of cycle type $ (d_1, d_2, \ldots, d_r) $. It follows immediately that the characteristic polynomial of $\rho(\operatorname{Frob})$ is

\begin{equation*} \det(t \operatorname{Id}-\rho(\operatorname{Frob}))=(t^{d_1}-1) \cdots (t^{d_r}-1) \in \mathbb{F}_2[t]. \end{equation*}

On the other hand, by lemma 5.2, there is a Galois-equivariant exact sequence

\begin{equation*} 0 \to \mathbb{F}_2 \to V \to J[2] \to 0, \end{equation*}

where the first map is given by $1\to P_1+P_2+\cdots P_{2g+1}$ and the action of $\operatorname{Frob}$ on the sum $P_1+\cdots+P_{2g+1}$ is trivial. This implies that

\begin{equation*} \det(t \operatorname{Id}-\rho(\operatorname{Frob}))=\det(t\operatorname{Id}-\rho_2(\operatorname{Frob})) (t-1), \end{equation*}

which, combined with our previous determination of the characteristic polynomial of $\rho(\operatorname{Frob})$, concludes the proof.

Proof. This is a direct application of lemma 5.3, combined with the fact that by corollary 5.1 we have $P_{C}(t) \equiv f_{C}(t)\ (\mathrm{mod}\ 2)$.

Proof of theorem 1.4 for p odd

The inequality $\dim_{\mathbb{Q}} L_g(\mathbb{F}_q) \leq g+1$ follows immediately from the symmetry relation $a_{g+i}=q^ia_{g-i}$ satisfied by the coefficients of the L-polynomials; it thus suffices to establish the lower bound $\dim_{\mathbb{Q}} L_g(\mathbb{F}_q) \geq g+1$. Consider the g + 1 curves $C_0,\ldots,C_g$ of corollary 5.4 (any choice of the irreducible polynomials $f_d(x)$ will work) and the corresponding L-polynomials $P_{C_0}(t),\ldots,P_{C_g}(t)$. Let $M \subseteq \mathbb{Z}[t]$ be the $\mathbb{Z}$-module generated by these polynomials; it is clear that in order to prove the theorem it suffices to show that $\operatorname{rank}_{\mathbb{Z}} M \geq g+1$. Notice that $M \otimes \mathbb{F}_2$ is in a natural way a vector subspace of $\mathbb{F}_2[t]$, and that

\begin{equation*} \operatorname{rank}_{\mathbb{Z}} M \geq \dim_{\mathbb{F}_2} (M \otimes \mathbb{F}_2). \end{equation*}

Let $N \subset \mathbb{F}_2[t]$ be the image of the linear map

\begin{equation*} \begin{array}{ccc} M \otimes \mathbb{F}_2 & \to & \mathbb{F}_2[t] \\ q(t) & \mapsto & (t-1)q(t). \end{array} \end{equation*}

The $\mathbb{F}_2$-vector space N is generated by the g + 1 polynomials $(t-1)P_{C_i}(t)$ for $i=0,\ldots,g$, hence, by corollary 5.4, by the g + 1 polynomials

\begin{equation*} t^{2g+1} +1 \quad\text{and}\quad t^{2g+1}+t^{2g+1-i}+t^i+1 \text{for }i=1,\ldots,g. \end{equation*}

It is immediate to check that these g + 1 polynomials are $\mathbb{F}_2$-linearly independent, which implies

\begin{equation*} \operatorname{rank}_{\mathbb{Z}} M \geq \dim_{\mathbb{F}_2} (M \otimes \mathbb{F}_2) = \dim_{\mathbb{F}_2} N=g+1. \end{equation*}

Proof of theorem 1.4 for p = 2

Fix $0\leq r\leq g$. Let $h(x) \in \mathbb{F}_q[x]$ be a separable polynomial of degree r such that $h(0)\neq 0$. Such a polynomial exists: for $r=0, 1$ we may take $h(x)=1$ or $h(x)=x+1$, respectively, and for $r \geq 2$ it suffices to take as h(x) the minimal polynomial of any element that generates $\mathbb{F}_{q^r}$ over $\mathbb{F}_q$.

Consider the affine curve defined by the equation $y^2+yh(x)=x^{2g+1-r}h(x)$. We claim that this curve is smooth. Indeed, an $\overline{\mathbb{F}_q}$-point $(x_0,y_0)$ on the curve is singular if and only if

\begin{equation*} \begin{cases} y_0^2+y_0h(x_0)=x_0^{2g+1-r}h(x_0)\\ h(x_0)=0\\ y_0h'(x_0)=(2g+1-r)x_0^{2g-r}h(x_0)+x_0^{2g+1-r}h'(x_0) \end{cases}. \end{equation*}

Here the second and third equations are given by the vanishing of the partial derivatives in y and x of the defining equation, respectively. By the second equation, x ₀ is a root of h. So, by the first one, $y_0=0$. Hence, the third equation becomes $x_0^{2g+1-r}h'(x_0)=0$: but $x_0\neq 0$ since $h(0)\neq 0$, and $h'(x_0)\neq 0$ since h is separable, so the above system has no solutions. Let $C/\mathbb{F}_q$ be the smooth projective curve given by the completion of the curve above. The curve C has genus g, because the degree of $x^{2g+1-r}h(x)$ is $2g+1$ and the degree of h(x) is at most g. In particular, $P_C(t)$ is an element of $\mathcal{P}_g(\mathbb{F}_q)$. We will show that the reduction of $P_C(t)$ modulo 2 has degree r.

Let $\ell$ be an odd prime and let $T_\ell J$ be the $\ell$-adic Tate module of the Jacobian J of C. Let $f_{C, \ell^\infty}(t):=\det(t\operatorname{Id}-\rho_{\ell^\infty}(\operatorname{Frob})\mid T_\ell J)$. If $\alpha\in \overline{\mathbb{F}_q}$ is a root of $f_{C, \ell^\infty}(t)$ with multiplicity d, then $q/\alpha$ is a root of $f_{C, \ell^\infty}(t)$ with multiplicity d. Hence, we can write $f_{C, \ell^\infty}(t)=t^gQ_C(t+q/t)$ with $Q_C(t)\in \mathbb{Z}[t]$ of degree g. Let r ₂ be the 2-rank of J, as defined in [Reference González24, Section 1]. By [Reference González24, proposition 3.1], r ₂ is equal to the sum of the multiplicities of the non-zero roots of $Q_C(t)$ modulo 2. Hence,

\begin{equation*} Q_C(t)\equiv t^{g-r_2}\tilde{Q}_C(t)(\mathrm{mod}\ 2) \end{equation*}

with $\tilde{Q}_C(t)\in \mathbb{F}_2[t]$ a polynomial of degree r ₂ such that $\tilde{Q}_C(0)\neq 0$ (in $\mathbb{F}_2$). In [Reference Castryck, Streng and Testa14, proof of theorem 23], the authors show that the 2-rank of J is equal to one less than the number of distinct projective points where $H_1(X, Z) := h(X/Z)Z^{g+1}$ vanishes (see also [Reference Elkin and Pries21]). In our case, since h(x) is separable, this implies $r_2 = \deg h(x) = r$. Hence, we have

\begin{equation*} Q_C(t)\equiv t^{g-r}\tilde{Q}_C(t)(\mathrm{mod}\ 2) \end{equation*}

with $\tilde{Q}_C(t)$ of degree r. As q is a power of 2, we obtain

\begin{equation*} f_{C, \ell^\infty}(t)\equiv t^gQ_C\left(t+\frac qt\right)\equiv t^gQ_C\left(t\right)\equiv t^{2g-r}\tilde{Q}_C(t)(\mathrm{mod}\ 2). \end{equation*}

By theorem 1.7,

(24)\begin{equation} P_C(t)\equiv t^{2g}f_{C, \ell^\infty}\left(t^{-1}\right)\equiv t^{2g}t^{-2g+r}\tilde{Q}_C\left(t^{-1}\right)\equiv t^r \tilde{Q}_C\left(t^{-1}\right)(\mathrm{mod}\ 2). \end{equation}

Since $\tilde{Q}_C(0)\not\equiv 0(\mathrm{mod}\ 2)$, we see that the reduction of $P_C(t)$ modulo 2 has degree r. So, for each $0\leq r\leq g$, we can find a smooth hyperelliptic curve C_r of genus g such that $P_{C_r}(t)$ modulo 2 has degree r. Therefore, the polynomials $\{P_{C_r}(t)\mid 0\leq r\leq g\}$ are linearly independent modulo 2. The result follows as in the proof of theorem 1.4.