A proof of the Hasse-Weil inequality for genus 2 \`a la Manin

TL;DR

This paper proves the Hasse-Weil inequality for genus 2 curves of a specific form, employing elementary methods inspired by Manin's 1956 proof for genus 1, advancing understanding of algebraic curves.

Contribution

It provides a new elementary proof of the Hasse-Weil inequality for genus 2 curves, extending Manin's approach from genus 1 to genus 2.

Findings

Established the Hasse-Weil inequality for genus 2 curves of the form y^2=f(x)

Demonstrated the effectiveness of elementary methods in higher genus cases

Extended classical proof techniques to more complex algebraic curves

Abstract

We prove the Hasse-Weil inequality for genus 2 curves given by an equation of the form y^2 = f(x) with f a polynomial of degree 5, using arguments that mimic the elementary proof of the genus 1 case obtained by Yu. I. Manin in 1956.

Equations

$$|\#C({\mathbb{F}}_q)-(q+1)|\le 2g\sqrt{q}.$$

$$0\leq d_{n}=\left\{\begin{array}[]{lr}\deg(u_{1,n})&(\mbox{if}\;\;\psi_{n}\neq 0)\\ 0&(\mbox{if}\;\;\psi_{n}=0)\end{array}\right.=n^{2}+(q+1-\#E(\mathbb{F}_{q}))n+q.$$

$$\kappa :\mathcal{J}\setminus \Theta$$

$$D$$

$$\mathcal{J}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathrm{deg}({\psi }_n)=[{\mathbb{F}}_q(x,y):{\mathbb{F}}_q(\frac{{\mu }_{1,n}(x)}{{\mu }_{2,n}(x)})]=[{\mathbb{F}}_q(x,y):{\mathbb{F}}_q(x)]\cdot [{\mathbb{F}}_q(x):{\mathbb{F}}_q(\frac{{\mu }_{1,n}(x)}{{\mu }_{2,n}(x)})]$$

$$\#\mathcal{J}({\mathbb{F}}_q)=\#\text{Ker}(\varphi -[1])=\#\text{Ker}(-[n+1])=(n+1{)}^4.$$

$$\#\mathcal{J}({\mathbb{F}}_q)=\frac{{\left(\#\mathcal{H}({\mathbb{F}}_q)\right)}^2+\#\mathcal{H}({\mathbb{F}}_{q^2})}{2}-q.$$

$$\#{\mathcal{J}}^{\text{TW}}({\mathbb{F}}_q)=\#\text{Ker}(\varphi +[1])=\#\text{Ker}(-[n]+1)=(n-1{)}^4.$$

$$\begin{array}{rl}\#{\mathcal{J}}^{\text{TW}}({\mathbb{F}}_q)& =\frac{\#{\mathcal{H}}^{\text{TW}}({\mathbb{F}}_q{)}^2+\#\mathcal{H}({\mathbb{F}}_{q^2})}{2}-q\\ & =\frac{(2n^2+2-\#\mathcal{H}({\mathbb{F}}_q){)}^2+\#\mathcal{H}({\mathbb{F}}_{q^2})}{2}-q=(n-1{)}^4\end{array}$$

$$\displaystyle\begin{split}\#\mathcal{H}(\mathbb{F}_{q})^{2}-(2n^{2}+2-\#\mathcal{H}(\mathbb{F}_{q}))^{2}&=2\big{(}(n+1)^{4}-(n-1)^{4}\big{)}\\ &=16n(n^{2}+1),\end{split}$$

$$-{\Psi }_n(x,y)$$

$$=[\infty -(x^q,y^q)]+n[\infty -(x,y)]$$

$$=[(x^q,-y^q)-\infty ]+n[(x,-y)-\infty ]={\Psi }_n(x,-y).$$

$${\psi }_{-1}(x,y)=\frac{x^{3q+2}+x^{2q+3}+2a_4{x}^{2q+2}+a_3(x^{2q+1}+x^{q+2})+2a_2{x}^{q+1}+a_1(x^q+x)+2a_0+2f(x{)}^{\frac{q+1}{2}}}{(x^q-x{)}^2}.$$

$$\nu (\alpha )=4f(\alpha ).$$

$$2({\alpha }^5+a_4{\alpha }^4+a_3{\alpha }^3+a_2{\alpha }^2+a_1\alpha +a_0)-2f(\alpha )=0.$$

$$\frac{d}{dx}\nu (x){\mid }_{\alpha }$$

$$=5{\alpha }^4+4a_4{\alpha }^3+3a_3{\alpha }^2+2a_2\alpha +a_1-f^{\prime }(\alpha )$$

$$=f^{\prime }(\alpha )-f^{\prime }(\alpha ).$$

$$=0.$$

$${\delta }_1=3(q+1)-\#\mathcal{H}({\mathbb{F}}_q).$$

$${\kappa }_4({\Psi }_1(x,y))=\frac{x^{3q+2}+x^{2q+3}+2a_4{x}^{2q+2}+a_3(x^{2q+1}+x^{q+2})+2a_2{x}^{q+1}+a_1(x^q+x)+2a_0-2f(x{)}^{\frac{q+1}{2}}}{(x^q-x{)}^2}.$$

$$2\Theta \bullet {\Theta }_n=\mathrm{deg}{\psi }_n=2{\Phi }_n^{\ast }\Theta \bullet \Theta .$$

$${\kappa }_4({\iota }_w(x,y))=\frac{2a_0+a_1(x+w)+2a_{2}xw+a_3(x+w)xw+2a_4(xw{)}^2+(x+w)(xw{)}^2}{(x-w{)}^2}$$

$${\kappa }_4({\Psi }_{0,w}(x,y))=\frac{2a_0+a_1(x^q+w)+2a_2{x}^{q}w+a_3(x^q+w)x^{q}w+2a_4(x^{q}w{)}^2+(x^q+w)(x^{q}w{)}^2}{(x^q-w{)}^2}$$

$$4=2\Theta \bullet \Theta =\left(\sum C_i\right)\bullet \Theta =\sum (C_i\bullet \Theta ).$$

$${\Phi }_{n,w}^{\ast }\Theta \bullet \Theta =\mathrm{deg}{\kappa }_4({\Psi }_{n,w}(x,y))=\max \{\mathrm{deg}{\mu }_{1,n}^w,\mathrm{deg}{\mu }_{2,n}^w\},$$

$$\delta_{n}:=\Bigg{\{}\begin{array}[]{lr}0&\text{if }\Psi_{n}:\mathcal{H}\to\mathcal{J}\text{ is constant};\\ \max\{\deg(\mu_{1,n}^{w}),\deg(\mu_{2,n}^{w})\}&\text{if }\Theta_{n}\in\text{Supp div}(\kappa_{4});\\ \max\{\deg(\mu_{1,n}),\deg(\mu_{2,n})\}&\text{otherwise.}\end{array}$$

$$(\alpha +\beta +\gamma {)}^{\ast }D-(\alpha +\beta {)}^{\ast }D-(\alpha +\gamma {)}^{\ast }D-(\beta +\gamma {)}^{\ast }D+{\alpha }^{\ast }D+{\beta }^{\ast }D+{\gamma }^{\ast }D\sim 0$$

$${\delta }_{n-1}+{\delta }_{n+1}=2{\delta }_n+4.$$

$$2{\Phi }_n^{\ast }\Theta -{\Phi }_{n+1}^{\ast }\Theta -{\Phi }_{n-1}^{\ast }\Theta +2\Theta \sim 0,$$

$$2{\Phi }_n^{\ast }\Theta +2\Theta \sim {\Phi }_{n-1}^{\ast }\Theta +{\Phi }_{n+1}^{\ast }\Theta .$$

Full text

††journal: : To be chosen

1 Manin’s proof of the Hasse inequality for genus one

Recall the following theorem:

Theorem 1.1** (Hasse-Weil).**

Let $C$ be an algebraic curve of genus $g$ over $\mathbb{F}_{q}$ then

[TABLE]

The Hasse-Weil inequality for an elliptic curve $E/\mathbb{F}_{q}$ (so the case of genus one, due to Hasse in a series of papers [7], [8], [9]) was obtained in an elementary way by Manin for $\text{char}\;\mathbb{F}_{q}\neq 2,3$ in [11] (see [12] for an English translation). Adjustments of these elementary arguments to the remaining genus one cases are presented in [3] and [16]. Our goal is to extend these ideas to the case of genus $2$ curves. To facilitate this, we very briefly summarize in this first section Manin’s argument in the genus one situation. Throughout the paper, we restrict to finite fields of odd characteristic.

Let $E/\mathbb{F}_{q}$ be an elliptic curve given by an equation $y^{2}=f(x)$ where $f$ is a polynomial of degree $3$. Consider $\phi,[n]\in\text{\rm End}_{\mathbb{F}_{q}}(E)$ where $\phi$ is the $q$-th Frobenius and $[n]$ the multiplication by $n$. Consider $\psi_{n}:=\phi+[n]\in\text{\rm End}_{\mathbb{F}_{q}}(E)$. If $\psi_{n}$ is non-trivial then it is of the form (x,y)\mapsto\big{(}\frac{u_{1,n}(x)}{u_{2,n}(x)},y\frac{v_{1}(x)}{v_{2}(x)}\big{)}, with $u_{1,n},u_{2,n},v_{1},v_{2}\in\mathbb{F}_{q}[x]$ such that $\text{gcd}(u_{1,n},u_{2,n})=1$ (see [17, §2.9]). The Hasse-Weil inequality for $E$ follows from the claim that $d_{n}:=\deg(\psi_{n})$ satisfies

[TABLE]

Here for $\psi_{n}\neq 0$ by definition $\deg(\psi_{n})=[\mathbb{F}_{q}(E):\psi_{n}^{*}\mathbb{F}_{q}(E)]$, and $\deg(u_{1,n})$ is the degree of the polynomial $u_{1,n}\in\mathbb{F}_{q}[x]$ (and $\deg(0):=0$).

The leftmost equality in (1) for $\psi_{n}\neq 0$ follows from the elementary observation $\deg(\psi_{n})=[\mathbb{F}_{q}(x):\mathbb{F}_{q}(\frac{u_{1},n(x)}{u_{2},n(x)})]=\max\{\deg(u_{1,n}(x)),\deg(u_{2,n}(x))\}$ (see [17, §2.9]) or [15, Lemma 6.2]), together with $\deg(u_{1,n}(x))>\deg(u_{2,n}(x))$. The latter follows from $\psi_{n}(\infty)=\infty$, implying $v_{\infty}(\tfrac{u_{1,n}(x)}{u_{2,n}(x)})<0$. The rightmost equality in (1) is shown by induction on $n$ using the basic identity $d_{n-1}+d_{n+1}=2d_{n}+2$ (see [2, Lemma 8.5]).

Finally the non-negativity of $d_{n}=n^{2}+(q+1-\#E(\mathbb{F}_{q}))n+q$ yields that the discriminant of this quadratic polynomial in $n$ is non-positive, implying the Hasse inequality.

In order to extend these ideas to genus $2$ curves, we introduce an analogous $\delta_{n}$ which also satisfies a basic identity, namely $\delta_{n-1}+\delta_{n+1}=2\delta_{n}+4$, in the genus $2$ case.

2 An analogous $\delta_{n}$ for genus $2$

Let $k:=\mathbb{F}_{q}$ be a finite field of odd cardinality $q$, and let $\mathcal{H}$ be a hyperelliptic curve of genus $2$ over $k$. Throughout, we assume that $\mathcal{H}$ is given by the equation $Y^{2}=X^{5}+a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}$. By $\mathcal{J}$ we denote the Jacobian variety associated to $\mathcal{H}$. The points of $\mathcal{J}$ correspond to divisor classes $[D]\in\text{Pic}^{0}(\mathcal{H})$.

Fix the point $\infty\in\mathcal{H}$ and consider the Abel-Jacobi map $\iota\in\text{Mor}_{k}(\mathcal{H},\mathcal{J})$ given by $P\mapsto[P-\infty]$. We have that $\Theta:=\text{Im}\;\iota$ is the theta divisor of $\mathcal{J}$ and $\Theta\cong\mathcal{H}$. Consider the $q$-th power Frobenius map $\phi\in\text{End}_{k}(\mathcal{J})$ and the morphism $\Phi:=\phi\circ\iota\in\text{Mor}_{k}(\mathcal{H},\mathcal{J})$. Since $\text{Mor}_{k}(\mathcal{H},\mathcal{J})\cong\mathcal{J}(k(\mathcal{H}))$ is an Abelian group, we define $\Psi_{n}:=\Phi+n\cdot\iota\in\text{Mor}_{k}(\mathcal{H},\mathcal{J})$ and $\Theta_{n}:=\text{Im}\;\Psi_{n}\subset\mathcal{J}$.

Similar to $d_{n}=\deg(\psi_{n})$ in (1), we will define the ‘complexity’ (height) of $\Psi_{n}$ and denote this by $\delta_{n}$.

Consider the generic point of $\mathcal{J}$ given by $\mathfrak{g}:=[(x_{1},y_{1})+(x_{2},y_{2})-2\infty]$ and the function field of $\mathcal{J}$ denoted by $k(\mathcal{J})\cong k(x_{1}+x_{2},x_{1}x_{2},\tfrac{y_{1}-y_{2}}{x_{1}-x_{2}},\tfrac{x_{2}y_{1}-x_{1}y_{2}}{x_{1}-x_{2}})$. The Riemann-Roch space $\mathcal{L}(2\Theta)\subset k(\mathcal{J})$ has dimension $4$; a basis is given by $\{\kappa_{1},\kappa_{2},\kappa_{3},\kappa_{4}\}$ with $\kappa_{1}:=1,\kappa_{2}:=x_{1}+x_{2},\kappa_{3}:=x_{1}x_{2},\kappa_{4}:=\tfrac{F_{0}(x_{1}+x_{2},x_{1}x_{2})-2y_{1}y_{2}}{(x_{1}-x_{2})^{2}}$ where $F_{0}(A,B):=2a_{0}+a_{1}A+2a_{2}B+a_{3}AB+2a_{4}B^{2}+AB^{2}$ (see [1, Chapter 2] and [4, Page 5]). This basis is used to define a singular surface $\mathcal{K}\!\subset\!\mathbb{P}^{3}$ birational to the Kummer surface associated to $\mathcal{J}$, as the closure of the image of

[TABLE]

As a remark, let $[-1]\in\text{Aut}(\mathcal{J})$ be the involution on $\mathcal{J}$ given by $[-1]\mathfrak{g}=[(x_{1},-y_{1})+(x_{2},-y_{2})-2\infty]=-\mathfrak{g}$. Note that $[-1]^{*}:k(\mathcal{J})\to k(\mathcal{J})$ is the trivial map on $\mathcal{L}(2\Theta)\subset k(\mathcal{J})$. Particularly for all $D\in\mathcal{J}\setminus\Theta$ we have that $\kappa_{i}(D)=\kappa_{i}(-D)$.

Using the previous remark, suppose that $\Theta_{n}:=\text{Im}\;\Psi_{n}\not\subset\Theta$ and $\Theta_{n}$ is not a zero of $\kappa_{4}$. Let $(x,y)\in\mathcal{H}$ be generic, then we have that $\psi_{n}(x,y):=(\kappa_{4}\circ\Psi_{n})(x,y)=(\kappa_{4}\circ\Psi_{n})(x,-y)$. Hence $\psi_{n}(x,y)\in k(x)$, that is, $\psi_{n}(x,y)=:\tfrac{\mu_{1,n}(x)}{\mu_{2,n}(x)}$ is a rational function in one variable $x$, which is the one that we will use to define $\delta_{n}$.

Our geometric situation is described in the following diagram:

[TABLE]

Here $\pi$ is a projection and $\kappa_{4}=\pi\circ\kappa$. Since $\Theta_{n}$ is not a zero or a pole of $\kappa_{4}$, we define $\delta_{n}:=\deg\mu_{1,n}$.

There are two situations left to define $\delta_{n}$ for every $n\in\mathbb{Z}$. The first is when $\Psi_{n}$ is constant (hence equal to the zero map). In this case $\Theta_{n}\subset\mathcal{J}$ is a point and we define $\delta_{n}:=0$. The second is when $\{0\}\neq\Theta_{n}$ is a curve which is a zero or a pole of $\kappa_{4}$, that is $\Theta_{n}\in\text{Supp div}(\kappa_{4})$. In the following section (Formula (11)) we will define $\delta_{n}$ for this special situation. We show that if $\Psi_{n}$ is non-constant but $\kappa_{4}(\Psi_{n}(x,y))=c$ is constant then $c$ can only be [math] or $\infty$, that is, the curve $\Theta_{n}=\text{Im}\Psi_{n}$ is a zero or a pole of $\kappa_{4}$ respectively (see Lemma 3.12).

Further, if $\Theta_{n}$ is not a zero or a pole of $\kappa_{4}$, we show in the next section that $\tfrac{\deg\psi_{n}}{2}=\max\{\deg\mu_{1,n},\deg\mu_{2,n}\}$. In the case $\Theta_{n}$ is a zero or a pole of $\kappa_{4}$, a similar equality will be shown taking a translation of $\Theta_{n}$ by a $2$-torsion point of $\mathcal{J}$ in order to avoid the pole and zero divisor of $\mathcal{J}$.

Finally, we show the basic identity $\delta_{n-1}+\delta_{n+1}=2\delta_{n}+4$. The same strategy also employed by Manin for genus one will then lead to a proof of the Hasse-Weil inequality in this case.

3 Proof of the Hasse-Weil inequality for genus $2$

We use the notations $\mathcal{H}/\mathbb{F}_{q}$, $\Theta,\Theta_{n}$ etc. introduced in the previous section.

Lemma 3.1**.**

*Let $(x,y)\in\mathcal{H}/\mathbb{F}_{q}$ be generic. Suppose that $\Theta_{n}\subset\mathcal{J}$ is a curve that is not a zero nor a pole of $\kappa_{4}$. Then $\psi_{n}(x,y)=\tfrac{\mu_{1,n}(x)}{\mu_{2,n}(x)}$ for some coprime polynomials $\mu_{1,n},\mu_{2,n}$.

Moreover $\tfrac{\deg\psi_{n}}{2}=\max\{\deg\mu_{1,n}(x),\deg\mu_{2,n}(x)\}$.*

Proof.

Since $\Theta_{n}$ is not a zero or a pole of $\kappa_{4}$ and $\kappa_{4}\in\mathcal{L}(2\Theta)$, the function $\psi_{n}(x,y)=\kappa_{4}(\Psi_{n}(x,y))$ is defined and non-zero. In the previous section we saw that $\psi_{n}(x,y)=\psi_{n}(x,-y)$, and hence $\psi_{n}(x,y)\in\mathbb{F}_{q}(x)$. This shows the existence of the coprime $\mu_{1,n},\mu_{2,n}\in\mathbb{F}_{q}[x]$.

Now,

[TABLE]

and the lemma follows. ∎

It can happen that $\Psi_{n}$ is the zero map, which implies $\Theta_{n}=\text{Im}\Psi_{n}$ is a point. For example consider the hyperelliptic curve $Y^{2}=X^{5}+5X$ over $\mathbb{F}_{49}$. An explicit MAGMA computation shows that $\psi_{7}:=\Phi+7\iota\in\text{\rm Mor}_{\mathbb{F}_{49}}(\mathcal{H},\mathcal{J})\cong\mathcal{J}(\mathbb{F}_{49}(\mathcal{H}))$ is the zero map:

 p := 7; q := p^2; F := FiniteField(q);> P<x> := PolynomialRing(F);> H := HyperellipticCurve(x^5+5x);> FH<X,Y> := FunctionField(H); HE := BaseExtend(H,FH);> JE := Jacobian(HE); M<t> := PolynomialRing(FH);> Phi := JE![t-X^q, Y^q];> GPt := JE![t-X,   Y];> -7GPt;(x + 6X^49, (X^120 + X^116 + 5X^112 + 6X^108 + X^92 + X^88 + 5X^84 + 6X^80 + 5X^64 +5X^60 + 4X^56 + 2X^52 + 6X^36 + 6X^32 + 2X^28 + X^24)Y, 1)> Phi;(x + 6X^49, (X^120 + X^116 + 5X^112 + 6X^108 + X^92 + X^88 + 5X^84 + 6X^80 + 5X^64 +5X^60 + 4X^56 + 2X^52 + 6X^36 + 6X^32 + 2*X^28 + X^24)Y, 1)> Phi+7GPt;(1, 0, 0)In this example $\mathcal{J}$ is isogenous to the square of a supersingular elliptic curve and the ground field has $p^{2}=49$ elements. The characteristic polynomial of Frobenius $\phi\in\text{\rm End}_{\mathbb{F}_{49}}(\mathcal{J})$ is given by $\chi_{\phi}(X):=({X}+7)^{4}$ which is the main reason of this behavior.

A general construction of curves having Jacobian isogenous to a square of a supersingular elliptic curve was achieved by Moret-Bailly in [14].

The following proposition and lemma isolates a special case for our final proof of the Hasse-Weil inequality for genus $2$. Note that the example discussed above illustrates this special case.

Proposition 3.2**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$, given by an equation $y^{2}=f(x)$ with $f$ of degree $5$. Let $\mathcal{J}$ be the Jacobian of $\mathcal{H}$ and $\iota\colon\mathcal{H}\to\mathcal{J}$ the map $P\mapsto[P-\infty]$. Suppose that there is an $n\in\mathbb{Z}$ such that $\Psi_{n}=(\phi+[n])\circ\iota\in\text{\rm Mor}_{\mathbb{F}_{q}}(\mathcal{H},\mathcal{J})$ is constant. Then $q$ is a perfect square and $\#\mathcal{H}(\mathbb{F}_{q})=q+1+4n=q+1\pm 4\sqrt{q}$.

Proof.

First, we show that if $\Psi_{n}=(\phi+[n])\circ\iota$ is constant, then $\phi=-[n]$.

We have that $\Psi_{n}=(\phi+[n])\circ\iota$ is constant and $0\in\text{Im}\Psi_{n}$, hence $\Psi_{n}=0$; this is equivalent to $(\phi+[n])(\Theta)=0$ since $\Theta=\iota(\mathcal{H})$. Moreover, $\Theta$ generates $\mathcal{J}$, that is $\mathcal{J}=\{D_{1}+D_{2}:D_{1},D_{2}\in\Theta\}$. So if any $\varphi\in\mbox{End}(\mathcal{J})$ is zero on $\Theta$ then it is the zero map. Hence $\phi=-[n]\in\text{End}(\mathcal{J})$.

Note that $\phi=-[n]$ implies $q^{2}=\deg(\phi)=\deg([-n])=n^{4}$. Hence $q=n^{2}$ is a perfect square and $n=\pm\sqrt{q}$.

Now we proceed to count $\#\mathcal{H}(\mathbb{F}_{q})$. Using that $\phi=-[n]$ we have that:

[TABLE]

(Here we used that $n+1$ is not a multiple of $\mbox{char}(\mathbb{F}_{q})$). Moreover, an easy counting argument (see [1, Chapter 8, §2]) shows:

[TABLE]

Consider the quadratic twist of $\mathcal{H}$ denoted by $\mathcal{H}^{\texttt{\tiny{TW}}}$ and its Jacobian $\mathcal{J}^{\texttt{\tiny{TW}}}$, then:

[TABLE]

Similarly as in (4) and using that $\#\mathcal{H}(\mathbb{F}_{q})+\#\mathcal{H}^{\texttt{\tiny{TW}}}(\mathbb{F}_{q})=2q+2=2n^{2}+2$ and $\mathcal{H}^{\texttt{\tiny{TW}}}(\mathbb{F}_{q^{2}})\cong\mathcal{H}(\mathbb{F}_{q^{2}})$, we have that:

[TABLE]

Subtracting (6) from (4) yields:

[TABLE]

which can be rewritten as $\#\mathcal{H}(\mathbb{F}_{q})=n^{2}+4n+1=q+1\pm 4\sqrt{q}$. ∎

Corollary 3.3**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$ given by an equation $Y^{2}=f(X)$ with $f$ of degree $5$. Let $\mathcal{J}$ be its Jacobian and suppose that $\Psi_{n}=(\phi+[n])\circ\iota$ is constant. Then $\text{\rm Im}\;\Psi_{n-1}=\text{\rm Im}\;\Psi_{n+1}=\Theta\in\text{\rm Div}(\mathcal{J})$.

Proof.

By the Proposition (3.2) $\phi=-[n]$, hence $\Psi_{n\pm 1}=\pm[1]\circ\iota=\pm\iota\in\text{Mor}_{\mathbb{F}_{q}}(\mathcal{H},\mathcal{J})$. Further, $\text{Im}\;\iota=\text{Im}-\!\iota=\Theta$ since $\Theta\cong\mathcal{H}$ (here we used that $\Theta$ is symmetric with respect of the hyperelliptic involution under $\iota$). ∎

Lemma 3.4**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$ given by $Y^{2}=f(X)$ with $\deg f(X)=5$ and let $\mathcal{J}$ be the Jacobian of $\mathcal{H}$. Let $(x,y)\in\mathcal{H}$ be generic, then $-\Psi_{n}(x,y)=\Psi_{n}(x,-y)$.

Proof.

Using $[(a,b)-\infty]=[\infty-(a,-b)]$ for any $(a,b)\in\mathcal{H}$ one finds

[TABLE]

∎

Now we calculate some values of $\delta_{n}$.

Proposition 3.5**.**

Let $\mathcal{H}/\mathbb{F}_{q}:\;Y^{2}=X^{5}+a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=:f(X)$ be a genus $2$ curve ($q$ odd). Then $\delta_{-1}=\#\mathcal{H}(\mathbb{F}_{q})+q+1$.

Proof.

For $(x,y)\in\mathcal{H}$ generic, $\Psi_{-1}(x,y)=[(x^{q},yf(x)^{\frac{q-1}{2}})+(x,-y)-2\infty]$. Lemma 3.1 shows that $\delta_{-1}$ equals the degree of $\psi_{-1}(x,y)=\kappa_{4}(\Psi_{-1}(x,y))\in\mathbb{F}_{q}(x)$. This degree is the maximum of the polynomial degrees of the numerator and the denominator, assuming these are coprime. Here

[TABLE]

Let $\nu(x)$ and $\eta(x)$ be respectively the numerator and denominator of $(\ref{k4fi})$ before cancellation of common factors. Note that $\deg(\eta)=2q$ and that every $\alpha\in\mathbb{F}_{q}$ is a double root of $\eta(x)$. Furthermore, $\deg(\nu(x))=3q+2>2q$, hence $\delta_{-1}=3q+2-\deg(\gcd(\nu(x),\eta(x)))$.

Since $\psi_{-1}(x,y)\in\mathbb{F}_{q}(\mathcal{H})$, the common factors $(x-\alpha)$ of $\nu$ and $\eta$ occur at the points $(\alpha,\beta)\in\mathcal{H}$ such that $\alpha\in\mathbb{F}_{q}$ and $\beta\in\mathbb{F}_{q}^{*}$ or $\beta\in\mathbb{F}_{q^{2}}^{*}\setminus\mathbb{F}_{q}^{*}$ or $\beta=0$. Hence, we have three possibilities for cancellations:

**Case $\beta\in\mathbb{F}_{q}^{*}$:

**In this case $(\alpha,\beta)\in\mathcal{H}(\mathbb{F}_{q})$ and therefore $f(\alpha)$ is a square in $\mathbb{F}_{q}^{*}$. Hence $f(\alpha)^{\frac{q-1}{2}}=1$. Moreover, $\alpha^{q}=\alpha$. Using this, the last term of $\nu(\alpha)$ is $2f(\alpha)^{\frac{q+1}{2}}=2f(\alpha)f(\alpha)^{\frac{q-1}{2}}=2f(\alpha)$ and

[TABLE]

Since $\beta\neq 0$ there is no cancellation of a factor $(x-\alpha)$ in this case.

**Case $\beta=0$:

**We have that $f(\alpha)=0$ and $\alpha^{q}=\alpha$, so the numerator of (8) is $2f(\alpha)=0$. Therefore $\nu(x)$ and $\eta(x)$ share the linear factor $x-\alpha$ with multiplicity one or two. The multiplicity in fact equals one since $\frac{d}{dx}\nu(x)\mid_{\alpha}=4f^{\prime}(\alpha)\neq 0$ as $f(x)$ does not have repeated zeros.

**Case $\beta\not\in\mathbb{F}_{q}$:

**In this case $f(\alpha)$ is nonzero and is not a square in $\mathbb{F}_{q}^{*}$. Therefore $f(\alpha)^{\frac{q-1}{2}}=-1$ by Euler’s criterion. Moreover $\alpha^{q}=\alpha$ and $\nu(\alpha)$ is in this case

[TABLE]

To find the multiplicity of $\alpha$ as a zero of $\nu(x)$, note that

[TABLE]

This tells us that the factor $(x-\alpha)^{2}$ appears in $\nu$ and then it cancels with the denominator.

Combining the cases, one concludes $\deg(\gcd(\nu(x),\eta(x)))=2q+1-\#\mathcal{H}(\mathbb{F}_{q})$ and therefore $\deg(\kappa_{4}(\Psi_{-1}(x,y)))=q+1+\#\mathcal{H}(\mathbb{F}_{q})$. ∎

Proposition 3.6**.**

With notations as in Proposition 3.5 one has

[TABLE]

Proof.

Note that

[TABLE]

This expression differs from (8) only at the sign of the last term of the numerator, namely $2f(x)^{\frac{q+1}{2}}$. An analogous argument as the one given in Proposition 3.5 proves the proposition. ∎

We will use the following definition in order to interpret $\delta_{n}$ in the case where $\Theta_{n}$ is a curve that is a zero or a pole of $\kappa_{4}$. The case $\Theta_{n}=\Theta$ (which is a pole of $\kappa_{4}$) occurs, e.g., for $n=0$.

Definition 3.7**.**

Let $D_{1},D_{2}\in\text{\rm Div}(\mathcal{J})$. By $D_{1}\bullet D_{2}$ we denote the intersection number of the divisors $D_{1}$ and $D_{2}$ on the surface $\mathcal{J}$.

For details and properties of this see, e.g., [6, Appendix C or Chapter V].

Lemma 3.8**.**

Suppose that $\text{\rm Im}\Psi_{n}=\Theta_{n}\not\subset\Theta$ and that $\psi_{n}=\kappa_{4}\circ\Phi_{n}\circ\iota\colon\mathcal{H}\to\mathbb{P}^{1}$ is nonconstant, where $\Phi_{n}:=\phi+[n]\in\text{\rm End}(\mathcal{J})$. Then

[TABLE]

Proof.

Let $(x,y)\in\mathcal{H}$ be generic. Since $\text{Im}\Psi_{n}=\Theta_{n}\not\subset\Theta$ and since $\kappa_{4}\in\mathbb{F}_{q}(\mathcal{J})$ has divisor $D-2\Theta$ for some effective divisor $D\in\text{Div}(\mathcal{J})$, we have that $\psi_{n}(x,y)=\kappa_{4}(\Psi_{n}(x,y))\in\mathbb{F}_{q}(x)$ by Lemma 3.1. Moreover by assumption this rational function is nonconstant. Therefore $\deg\psi_{n}=\deg\left(({\kappa_{4}}_{\mid{\Theta_{n}}})^{*}\infty\right)=2\Theta\bullet\Theta_{n}$ which shows the first equality.

For the second, note that $\Phi_{n}^{-1}(\Theta)=\{D\in\mathcal{J}:\Phi_{n}(D)\in\Theta\}$ and $\Theta$ is the locus where $\kappa_{4}$ has a pole (in fact a double pole). Since $\deg(\psi_{n})=\deg({\kappa_{4}\circ\Phi_{n}}_{\mid\Theta})$ we conclude $\deg\psi_{n}=2\Phi_{n}^{-1}(\Theta)\bullet\Theta$. Applying [5, Lemma 1.7.1] this equals $2\Phi_{n}^{*}\Theta\bullet\Theta$. ∎

We will deal with the cases where $\Theta_{n}=\Theta$ by using a linear equivalent divisor $\Theta_{n}^{\prime}\in\text{Div}(\mathcal{J})$ and Lemma 3.8.

First we show that $\Theta\bullet\Theta=2$ using the pole divisor of $\kappa_{4}$. To achieve this we use (see [6, Chapter V. Theorem 1.1]) that if $\Theta\sim\Theta^{\prime}$ as divisors ($\sim$ denoting linear equivalence), then $\Theta\bullet\Theta=\Theta\bullet\Theta^{\prime}$. A suitable divisor $\Theta^{\prime}$ will be constructed as a symmetric translation of $\Theta\subset\mathcal{J}$ (with respect of $[-1]\in\text{Aut}(\mathcal{J})$).

Remark 3.9**.**

The following lemmas use that for divisors $D,D^{\prime}$ on $\mathcal{J}$ and any point $\xi\in\mathcal{J}$, denoting by $t_{\xi}$ the translation over $\xi$, we have $D\bullet D^{\prime}=D\bullet t_{\xi}^{*}D^{\prime}$. Indeed, the fact that $D^{\prime}$ and $t_{\xi}^{*}D^{\prime}$ are algebraically equivalent is a special case of [10, I §2 Proposition 2], and the fact that algebraically equivalent cycles are numerically equivalent can be found in [5, 19.3.1].

Lemma 3.10**.**

*Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$ given by $Y^{2}=X^{5}+a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=:f(X)$ and consider its Jacobian $\mathcal{J}$.

Then $\Theta=\iota(\mathcal{H})\subset\mathcal{J}$ satisfies $\Theta\bullet\Theta=2$.*

Proof.

Let $(w,0)\in\mathcal{H}(\overline{\mathbb{F}}_{q})$ be a Weierstrass point and consider $\iota_{w}\in\text{Mor}(\mathcal{H},\mathcal{J})$ given by $P\mapsto[P+(w,0)-2\infty]$. Let $(x,y)\in\mathcal{H}$ be the generic point. We have that $\Theta^{\prime}:=\text{Im}\;\iota_{w}\subset\mathcal{J}$ is a translation of $\Theta$, and therefore $\Theta^{\prime}\sim\Theta$ in $\text{Div}(\mathcal{J})$. Since $\kappa_{4}\circ[-1]=\kappa_{4}$ and $[(x,-y)+(w,0)-2\infty]=2\infty-(x,y)-(w,0)]$ it follows that $\kappa_{4}(\iota_{w}(x,y))\in\mathbb{F}_{q}(x)$. Analogous to the proof of Lemma 3.8 one obtains $\Theta\bullet\Theta=\Theta^{\prime}\bullet\Theta=\deg\kappa_{4}(\iota_{w}(x,y))$ where deg denotes the degree of the given element of $\mathbb{F}_{q}(x)$ (which is half the degree of the map $\kappa_{4}\circ\iota_{w}\colon\mathcal{H}\to\mathbb{P}^{1}$). Note that

[TABLE]

One observes that both the numerator and denominator here are divisible by $(x-w)$ and the derivative w.r.t. $x$ of the numerator, evaluated at $x=w$, equals $f^{\prime}(w)\neq 0$. Hence $\deg\kappa_{4}(\iota_{w}(x,y))=2$ which proves the lemma. ∎

Lemma 3.10 also follows using the adjunction formula ([6, Chapter V, 1.5]).

Let $\Phi_{n}:=\phi+[n]\in\text{End}(\mathcal{J})$. Using an analogous argument, we calculate $(\Phi_{0})_{*}\Theta\bullet\Theta=\Theta\bullet\Phi_{0}^{*}\Theta$; the equality of these intersection numbers is a consequence of the ‘projection formula’ [5, Proposition 2.3(c)]. In fact, in the present case equality is also established by computing both numbers directly.

Lemma 3.11**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$ given by $Y^{2}=X^{5}+a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=:f(X)$ and consider its Jacobian $\mathcal{J}$. With notations as before, we have $(\Phi_{0})_{*}\Theta\bullet\Theta=2q=\Theta\bullet\Phi_{0}^{*}\Theta$.

Proof.

As $\Phi_{0}\colon\mathcal{J}\to\mathcal{J}$ is the $q$th power Frobenius morphism, its restriction to $\Theta$ maps $\Theta$ to itself and has degree $q$. As a consequence $(\Phi_{0})_{*}\Theta=q\Theta$ and therefore Lemma 3.10 implies the first equality. For the second equality, let $(x,y)\in\mathcal{H}$ be generic. By a similar argument the one presented in Lemma 3.10 we translate $\Psi_{0}\in\text{Mor}(\mathcal{H},\mathcal{J})$ by $\iota(w,0)\in\mathcal{J}(\overline{\mathbb{F}}_{q})$, namely $\Psi_{0,w}(x,y)=[(x^{q},yf(x)^{\tfrac{q-1}{2}})+[(w,0)-2\infty]$. Then, since $\text{Im}\Psi_{0}=\Phi_{0}(\Theta)$ we define $\Phi_{0,w}:=\Phi_{0}+\iota(w,0)$ and we have that $\Theta\bullet\Phi_{0}^{*}\Theta=\Theta\bullet\Phi_{0,w}^{*}\Theta$. The latter intersection number equals the degree of

[TABLE]

(considered as a morphism $\mathbb{P}^{1}\to\mathbb{P}^{1}$). Take $v\in\overline{\mathbb{F}}_{q}$ with $v^{q}=w$, then the denominator in the right-hand-side of (10) equals $(x-v)^{2q}$. The numerator equals $(2a_{0}+a_{1}(x+v)+2a_{2}xv+a_{3}(x+v)xv+2a_{4}(xv)^{2}+(x+v)xv)^{q}$. Evaluating the numerator at $x=v$ yields $(2f(v))^{q}=0$, hence the numerator is divisible by $(x-v)^{q}$. Since the derivative of $2a_{0}+a_{1}(x+v)+2a_{2}xv+a_{3}(x+v)xv+2a_{4}(xv)^{2}+(x+v)xv$ evaluated at $x=v$ equals $f^{\prime}(v)\neq 0$ it follows that the rational function (10) has degree $2q$. ∎

Lemma 3.12**.**

Suppose that $\Theta_{n}\subset\mathcal{J}$ is a curve. If $\kappa_{4}(\Psi_{n}(x,y))=c$ is constant then $c\in\{0,\infty\}$.

Proof.

Let $\Theta_{n}=\text{Im}\Psi_{n}\in\text{Supp}\text{ div}(\kappa_{4})$ then $\kappa_{4}(\Psi_{n}(x,y))\in\{0,\infty\}$ depending on $\Theta_{n}$ being a zero or a pole of $\kappa_{4}$.

Suppose that $\kappa_{4}(\Psi_{n}(x,y))=c\in{\mathbb{F}}_{q}^{*}$. Since $\kappa_{4}\in\mathcal{L}(2\Theta)$ and $\deg\kappa_{4}(\Psi_{n}(x,y))=\Theta_{n}\bullet\Theta=0$ (note that here we use $c\neq 0,\infty$), this contradicts the fact that the curves $\Theta_{n}$ and $\Theta$ intersect in $0\in\mathcal{J}$. With this we have that $c\in\{0,\infty\}$. ∎

Lemma 3.12 (rather, its proof) provides a geometric reason for the fact that $\kappa_{4}\in\mathcal{L}(2\Theta)$ cannot be constant $\neq 0,\infty$ when restricted to the curves $\Theta_{n}$: these curves will always intersect $\Theta$ and therefore they will have a positive intersection number (degree of $\kappa_{4}(\Psi_{n}(x,y)$). However, if the curve $\Theta_{n}$ equals $\Theta$ or is contained in the zero-locus of $\kappa_{4}$, then of course $\kappa_{4}|_{\Theta_{n}}$ is the constant map $\infty$ or [math], respectively.

Note that when $\Theta_{n}$ has dimension zero, then $\Theta_{n}=\{0\}\subset\mathcal{J}$ (compare the proof of Proposition 3.2). This case is treated separately in the definition of $\delta_{n}$ given below.

Using Lemma 3.12, it follows that the only cases where $\delta_{n}$ is not defined yet, is the situation where $\Theta_{n}$ is a pole or a zero of $\kappa_{4}$. The following property of $\kappa_{4}$ will be useful.

Lemma 3.13**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve given by an equation $Y^{2}=f(X)$ where $f$ has degree $5$ and consider its Jacobian $\mathcal{J}$. Then the function $\kappa_{4}$ on $\mathcal{J}$ has a divisor of zeros $D_{0}=\textrm{ \rm div}_{0}(\kappa_{4})$ such that its support consists of at most four irreducible curves.

Proof.

Let $\text{div}(\kappa_{4})=D_{0}-2\Theta$ with $D_{0}$ effective. We have that $D_{0}=\sum C_{i}$ where the $C_{i}\subset\mathcal{J}$ are irreducible curves. Then

[TABLE]

As $\Theta$ is ample, $C_{i}\bullet\Theta>0$ (see [6, Chapter V, Theorem 1.10]) which implies that the support of the zero divisor of $\kappa_{4}$ consists of four or less irreducible curves in $\mathcal{J}$. ∎

Lemma 3.14**.**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve of genus $2$ given by $Y^{2}=X^{5}+a_{4}X^{4}+a_{3}X^{3}+a_{2}X^{2}+a_{1}X+a_{0}=f(X)$ and let $\mathcal{J}$ be its Jacobian. Assume $\text{\rm Im}\;\Psi_{n}=\Theta_{n}\subset\mathcal{J}$ is a curve that is a zero or a pole of $\kappa_{4}$. Let $(w,0)\in\mathcal{H}(\overline{\mathbb{F}}_{q})$ be some Weierstrass point. Define $\Phi_{n,w}:=\Phi_{n}+\iota(w,0)$ where $\Phi_{n}=\phi+[n]\in\mbox{End}(\mathcal{J})$ and consider $\Psi_{n,w}:=\Psi_{n}+\iota(w,0)\in\text{\rm Mor}_{\overline{\mathbb{F}}_{q}}(\mathcal{H},\mathcal{J})$. Take the generic point $(x,y)\in\mathcal{H}$, then $\kappa_{4}(\Psi_{n,w}(x,y))=:\tfrac{\mu_{1,n}^{w}(x)}{\mu_{2,n}^{w}(x)}\in\overline{\mathbb{F}}_{q}(x)^{*}$ for coprime $\mu_{1,n}^{w},\mu_{2,n}^{w}\in\overline{\mathbb{F}}_{q}[x]$ and ${\Phi_{n,w}}^{*}\Theta\bullet\Theta=\max\{\deg\mu_{1,n}^{w},\deg\mu_{2,n}^{w}\}$.

Proof.

Suppose that $\Psi_{n}(\mathcal{H})=\Theta_{n}=\Theta$ (is a pole of $\kappa_{4}$), it follows that $\Theta_{n}^{w}:=\text{Im}\Psi_{n,w}\not\subset\Theta$. We want $\Theta_{n}^{w}$ to avoid the support of $\kappa_{4}$. We do this in order to have a well defined degree of $\kappa_{4}$ restricted to the symmetric divisor $\Theta_{n}^{w}$ (see Remark 3.9).

If the curve $\Theta_{n}^{w}\subset\mathcal{J}$ would be a zero of $\kappa_{4}$, then by Lemma 3.13 there are at most four possible curves $\{C_{1},C_{2},C_{3},C_{4}\}$ in the zero locus $\kappa_{4}$. Further, there are exactly five affine Weierstrass points in $\mathcal{H}$, and then for at least one of them say $(\hat{w},0)\in\mathcal{H}$ we have that $\Theta_{n}^{\hat{w}}\not\in\{C_{1},C_{2},C_{3},C_{4}\}$.

As a result, $\kappa_{4}(\Psi_{n,{\hat{w}}}(x,y))$ is well defined and non-constant since $\Theta_{n}^{\hat{w}}\not\in\text{Supp}\text{ div}(\kappa_{4})$ (see Lemma 3.12). Further, $\text{Im}\Psi_{n,\hat{w}}$ is symmetric with respect to $[-1]\in\text{Aut}(\mathcal{J})$ since $\Psi_{n}(x,y)+\iota({\hat{w}},0)$ is. Hence $\kappa_{4}(\Psi_{n,{\hat{w}}}(x,y))=\kappa_{4}(\Psi_{n,{\hat{w}}}(x,-y))\in\overline{}\mathbb{F}_{q}(x)^{*}$.

By Remark 3.9 and Lemma 3.8, as in the previous lemmas,

[TABLE]

which proves the lemma for $\Theta_{n}$ a pole of $\kappa_{4}$. Similarly if $\Theta_{n}$ is a zero of $\kappa_{4}$ one can translate $\Theta_{n}$ by some Weierstrass point of $\mathcal{H}$ in order to avoid the zero or the pole divisor of $\kappa_{4}$. Therefore the degree of $\kappa_{4}(\Psi_{n}^{w}(x,y))$ is well defined. ∎

Using the previous Lemmas, we know that there is always a $(w,0)\in\mathcal{H}$ such that the value $\Theta_{n}\bullet\Theta$ can be obtained using the degree of the rational function $\kappa_{4}$ restricted to the generic point of $\Theta_{n}^{w}$. Further, we know that if $\Psi_{n}$ is non-constant but $\kappa_{4}(\Psi_{n}(x,y))$ is constant then it must be [math] or $\infty$ and $\Theta_{n}=\text{Im}\Psi_{n}\in\text{Supp}\text{ div}(\kappa_{4})$ by Lemma 3.12. Hence we have a definition of $\delta_{n}$ for all $n\in\mathbb{Z}$:

[TABLE]

Before proving our basic identity for genus $2$, namely $\delta_{n-1}+\delta_{n+1}=2\delta_{n}+4$ to obtain the Hasse-Weil inequality in this case à la Manin, we recall an additional result.

Theorem 3.15** (Theorem of the cube for Abelian varieties).**

Let $\mathcal{A}$ be an Abelian variety, $\alpha,\beta,\gamma\in\text{\rm End}_{k}(\mathcal{A})$ and $D\in\text{Div}(\mathcal{A})$, then

[TABLE]

where $\sim$ denotes linear equivalence.

Proof.

See [13, Corollary 5.3]. ∎

Corollary 3.16**.**

Let $\Theta\in\text{\rm Div}(\mathcal{J})$ and consider the multiplication-by-$n$ map $[n]\in\text{\rm End}(\mathcal{J})$. Then $[n]^{*}\Theta\sim n^{2}\Theta\in\text{\rm Div}(\mathcal{J})$.

Proof.

This is an application of Theorem 3.15 with $\alpha=[1],\beta=[-1]$ and $\gamma=[n]$ together with induction w.r.t. $n$. For details, see [13, Corollary 5.4] and use that in the present case $[-1]^{*}\Theta=\Theta$. ∎

Theorem 3.17**.**

Let $\mathcal{H}$ be a hyperelliptic curve of genus $2$ over $\mathbb{F}_{q}$ with one rational point at infinity, then:

[TABLE]

Moreover, $\delta_{n}=2n^{2}+n(q+1-\#\mathcal{H}(\mathbb{F}_{q}))+2q$.

Proof.

As before, we denote $\Phi_{m}:=\phi+[m]\in\text{End}(\mathcal{J})$ where $\phi$ is the $q$-th Frobenius map. We begin with some cases when either $\Psi_{n}$ or $\Psi_{n\pm 1}$ is constant.

Suppose that $\Psi_{n}$ is constant, then by definition (11) we have that $\delta_{n}=0$. By Corollary 3.3 we have that $\Theta_{n\pm 1}=\Theta$ and $\Psi_{n\pm 1}=\pm[1]\circ\iota$. It follows by Lemma 3.10 and the symmetry of $\Theta$ with respect to $[-1]$ that $\delta_{n\pm 1}=[\pm 1]^{*}\Theta\bullet\Theta=\Theta\bullet\Theta=2$ and the lemma follows.

Now suppose that $\Psi_{n-1}$ is constant. Then $\delta_{n-1}=0$ and by Corollary 3.3 we have that $\Theta_{n}=\Theta$ and $\Psi_{n}=\iota$. An analogous argument as in the previous case shows that $\delta_{n}=2$.

To prove that $\delta_{n+1}=8$, note that by Proposition 3.2 we have that $\phi=-[n-1]\in\text{End}(\mathcal{J})$. Therefore $\Phi_{n+1}:=\phi+[n+1]=[2]$ which means that $\Psi_{n+1}=\Phi_{n+1}\circ\iota=[2]\circ\iota$. Hence by Corollary 3.16 and Lemma 3.8 we obtain $\delta_{n+1}=\Phi_{n+1}^{*}\Theta\bullet\Theta=[2]^{*}\Theta\bullet\Theta=4\Theta\bullet\Theta=8$.

The case that $\Psi_{n+1}$ is constant is similar to the previous case: one uses the symmetry of $\Theta$ with respect to $[-1]\in\text{End}(\mathcal{J})$ to obtain $\delta_{n-1}=[-2]^{*}\Theta\bullet\Theta=[2]^{*}\Theta\bullet\Theta=8$.

Now we assume that $\Psi_{n\pm 1}$ and $\Psi_{n}$ are non-constant. In the case that $\Theta_{n}\in\text{Supp}\text{ div}(\kappa_{4})$, by Lemma 3.14 there is a $(w,0)\in\mathcal{H}$ such that $\delta_{n}=\Phi_{n,w}^{*}\Theta\bullet\Theta=\deg\kappa_{4}(\Psi_{n}^{w}(x,y))$. Using Remark 3.9 and Lemma 3.8, the latter integer equals $\Phi_{n}^{*}\Theta\bullet\Theta$. Note that also in the case that $\Theta_{n}\not\in\text{Supp}\text{ div}(\kappa_{4})$ we have $\delta_{n}=\Phi_{n}^{*}\Theta\bullet\Theta$ (see Lemma 3.8). We will finish the proof by studying these intersection numbers.

Using Theorem 3.15, let $D:=\Theta\in\text{Div}(\mathcal{J})$ and take $\alpha:=\phi+[n],\beta=[1],\gamma:=-[1]\in\text{\rm End}_{\mathbb{F}_{q}}(\mathcal{J})$. The Theorem of the cube (3.15) implies:

[TABLE]

or equivalently:

[TABLE]

Intersecting both sides of the equivalence with $\Theta$ proves the first part of the theorem. To be more precise, we use Lemma 3.8 together with Lemma 3.10 to deduce $2\delta_{n}+4=\delta_{n-1}+\delta_{n+1}$.

The explicit formula for $\delta_{n}$ now follows by induction, noting (Proposition 3.6) that $\delta_{1}=3(q+1)-\#\mathcal{H}(\mathbb{F}_{q})$ and (Lemma 3.11) that $\delta_{0}=2q$. ∎

Corollary 3.18** (Hasse-Weil for $g\!=\!2$).**

Let $\mathcal{H}/\mathbb{F}_{q}$ be a hyperelliptic curve with one rational point at infinity and $\mbox{char}(\mathbb{F}_{q})\neq 2$, then:

[TABLE]

Proof.

Consider the polynomial in $n$ appearing in the previous Theorem 3.17. The polynomial has the form $\delta(x):=2x^{2}+Tx+2q$ with $T:=q+1-\#\mathcal{H}(\mathbb{F}_{q})$. Its discriminant is

[TABLE]

We want to prove that $\Delta_{\delta}\leq 0$ since that would imply that $|T|\leq 4\sqrt{q}$, which is exactly the statement of the Hasse-Weil inequality for $g=2$.

We already proved in Proposition 3.2 that if $n\in\mathbb{Z}$ exists such that $\Psi_{n}\in\text{Mor}_{\mathbb{F}_{q}}(\mathcal{H},\mathcal{J})$ is constant, then $\Psi_{n}=0$, $q=n^{2}$ is a perfect square and $\#\mathcal{H}(\mathbb{F}_{q})=q+1\pm 4\sqrt{q}$. Hence from the existence of such $n$, the Hasse-Weil inequality over $\mathbb{F}_{q}$ for the curve in question follows. So from now on we will suppose that $\Psi_{n}$ is non-constant for every $n\in\mathbb{Z}$. By Theorem 3.17 this implies that $\delta_{n}=\Phi_{n}^{*}\Theta\bullet\Theta$ for all $n\in\mathbb{Z}$.

It is clear that $\delta_{n}>0$ for all $n\in\mathbb{Z}$ by definition. This is since $\Psi_{n}$ is non-constant, hence $\Theta_{n}\subset\mathcal{J}$ is a curve, implying that $\delta_{n}$ is the degree of the rational function $\kappa_{4}(\Psi_{n,w}(x,y))\in\mathbb{F}_{q}(x)^{*}$ or $\kappa_{4}(\Psi_{n}(x,y))\in\mathbb{F}_{q}(x)^{*}$ depending on $\Theta_{n}$ being in $\text{Supp div}(\kappa_{4})$ or not.

Another fast and not very elementary argument for this uses that the divisor $\Theta$ is ample, hence by the Nakai-Moishezon criterion for ampleness on surfaces (see [6, Chapter V, Theorem 1.10]), its intersection number with any curve is positive.

Now from Theorem 3.17 we have that $\delta_{n}=2n^{2}+(q+1-\#\mathcal{H}(\mathbb{F}_{q}))n+2q$. Consider $\delta(x)=2x^{2}+(q+1-\#\mathcal{H}(\mathbb{F}_{q}))x+2q$. We claim that $\delta(x)$ is non-negative for all $x\in\mathbb{R}$, hence it has non-positive discriminant $\Delta_{\delta}$. This will imply the Hasse-Weil inequality for this case.

Suppose that the Hasse-Weil inequality for genus $2$ is false. This is equivalent to the statement $\Delta_{\delta}>0$. In this case $\delta(x)$ has two different real zeros $\alpha<\beta$. We have that $\Delta_{\delta}$ in terms of $\alpha$ and $\beta$ is given by:

[TABLE]

The integer $\Delta_{\delta}$ is assumed to be positive, so we conclude $4(\alpha-\beta)^{2}\geq 1$. Moreover, recall $\delta(n)>0$ for every $n\in\mathbb{Z}$. Since for any $x_{0}\in(\alpha,\beta)$ we have that $\delta(x_{0})<0$, it follows that $(\alpha,\beta)$ contains no integers. This implies that $\beta-\alpha<1$ and then $1\leq 4(\alpha-\beta)^{2}<4$.

So we have just three situations for positive discriminant: $T^{2}-16q\in\{1,2,3\}$. Each of these possibilities results in a contradiction as we will see below.

Case $T^{2}-16q=3$: There are no integers $(T,q)$ satisfying this, as one checks by reducing modulo $4$.

Case $T^{2}-16q=2$: Again reducing modulo $4$ implies that no integral solutions $(T,q)$ exist.

Case $T^{2}-16q=1$: Then

Subcase (i): $T=8w+1=q+1-\#\mathcal{H}(\mathbb{F}_{q})$, $q=4w^{2}+w=p^{n}$.

Since $p$ is the only prime dividing $q=w(4w+1)$ and since $\text{gcd}(w,4w+1)=1$, it follows that $w=\pm 1$ or $4w+1=\pm 1$. We proceed to check all possibilities.

If $4w+1=+1$ then $w=0$ and $q=0$ which is not possible.

If $4w+1=-1$ then $w=-\tfrac{1}{2}$ which is absurd since $w$ is an integer.

If $w=+1$ then $q=5$ and $T=9$. However $9=5+1-\#\mathcal{H}(\mathbb{F}_{5})$ is impossible.

If $w=-1$ then $q=3$ and $T=-7$. However $\mathcal{H}(\mathbb{F}_{3})$ has at most $2\cdot 3+2$ rational points, hence $T\geq 3+1-8=-4$.

Subcase (ii): $T=8w+7=q+1-\#\mathcal{H}(\mathbb{F}_{q})$, $q=4w^{2}+7w+3=p^{n}$.

Again $p$ is the only prime dividing $q=4w^{2}+7w+3=(w+1)(4w+3)$. Moreover these two factors are coprime since $4(w+1)-(4w+3)=1$. Therefore one of the factors must be $\pm 1$. Again we check all possibilities.

If $w+1=1$ then $q=3$ and $T=7$. However any curve $C/\mathbb{F}_{3}$ has at least [math] rational points, hence $T=3+1-\#C(\mathbb{F}_{3})\leq 4$.

If $w+1=-1$ then $q=5$ and $T=-9$. Any hyperelliptic $\mathcal{H}/\mathbb{F}_{5}$ satisfies $\#\mathcal{H}(\mathbb{F}_{5})\leq 2\cdot(5+1)$, hence $T\geq 6-12=-6$.

The case $4w+3=1$ is impossible since $w$ is assumed to be an integer.

Finally, $4w+3=-1$ leads to $q=0$ which is absurd.

This shows that assuming $\Delta_{\delta}=T^{2}-16q>0$ leads to a contradiction. Therefore $|T|\leq 4\sqrt{q}$ which is the Hasse-Weil inequality for this case. ∎

Remark. Note that our definition of the integers $\delta_{n}$ is elementary and completely analogous to the definition by Manin of the integers $d_{n}$. However, whereas Manin also succeeded in presenting a completely elementary proof of the basic identity for the $d_{n}$, we used the interpretation of the $\delta_{n}$ as intersection numbers in order to show an analogous basic identity in the genus two case. To obtain a fully elementary proof also in genus two, it therefore suffices to replace this intersection theory argument by a calculation in the spirit of what Manin did. We do not know whether this is a feasible task.