Hasse Weil Bound

宝宝巴士第一集: 学会Riemann Hypothesis.
宝宝巴士第二集: 证明Hasse Weil Bound.
宝宝巴士第三集: 证明nonsingular curve上的Riemann Hypothesis.
宝宝巴士第四集: 证明Weil Conjecture.

Hasse Weil Bound

Let $X$ be a projective nonsingular curve over $\mathbb{F}_{p}$ . The genus of $X$ is $g$ . Define the set of $\mathbb{F}_{q}$ -points $X(\mathbb{F}_{q})$ to be $\mathrm{Hom}(\mathrm{Spec}(\mathbb{F}_{q}),X)$ and denote the size of the set by $N$ , here $q=p^{m}$ .

Theorem 1 (Hasse-Weil). $$\left| N-1-q \right| \leq 2g\sqrt{q}AG$$

We will prove the theorem in several steps. The first thing we need to do is relate $N$ with the intersection multiplicity.

Definition 1. The Galois group $\mathrm{Gal}(\bar{\mathbb{F}_{q}} /\mathbb{F}_{q})$ action on $X$ as follows: For each $\sigma\in \mathrm{Gal}(\bar{\mathbb{F}_{q}} /\mathbb{F}_{q})$ , $f:X_{\bar{\mathbb{F}_{q}}}\to X_{\bar{\mathbb{F}_{q}}}$ is the fibre product $\sigma\times id_{X}$ .

The geometric Frobenius morphism $\varphi$ is the automorphism associated to the Frobenius on $\mathbb{F}_{q}$ .

Lemma 1. Let $X$ is reduced $k$ -scheme and $Y$ is separated $k$ -scheme. $f:X\to Y$ is a morphism. Let $\Gamma_{f}(X)$ be the closed subscheme with reduced subscheme structure. Then $\Gamma_{f}(X)\cong X$ .

Proof

Proof. Note that $\Gamma_{f}$ is obtained by pulling back the diagonal:

$\Delta$ is closed immersion so $\Gamma_{f}$ is a closed immersion, hence $\Gamma_{f}$ is an isomorphism to the schematic image $\overline{\Gamma_{f}(X)}$ . Since $X$ is reduced, we may factor $\Gamma_{f}$ through $\Gamma_{f}(X)$ , which induces an morphism from $\overline{\Gamma_{f}(X)}$ to $\Gamma_{f}(X)$ . From the definition of reduced subscheme, $\overline{\Gamma_{f}(X)}\cong \Gamma_{f}(X)$ . ◻

Theorem 2. Denote the image of $\Gamma_{f}$ and $\Delta$ by $\Gamma_{f}$ and $\Delta$ . Then the intersection multiplicity $\Gamma_{f}\cdot \Delta= N$ .

Proof

Proof. The fixed point of $X$ under Galois group action $\mathrm{Gal}(\bar{\mathbb{F}_{q}} /\mathbb{F}_{q})$ is exactly the $\mathbb{F}_{q}$ -points $X(\mathbb{F}_{q})$ . So $x$ is $\mathbb{F}_{q}$ -point if and only if it’s in the intersection of $\Gamma_{f}\cdot \Delta$ . It suffices to show each point has multiplicity 1. From the lemma, $\Gamma_{f}$ and $\Delta$ are all isomorphic to $X$ , so they are nonsingular. For every point $(\varphi(x),x)\in \Gamma_{f}\cap \Delta$ , the tangent space is the product of corresponding tangent space $X$ . Note $(d\varphi)_{x}=0$ , $\Gamma_{f}$ and $\Delta$ intersect transversally, so each point has single multiplicity. ◻

To get the estimation of $\Gamma_{f}\cdot \Delta$ , we need the Hodge index theorem. Recall some facts in the intersection theory: Fix an very ample line sheaf $\mathcal{O}_{X\times X}(1)$ , and denote the corresponding hyperplane section by $H$ .

Definition 2. Let $\mathcal{L}$ be a line sheaf on $X$ . The Hilbert polynomial $\chi(\mathcal{L}(n))$ is a quadratic function $\frac{\alpha_{2}(\mathcal{L})}{2}n^{2}+\alpha_{1}(\mathcal{L})n+\alpha_{0}(\mathcal{L})$ . The degree of $\mathcal{L}$ is defined to be $\alpha_{1}(\mathcal{L})-\alpha_{1}(\mathcal{O}_{X})$ .

If $D$ is a divisor of $X$ , by Hirzebruch-Riemann-Roch theorem, $D\cdot H= \deg(\mathcal{O}_{X}(D))$ , we define the number be the degree of the divisor $D$ . The degree of $D$ is a additive function on the divisor group $\mathrm{Div}(X)$ since the first chern class $c_{1}$ is additive on tensor products.

Definition 3. Let $f:X\to Y$ be a proper morphism of varieties. Define the degree of $f$ be the number $[K(X),K(f(Y))]$ . If $\dim f(Y)<\dim X$ , define the degree be 0.

Using the projection formula $f_{*}(x\cdot f^{*}y)=f_{*}x\cdot y$ , we can show $f^{*}C\cdot f^{*}D=\deg(f)C\cdot D$ . For any line bundle $\mathcal{L}$ , $\deg (f^{*}\mathcal{L})=\deg (f)\deg (\mathcal{L})$ . Note the degree is always a finite number.

Lemma 2. Let $\mathcal{L}_{1}$ , $\mathcal{L}_{2}$ be two line sheaves on a variety $X$ . Then $\mathrm{Hom}(\mathcal{L}_{1},\mathcal{L}_{2})=0$ provided $\deg\mathcal{L}_{1}>\deg\mathcal{L}_{2}$ .

Proof

Proof. Since $\mathrm{rk}(\mathcal{L}_{1})=\frac{\alpha_{2}(\mathcal{L}_{1})}{\alpha_{2}(\mathcal{O}_{X})}$ and $\mathrm{rk}(\mathcal{L}_{2})=\frac{\alpha_{2}(\mathcal{L}_{2})}{\alpha_{2}(\mathcal{O}_{X})}$ , we have $\alpha_{2}(\mathcal{L}_{1})=\alpha_{2}(\mathcal{L}_{2})$ . The Hilbert polynomial $P(\mathcal{L}_{1})(n)<P(\mathcal{L}_{2})(n)$ . Suppose there is a morphism $\varphi:\mathcal{L}_{1}\to \mathcal{L}_{2}$ . Let $\mathcal{K}$ , $\mathcal{F}$ be the kernel and the image of $\varphi$ , respectively.

We first show the support of $\mathcal{F}$ is $X$ . Assume there is a point $x\in X$ that is not in the support of $\mathcal{F}$ , since $\mathrm{Supp}\mathcal{F}$ is closed, theres’s an open neighborhood $U$ that has empty intersection with $\mathrm{Supp}\mathcal{F}$ . We can choose an affine open set $W=\mathrm{Spec}A$ that has nonempty intersection with $X-U$ and $U$ . Pick an nonzero global section $s$ supported on $X-U$ and $f$ is an element in ideal which define $W\cap \mathrm{Supp}s$ . Then $s$ is zero in the localization of $D(f)\subset W$ , so $sf^{n}=0$ for some $n$ . However, $s$ is also in $\mathcal{L}_{1}(W)$ , which is torsion-free, contradict to the above argument.

The Hilbert polynomial of $\mathcal{F}$ is also a quadratic function. Since $X$ is integral, the rank of $\mathcal{F}$ is integer, and it is 1. Thus $\mathrm{rk}(\mathcal{K})=0$ , which means $\alpha_{2}(\mathcal{K})=0$ , and $\dim \mathrm{Supp}\mathcal{K}\leq 1$ . Using a similar argument, we can prove that $\mathcal{K}=0$ . Therefore, $P(\mathcal{L}_{1})(n)=P(\mathcal{F})(n)>P(\mathcal{L}_{2})(n)$ . This contradicts the fact that $\mathcal{F}$ is a subsheaf of $\mathcal{L}_{2}$ . ◻

Lemma 3. Let $\{D_{i}\}$ be a set of divisor on $X\times X$ with bounded degree. Then $\{h^{0}(\mathcal{O}_{X\times X}(D_{i}))\}$ is bounded.

Proof

Proof. Assume $\deg D_{i}< m$ . We first reduce to the case $\deg D_{i}$ is a negative fixed number. For any divisor $D$ in the family and the corresponding line sheaf $\mathcal{O}_{X}(D)=\mathcal{L}$ , consider the exact sequence

0\to \mathcal{L}(-m)\xrightarrow{\cdot s} \mathcal{L}\to \mathcal{L}|_{C}\to 0

where $s$ is a regular section of $\mathcal{L}$ and $C$ is the support of $s$ , by Bertini theorem, such $s$ exists. Then $h^{0}(\mathcal{L})\leq h^{0}(\mathcal{L}(-m))+h^{0}(\mathcal{L}|_{C})$ . Note that $\deg_{C}(\mathcal{L}|_{C})=\deg_{C}(D\cap C)=\deg_{X\times X}(D)$ . The genus $g_{C}$ is independent on the choice of $C$ , and is given by the adjunction formula $g_{C}=1+\frac{1}{2}(K\cdot C+C\cdot C)$ .

If $\deg_{C}(\mathcal{L}|_{C})>2g-2$ , then $h^{0}(\mathcal{L}|_{C})=\deg_{C}(\mathcal{L}|_{C})+1-g\leq m+1-g$ .

If $\deg_{C}(\mathcal{L}|_{C})\leq 2g-2$ , then $h^{0}(\mathcal{L}|_{C})\leq h^{0}(\mathcal{L}|_{C}(2g-1-m))=g$ .

Then we have $h^{0}(\mathcal{L}|_{C})$ bounded and we reduce to the case $\deg(D_{i})$ is negative constant by twisting a small number. Since a global section in $h^{0}(\mathcal{O}_{X\times X}(D))$ will induce a morphism from $\mathcal{O}_{X\times X}$ to $\mathcal{O}_{X\times X}(D)$ , by above lemma, $\mathcal{O}_{X\times X}(D_{i})$ has no global sections. This finishes the proof. ◻

Theorem 3 (Hodge Index Theorem). Let $D$ be a divisor on $X\times X$ . If $\deg (D)=0$ , then $D\cdot D<0$ .

Proof

Proof. By above lemma, $\{h^{0}(nD)\}$ and $\{h^{0}(K-nD)\}$ are all bounded. Using Riemann-Roch theorem on surface,

h^{0}(nD)-h^{1}(nD)+h^{0}(K-nD)=\frac{1}{2}nD(nD-K)+1+p_{a}

For large $n$ , lhs is negative, so $n^{2}D\cdot D<0$ . ◻

We next compute the self intersection number of $\Gamma_{f}$ . Let $\deg(f)=d$ . Using the adjunction formula, $\Gamma_{f}\cdot (\Gamma_{f}+K)=2g-2$ . Since $\omega_{X\times X}=p_{1}^{*}\omega_{X}\otimes p_{2}^{*}\omega_{X}$ , $K_{X\times X}=p_{1}^{*}K_{X}+p_{2}^{*}K_{X}$ . $\Gamma_{f}\cdot p_{1}^{*}K_{X}$ is the same as the $\deg_{\Gamma_{f}}(p_{1}^{*}\omega_{X}\otimes \mathcal{O}_{\Gamma_{f}})=\deg_{\Gamma_{f}}(\Gamma_{f}^{*}p_{1}^{*}\omega_{X})$ , since $p_{1}\circ \Gamma_{f}=f$ , it is also the same as $\deg_{\Gamma_{f}}(f^{*}\omega_{X})=\deg (f)\deg(\omega_{X})=d(2g-2)$ . Similarly, $\Gamma_{f}\cdot p_{2}^{*}K_{X}=2g-2$ . So $\Gamma_{f}\cdot \Gamma_{f}=-d(2g-2)$ .

Let $f=id$ , the self intersection number $\Delta\cdot \Delta=2-2g$ .

Let $d_{1}(D)=D\cdot (P\times X)$ and $d_{2}(D)=D\cdot (X\times P')$ for some $P$ , $P'$ closed points in $X$ . Applying Bezout theorem, $d_{1}$ and $d_{2}$ are independent of the choice of $P$ and $P'$ .

Theorem 4. $$D\cdot D\leq 2d_{1}(D)d_{2}(D)$$

Proof

Proof. Let $V$ be the vector space generated by basis $P\times X$ , $X\times P'$ and $D$ . The intersection product defines a quadratic form

M=\begin{pmatrix} 0&1&d_{1}(D)\\ 1&0&d_{2}(D)\\ d_{1}(D)&d_{2}(D)&D\cdot D\end{pmatrix}

$\det(M)=2d_{1}(D)d_{2}(D)-D\cdot D$ . If $\det (M)<0$ , since $M$ is indefinite, we may choose $A_{1},A_{2},A_{3}$ such that $A_{i}\cdot A_{j}=0$ , and $A_{i}^{T}MA_{i}=a_{i}$ such that $a_{1}, a_{2}>0$ , $a_{3}<0$ . Then appropriate linear combination can give a degree 0 divisor with positive self intersection. Contradict to Hodge index theorem. ◻

By the above theorem, if we define $D*E=d_{1}(E)d_{2}(D)+d_{1}(D)d_{2}(E)-D\cdot E$ , then $D*D\geq 0$ . Using the Cauchy inequality for bilinear forms, we have $\left| \Gamma_{f}*\Delta \right|\leq \sqrt{(\Delta*\Delta)(\Gamma_{f}*\Gamma_{f})}$ , that is

\left| \Gamma_{f}\cdot \Delta-1-\deg(f) \right|\leq 2g\sqrt{\deg (f)}

It remains to show the degree of geometric Frobenius is $q$ . The degree is a generic condition so we may focus on an affine open set. The case $K(X)=\mathbb{F}_{q}(t)$ is clear, the morphism is given by $\sigma: \mathbb{F}_{q}(t)\to \mathbb{F}_{q}(t^{q})$ . For the general case, $K(X)$ is finite algebraic over $\mathbb{F}_{q}(t)$ . Since the action on $K(X)$ restrict on $k(t)$ is the action on $k(t)$ , $[K(X):k(t)]=[\sigma K(X):\sigma k(t)]$ , and the results follow from the multiplicity of extension degree.