[Abelian Varieties VI] Applications of HRR and Intersection Theory

We now come to some applications of the theorem of Hirzebruch-Riemann-Roch for abelian varieties.

1. Warm-up with Euler's exact sequence.

First, we recall the Euler's exact sequence (Chap VII, Gathmann's note).

$$0\to \mathscr{O}_{\mathbb{P}^n}\to \mathscr{O}_{\mathbb{P}^n}(1)^{\oplus n+1}\to T_{\mathbb{P}^n}\to 0$$

where $T_{\mathbb{P}^n}$ is the tangent sheaf of ${\mathbb{P}^n}$. If we denote $H:=\mathscr{O}_{\mathbb{P}^n}(1)$ the hyperplane divisor on ${\mathbb{P}^n}$, then this yields by axioms of Chern's characters that

$$c_t(T_{\mathbb{P}^n})=(1+H)^{n+1}$$

Example 41. $n=2$, then $c_t(T_{\mathbb{P}^2})=1+3Ht+3H^2t+t^3$, and hence, Chern characters of $T_{\mathbb{P}^2}$ is $c_1:=3H, c_2:=3H^2$. And this yields $Td({\mathbb{P}^2})=(1, \frac{1}{2}c_1, \frac{1}{12}(c_1^2+c_2))$. For trivial line bundle $\mathscr{O}_{\mathbb{P}^2}$, $ch(\mathscr{O}_{\mathbb{P}^2})=1$, HRR implies that

$$\chi(\mathbb{P}^2)=\deg_2 Td({\mathbb{P}^2})=\deg \frac{1}{12}(9H^2+3H^2)=1$$

Moreover, using the adjunction sequence, if $X$ is a smooth hypersurface of degree $d$ in ${\mathbb{P}^n}$, and $i: X\hookrightarrow {\mathbb{P}^n}$ the inclusion, then we have the short exact sequence

$$0\to T_X\to i^*T_{\mathbb{P}^n}\to N\to 0$$

And $N$ is called the normal bundle of $X$ in $\mathbb{P}^n$. In this case, it is a line bundle. This yields $c_t(T_{i^*\mathbb{P}^n})/c_t(N)=c_t(T_X)$. Because Chern characters commutes with the pull-back, we have $c_j(i^*\mathbb{P}^n)=i^*c_j(\mathbb{P}^n)$. Hence, if we denote $h:=i^*H$ the pullback of hyperplane divisor on ${\mathbb{P}^n}$, we have $c_t(T_{i^*{\mathbb{P}^n}})=(1+ht)^{n+1}$. And because $[X]$ can be consider as a divisor on ${\mathbb{P}^n}$, with $X\sim dH$ (because $A^k({\mathbb{P}^n})$ is generated by the class of $k$-dimensional linear subspace), we have $c_1(N)=i^*[X]=dH$. And $N$ is a line bundle implies that $c_n(N)$ vanishes, for $n>1$. This yields $c_t(N)=1+dHt$. Combining these things, we get

$$c_t(T_X)=(1+ht)^{n+1}/(1+dht)=(1+ht)^{n+1}(1-dht+d^2h^2t^2-...)$$

Remark 42. If $i$ is an embedding of codimension $d$, i.e. there exists $d$ divisors $D_1,...,D_d$ on $\mathbb{P}^n$ such that $D_1\cap D_2\cap...\cap D_d=X$, then the normal bundle is a vector bundle of rank $d$ in this case, and its Chern roots will be $i^*[D_i]$, which are $i^*d_iH$, for some hyperplane $H$ in $\mathbb{P}^n$.

Example 43. We will use our recent argument to compute the genus of a smooth projective plane curve of degree $d$. In this case, $c_t(T_X)=(1, \frac{1}{2}(3-d)h,...)$, where $h$ is a divisor on $X$ with $\deg h=d$. This yields by HRR that $\chi(X)=\frac{1}{2}d(3-d)$. Because $\chi(X)=1-g$, we obtain $g=\frac{1}{2}(d-2)(d-1)$.

Example 44. We will compute the Euler's characteristic of the line bundle $\mathscr{O}_{\mathbb{P}^n}(d)$ on $\mathbb{P}^n$. In this case

$$Td(T_X)=\frac{H^{n+1}}{(1-\exp(-H))^{n+1}}$$

And $ch(X)=\exp(dH)$. In particular, we want to compute the $n$-th coefficient of the power series

$$\frac{\exp(dH)H^{n+1}}{(1-\exp(-H))^{n+1}}$$

In terms of complex analysis, it is the residue at 0 of

$$\frac{\exp(dH)}{(1-\exp(-H))^{n+1}}dH$$

Let $x:=1-\exp(-H)$, then $\exp(H)=\frac{1}{1-x}$, and $\frac{dH}{dx}=\frac{1}{1-x}$. And it is sufficient for us to compute the residue at 0 of $\frac{(1-x)^{-d-1}}{x^{n+1}}$, and it is the $n$-th coefficient of $(1-x)^{-d-1}$, which is $(-1)^n{-d-1\choose n}={n+d\choose n}$. In short,

$$\chi(\mathbb{P}^n, \mathscr{O}_{\mathbb{P}^n}(d))={n+d\choose n}$$

Using the same method, we can also compute the Euler's characteristic of a smooth hypersurface in $\mathbb{P}^n$.

2. Embedding of abelian varieties. We will prove in this section that an abelian variety of dimension $g$ cannot be embedded into $\mathbb{P}^{2g-1}$, and it is embedded into $\mathbb{P}^{2g}$ iff it is an elliptic curve, or an abelian surface of degree 10 in $\mathbb{P}^4$. This shows, it is not easy to write explicitly equations that define an abelian variety in $\mathbb{P}^n$.

Now, let us denote $m$ the smallest integer such that $A$ can be embedded to $\mathbb{P}^m$. It is an embedding of codimension $m-g$. Hence, the normal bundle of $A$ in $\mathbb{P}^m$ is of rank $m-g$. Let us denote the Chern characters of $N$ as $c_1,...,c_{m-g}$. This yields by the adjunction sequence

$$0\to T_X\to i^*T_{\mathbb{P}^m}\to N\to 0$$

that $c_t(T_X)c_t(N)=c_t(i^*T_{ \mathbb{P}^m})$. For abelian varieties, the tangent sheaf is trivial, and hence

$$1+\sum_{i=1}^{m-g}c_it^i=c_t(N)=c_t(i^*T_{\mathbb{P}^m})=(1+ht)^{m+1}$$

From this, one can see for $n\ge m+1-g$, $h^n$ vanish. But when $n=g$, we have $\deg h^g$ is exactly the degree of $A$, which is never zero. Hence, $m+1-g\ge g+1$. This yields $m\ge 2g$.

Now, assume that $m=2g$. Look at the $g$-th coefficient in the last identity, we can see

$$c_g={2g+1\choose g}h^g$$

Taking the degree, we have $\deg c_g={2g+1\choose g}d$. We now make use of a following

Theorem 45 [A. Van de Ven - On the embedding of abelian varieties in Projective Spaces]. Let $X$ be a complete non-singular projective variety of dimension $2d$, and $Y$ is a non-singular projective subvariety of $X$ of dimension $d$. If $N$ is the normal bundle of $Y$ in $X$, and $c_g(N)$ the $g$-th Chern character of $N$, then $c_g(N)$ is equal to the self-intersection number of $Y$ on $X$.

As one can see from this, $\deg c_g$ is the self-intersection number of $A$ on $\mathbb{P}^m$. As we mentioned earlier, the $A^k(\mathbb{P}^m)$ is generated by $k$-th dimensional linear subspace of $\mathbb{P}^m$, we have $A\sim dH$ for some $g$-dimensional linear subspace $H$ on $\mathbb{P}^m$. This yields $\deg A.A=d^2=\deg c_g$. This follows $d={2g+1\choose g}$.

Now, let $L$ be any line bundle on $A$, the Riemann-Roch theorem for abelian variety reads

$$\chi(A,L)=\frac{1}{g!}\deg c_1(L)^g$$

Take $L:=h$, we have $\deg c_1(L)^g=\deg h^g = d = {2g+1\choose g}$. And because the Euler characteristic is an integer, we have $g!\mid {2g+1\choose g}$. This can happen only  if $g=1,2$. When $g=1$, it is an elliptic curve, embedded into $\mathbb{P}^2$, and of degree 3. If $g=2$, we have $d=10$, and it is an abelian surface of degree 10, embedded into $\mathbb{P}^4$. Such an abelian surface exists, due to the result of Mumford.

Remark 46. It also follows from the Riemann-Roch's theorem that $\chi(A)=0$.

3. HRR for smooth surfaces in $\mathbb{P}^4$. 

We will first visit HRR for smooth surfaces $X$. The tangent bundle of $X$ is of rank 2, with Chern characters $c_1(T_X), c_2(T_X)$. We have $c_1(T_X)=-c_1(\Omega_X)$, where $\Omega_X$ is the cotangent bundle of $X$. It is also of rank $2$. Recall that if $E$ is a vector bundle on $X$ of rank $n$, with Chern roots $\alpha_i$, then $\bigwedge^r E$ has Chern roots $\sum_{1\le i_1<...<i_r\le n}\alpha_{i_1}+...+\alpha_{i_r}$. From this, one can see $c_1(\Omega_X)=c_1(\bigwedge^2\Omega_X)=K$, where $K$ is the canonical bundle of $X$. This yields $c_1(T_X)=-K$. We can compute $c_2$ via the genus of $X$. The HRR for smooth surfaces reads

$$\chi(X)= \frac{1}{12}(K^2+c_2(T_X))$$

This yields $c_2(T_X)=12\chi(X)- K^2$. Now, if $X$ can be embedded into $\mathbb{P}^4$, we have by Theorem 45, the adjunction sequence yields

$$c_t(T_X)c_t(N)=(1+ht)^5=1+5ht+10h^2t^2+...$$

We know that $c_t(T_X)=1-Kt+c_2(T_X)t^2$ and $c_t(N)=1+c_1(N)+c_2(N)$. Equating both sides of the last identity, we get

$$c_1(N)=5h+K$, c_2(N)-Kc_1(N)+c_2(T_X)=10h^2$$

We know from Theorem 45 that $\deg c_2(N)=d^2$, $\deg h^2=d$, $c_2(T_X)=12\chi(X)-K^2$. So, taking the degree of the 2-nd graded part, we have

$$d^2 - 10d + 5hK - 2K^2 + 12\chi(X)=0$$

Now, for an abelian surface that can be embedded into $\mathbb{P}^4$, because the canonical sheaf is just trivial, and $\chi(X)=0$, by Remark 46, we have $d^2-10d=0$. This yields $d=10$. And we obtain a part of our result in the previous section.

4. An exercise in the book of Van der Geer and Ben Moonen.

We will prove in this exercise that an abelian variety $A$ of dimension $g$ cannot be embedded into $P:=\mathbb{P}_1^{2g-1}$. Assume that we have such an embedding $i: A\to P$, then the adjunction sequence yields

$$0\to T_A\to i^*T_P\to N\to 0$$

And in this case, again, we have $c_t(N)=c_t(i^*T_P)$. If we denote $pr_i: P\to \mathbb{P}^1$ the $i$-th projection, then it follows that $T_P=\oplus pr_i^*T_{\mathbb{P}_1}$, and because $c_t(P_1)=(1+H)^2=1+2Ht$, where $H$ is a point in $\mathbb{P}^1$, we have $c_t(T_P)=\prod_{i=1}^{2g-1}(1+2h_i)$, where $h_i$ is the pullback of $H_i$ via $pr_i$. Because $N$ is a vector bundle on $A$ of rank $g-1$, we have $c_g(N)=0$. And this yields the $g$-th part of the product $\prod_{i=1}^{2g-1}(1+2h_i)$ vanish, i.e. all the term of the form $h_{i_1}...h_{i_g}$ vanish, and there are ${2g-1\choose g}$ such terms.

But if we look at ${2g-1\choose g}$ (with index $i_1,...,i_g$ as above) projections from $P$ to $P_g:=\mathbb{P}_1\times ... \times \mathbb{P}^1$ ($g$ times), and $A\to P_g$ the composition, we will have at least one of them is surjective, since $\dim A\ge g$, and that means the pullback of points on $P_g$ via this map is never empty. And this yields the corresponding product $h_{i_1}...h_{i_g}$ is non zero, because they are just the pullback of points in each $\mathbb{P}^1$. This is a contradiction. Hence, an abelian variety of dimension $g$ cannot be embedded into $\mathbb{P}_1^{2g-1}$.

The later question is also interesting, that is can an abelian variety be embedded into $\mathbb{P}_1^{2g}$? First, if we are able to prove that there exists an elliptic curve $E$ that can be embedded into $\mathbb{P}^1\times \mathbb{P}^1$, then $E\times E\times ...\times E$ ($g$-times) is an abelian variety of dimension $g$, and it can be embedded into $\mathbb{P}_1^{2g}$. This rest of this section is devoted for the proof. We refer to the work of Peter Bruin. In this paper, he shows the existence of non-singular curve of type $(a,b)$, for $a,b>0$ in $\mathbb{P}^1\times \mathbb{P}^1$, and the genus of this curve is $(a-1)(b-1)$. If we choose $a=b=2$, we get the curve of type $(2,2)$ in $\mathbb{P}^1\times \mathbb{P}^1$, and it is an elliptic curve. It is actually the Edwards curve, and it turns out to be useful in cryptography. Now, let us make a sketch of his proof.

First, it is known that if $Z:=\mathbb{P}^1\times \mathbb{P}^1$, then $Pic(Z)$ is generated by $L_1, L_2$, where $L_1$ is the pull-back $p_1^*\mathscr{O}_{\mathbb{P_1}}(1)$, and $L_2$ is the pullback $p_2^*\mathscr{O}_{\mathbb{P}_1}(1)$. This yields, if $L$ is a divisor on $Z$, then $L=aL_1+bL_2$, for some $a,b\in \mathbb{Z}$. We call this line bundle is of type $(a,b)$. From this decomposition, we get

$$L=aL_1+bL_2\cong p_1^*(\mathscr{O}_{\mathbb{P}_1}(a))\otimes p_2^*(\mathscr{O}_{\mathbb{P}_1}(b))$$

From this, any line bundle of type $a(,b)$ on $Z$ is isomorphic, and we denote them $\mathscr{O}_Z(a,b)$ Now, making use of Kunneth's formula, we have

$$H^n(Z,\mathscr{O}_Z(a,b))\cong \bigoplus_{p+q=n}H^p(\mathbb{P}^1, \mathscr{O}_{\mathbb{P}_1}(a))\otimes H^q(\mathbb{P}^1, \mathscr{O}_{\mathbb{P}_1}(b))$$

Using Riemann-Roch, when $a,b\ge 0$, because $\deg K_{\mathbb{P}_1}=-2$, the higher cohomology groups of $H^*(\mathbb{P}_1,\mathscr{O}_{\mathbb{P}_1}(a))$ vanish, hence $h^0(\mathbb{P}_1,\mathscr{O}_{\mathbb{P}_1}(a))=a + 1$, and similarly for $b$. This yields $H^i(Z,\mathscr{O}_Z(a,b))$ vanish for $i=1,2$, and $h^0(Z, \mathscr{O}_Z(a,b))=(a+1)(b+1)$. Similarly, if $a,b<0$, we get $H^i(Z, \mathscr{O}_Z)$ vanish for $i=0,1$, and $h^2(Z,\mathscr{O}_Z(a,b))=(a+1)(b+1)$.

Now, if we assume that there exists a smooth curve $Y$ of type $(a,b)$ on $Z$, i.e. if we denote homogeneous coordinate of $Z$ as $(x_0:x_1:y_0:y_1)$, then $Y$ is defined by homogeneous polynomial of the form $\sum_{i,j}c_{i,j}x_0^ix_1^{a-i}y_0^jy_1^{b-j}$. We call $Y$ is bi-degree of the type $(a,b)$. The closed immersion $Y\to Z$ induces the short exact sequence of sheaves

$$0\to \mathscr{O}_Z(-Y)\to \mathscr{O}_Z\to i_*\mathscr{O}_Y\to 0$$

And this induces the following long exact sequence

$$0\to H^0(Z, \mathscr{O}_Z(-Y))\to H^0(Z, \mathscr{O}_Z)\to H^0(Z,i_*\mathscr{O}_Y)\to $$

$$\to H^1(Z, \mathscr{O}_Z(-Y))\to H^1(Z, \mathscr{O}_Z)\to H^1(Z, i_*\mathscr{O}_Y)\to $$

$$\to H^2(Z, \mathscr{O}_Z(-Y))\to H^2(Z, \mathscr{O}_Z)\to H^2(Z, i_*\mathscr{O}_Y)\to 0$$

Now, if $Y$ is bi-degree of the type $a,b$, with $a,b>0$, we have by our earlier arguments

$$0\to H^0(Z,\mathscr{O}_Z(-a,-b))\to k\to H^0(Z, i_*\mathscr{O}_Y)\to H^1(Z, \mathscr{O}_Z(-a,-b))\to 0$$

$$H^1(Z, i_*\mathscr{O}_Y)\cong H^2(Z, \mathscr{O}_Z(-a,-b))$$

And this yields $H^0(Z, i_*\mathscr{O}_Y)=k, h^1(Z, i_*\mathscr{O}_Z)=(a-1)(b-1)$. However, since $i$ is a closed immersion, we have $H^*(Z, i_*\mathscr{O}_Y)\cong H^*(Y, \mathscr{O}_Y)$. And this implies $H^0(Y, \mathscr{O}_Y)=k$, i.e. $Y$ is connected, and $h^1(Y,\mathscr{O}_Y)=(a-1)(b-1)$, i.e. the genus of $Y$ is $(a-1)(b-1)$. So, if we choose $a=b=2$, $Y$ is of genus 1, and it is an elliptic curve!

And everything will be done if we point out the existence of smooth irreducible curve of type $(a,b)$ in $\mathbb{P}_1\times \mathbb{P}_1$. This can be done by Bertini's theorem and Serge's embedding.

Theorem 47 [Bertini's theorem]. Let $X$ be a non-singular variety on $\mathbb{P}^n$, then there exists a hyperplane $H$ in $\mathbb{P}^n$, not containing $X$, such that $H\cap X$ is a regular scheme.

Making use of this theorem, we can embed $Z$ into $\mathbb{P}^n$, where $n=ab+a+b$ as follows. First, we embedded $\mathbb{P}^1$ into $\mathbb{P}^a$ by sending $(x_0:x_1)$ to $(x_0^a:x_0^{a-1}x_1:...:x_1^a)$, and $\mathbb{P}^1$ into $\mathbb{P}^b$ by sending $(y_0:y_1)$ to $(x_0^b:x_0^{b-1}x_1...:x_1^b)$. And we then embed $\mathbb{P}^a\times \mathbb{P}^b$ into $\mathbb{P}^m$ via Serge's embedding. This sends $((s_0:...:s_a),(t_0:...:t_b))$ to $(...:s_it_j:...)$. Now using Bertini's theorem, let $H$ be the hyperplane in the theorem above, $H\cap \mathbb{P}_1\times \mathbb{P}_1$ is a curve in $\mathbb{P}^n$ defined by a bi-degree equation in terms of $x_0,x_1, y_0, y_1$ as $\sum_{i,j}c_{i,j}x_0^ix_1^{a-i}y_0^iy_1^{b-j}$. And due to the Bertini's theorem, it is regular, i.e. the local ring at every point is the regular local ring, this yields our curve is actually irreducible, and smooth.

[Abelian Varieties V.3] Riemann-Roch and Vanishing Theorem

(We will update the new index of theorems, etc..., as in my new PDF file, link here).

Before coming to the theorem of Riemann-Roch for abelian varieties, let us take a look on the cohomology groups of Poincare's bundle and Mumford's bundle. The circle of ideas mentioned in our previous notes still works, since for Poincare's bundle, $P|_{A\times \{L\}}=L$ for all $L\in Pic^0(A)$, and the restriction is never trivial unless $L$ is trivial. This yields the support of the higher direct image of the push-forward is of dimension 0, and hence, the higher cohomology group of this sheaf vanish. The same holds for Mumford's line bundle $\Lambda(L)$ in the case $L$ is ample. And the two line bundles related via the pull-back, due to our construction.

1. Cohomology groups of Poincare and Mumford's line bundles.

Let $P$ be the Poincare's bundle on $A\times \hat{A}$, and $p_2': A\times \hat{A}\to \hat{A}$ the projection on the second coordinate, by our earlier argument, for any line bundle $L\in Pic^0(A)$, and $L$ is not trivial, $P|_L=L$ is also not trivial. Using our Proposition 30, the cohomology groups of $P|_L$ vanish. And by Theorem 31, $Supp(R^qp'_{2,*}P)$ is of dimension 0, because it just has at most one point. This implies $H^p( \hat{A}, R^qp'_{2,*}P)$ vanish for all $p>0$. Now, making use of the Leray spectral sequence, we have

$$E_2^{p,q}=H^p( \hat{A}, R^qp'_{2,*}P)\Rightarrow H^{p+q}(A\times \hat{A}, P)$$

When $p=0$, this yields $E_2^{0,q}=R^qp'_{2,*}P\Rightarrow H^{q}(A\times \hat{A}, P)$. Look at the arrows of page 2 in our spectral sequence, we can see $0=H^1( \hat{A}, R^{q-2}p'_{2,*}P)\to E_2^{p,q}\to 0$, and hence $E_2^{p,q}$ is actually stable. This yields $R^qp'_{2,*}P\cong H^q(A\times \hat{A}, P)$, for all $q$.

Now, if $q>g$, the dimension of $A$, $R^qp'_{2,*}P$ vanishes. This yields $H^q(A\times \hat{A}, P)$ also vanishes whenever $q>g$. Making use of Serre's duality, we have $H^{2g-q}(A\times \hat{A}, P)\cong H^q(A\times \hat{A}, P^{-1})$, because the dualizing sheaf is isomorphic to the canonical bundle, which is trivial in the case $A$ is abelian. So, for all $q\ne g$, we have $H^q(A\times \hat{A}, P)=0$. This is the easy first part of the following

Theorem 36. Let $A$ be an abelian variety of dimension $g$, and $P$ the Poincare's sheaf of $A$, then $H^q(A\times \hat{A}, P)=0$, for $q\ne g$, and $H^g(A\times \hat{A}, P)=i_0(k)$, where $i_0(k)$ is the skyscrapper sheaf at $0$.

Proof. Theorem 9.1 in the book of G. Van der geer and Ben Moonen.

Now, we remember that if $L$ is an ample line bundle on $A$, then $A\times A\xrightarrow{(id, \lambda_L)}A\times \hat{A}$ is an isogeny, and $(id,\lambda_L)^*P=\Lambda(L)$, where $\Lambda(L)$ is the Mumford's line bundle of $L$. Take a look on the following commutative diagram

(DIAGRAM)

where $A\times \hat{A}\to \hat{A}$ is a proper morphism, and hence, separated of finite type, and $\lambda_L$ is an isogeny, which is flat. These things combined allows us to use the flat base change [HAG-...], which states

 $$\lambda_L^*R^ip'_{2,*}P=R^ip_{2,*}(id,\lambda_L)^*P=R^ip_{2,*}\Lambda(L)$$

And from Theorem 36, we have $R^ip_{2,*}\Lambda(L)=0$ if $i\ne g$, otherwise, its dimension is $\deg\lambda_L$. Because $\Lambda(L)|_{A\times \{a\}}=\lambda_L(a)$, which is trivial iff $a\in K(L)$, which is finite. Hence, using Theorem 31, $Supp(R^ip_{2,*}\Lambda(L))\subset K(L)$, which is of dimension $0$. This yields the higher cohomology groups of this sheaf vanish.

Making use of Leray spectral sequence as above, we have $R^ip_{2,*}\Lambda(L)\cong H^i(A\times A, \Lambda(L))$, which is $0$ if $i\ne g$, and is of dimension $\deg \lambda_L$ if $i=g$. We have proved the following

Proposition 37. Let $L$ be an ample line bundle on an abelian variety $A$ of dimension $g$, and $\Lambda(L)$ the Mumford's line bundle on $A\times A$, then $H^i(A\times A, \Lambda(L))=0$ if $i\ne g$, and $h^g( \Lambda(L))=\deg \lambda_L$. Furthermore, $\chi(\Lambda(L))=(-1)^gh^g(\Lambda(L))=(-1)^g\deg\lambda_L$.

2. Another computation for $\chi(\Lambda(L))$.

We will compute $\chi(\Lambda(L))$ by another way. Note that since $\Lambda(L)=m^*L\otimes p_1^*L^{-1}\otimes p_2^*L^{-1}$, we can make use of projection formula to make it have a simpler form. The map $f: A\times A\to A$ is a morphism, this yields by the projection formula

$$R^np_{2,*}(\Lambda(L))\otimes L\cong R^np_{2,*}(\Lambda(L)\otimes p_2^*L)\cong R^np_{2,*}(m^*L\otimes p_1^*L^{-1})$$

Because $Supp(R^np_{2,*}\Lambda(L))\subset K(L)$, which is finite, and $L$ can be trivialized on $K(L)$, this yields

$$R^np_{2,*}(m^*L\otimes p_1^*L^{-1})\cong R^np_{2,*}(\Lambda(L))\otimes L\cong R^np_{2,*}\Lambda(L)$$

If we restrict $m^*L\otimes p_1^*L^{-1}$ on $A\times \{a\}$, we will get $\lambda_L(a)$, which is trivial on $Pic^0(A)$ iff $a\in K(L)$, and hence, the support of this sheaf is of dimension 0, and the higher cohomology groups then vanish. Making use of Leray spectral sequence, we have $H^n(A\times A, m^*L\otimes p_1^*L^{-1})\cong R^np_{2,*}(m^*L\otimes p_1^*L^{-1})$. And hence, by our previous results, $H^n(A\times A,\Lambda(L))\cong H^n(A\times A, m^*L\otimes p_1^*L^{-1})$.

Now, the map $(m,p_1): A\times A\to A\times A$ sending $(a,b)$ to $(a+b,a)$ is an isomorphism. If we pullback $p_1^*L\otimes p_2^*L^{-1}$ via this map, we obtain $m^*L\otimes p_1^*L^{-1}$. To see this, we use the see-saw principle, for any line bundle $M$ on $A\times A$, we have $(m,p_1)^*M|_{\{a\}\times A}=\tau_a^*(M|_{A\times \{a\}})$, hence $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})|_{\{a\}\times A}=\tau_a^*L$. Also, $m^*L\otimes p_1^*L^{-1}|_{\{a\}\times A}=\tau_a^*L$, for all $a\in A$. Furthermore, $(m,p_1)^*M|_{A\times \{0\}}=M|_{\Delta}$, where $\Delta:=\{(a,a)\in A\times A\}$. From this, $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})|_{\Delta}$ is trivial. Also, $m^*L\otimes p_1^{*}L^{-1}|_{A\times \{0\}}$ is also trivial. This yields by the see-saw principle that $(m,p_1)^*(p_1^*L\otimes p_2^*L^{-1})\cong m^*L\otimes p_1^*(L^{-1})$. And hence, their cohomology groups agree. This yields

$$H^n(A\times A,\Lambda(L))\cong H^n(A\times A, p_1^*L\otimes p_2^*L^{-1})\cong \bigoplus_{p+q=n}H^p(A,L)\otimes H^q(A,L^{-1})$$

by the Kunneth's formula. From this,

$$h^n(\Lambda(L))=\sum_{p+q=n}h^p(L)h^q(L^{-1})$$

Using the result of Proposition 37, $h^n(\Lambda(L))=0$ for all $n\ne g$ yields for all $p$, $h^p(L)=0$ or $h^{n-p}(L)=0$ for $n\ne g$. On the other hand, $h^g(\Lambda(L))=\sum_{p=0}^nh^p(L)h^{n-p}(L^{-1})$. We will prove that there exists a unique index $i\le g$ such that both $h^i(L)$ and $h^{g-i}(L^{-1})$ are not zero. The existence of the index $i$ is clear. Assume that there exists $j\ne i$ such that both $h^j(L)$ and $h^{g-j}(L^{-1})$ are not zero. If $i<j$, then $g-j+i<g$, and both $h^i(L)$ and $h^{g-j}(L^{-1})$ are not zero, this is a contradiction. The case $i>j$ is similar. By Serre's duality,  for all $k$, we have $h^k(L)=h^{g-k}(L^{-1})$, and this yields there exists a unique $i\le g$ such that $h^i(L)\ne 0$. Such an $i$ is called the index of $L$, which is denoted $i(L)$. This yields $\chi(L)=(-1)^{i(L)}h^{i(L)}(L)$, and hence, $h^g(\Lambda(L))=h^{i(L)}(L)^2=\chi(L)^2$. And by Proposition 37, $\deg(\lambda_L)=\chi(L)^2$.

Combining things together, we get the following important

Theorem 38 (Vanishing Theorem). Let $L$ be an ample line bundle on an abelian variety $A$, then there exists a unique integer $i$ such that $h^i(L)\ne 0$, and $\deg(\lambda_L)=\chi(L)^2$.

Example 39. Again, we will prove the vanishing theorem for elliptic curves by direct computation. A divisor $D\in Div(E)$ is ample iff $\deg E>0$. This yields by R-R that $h^0(D)\ne 0$, and it is $\deg D$. Because the genus of $E$ is 1, it yields by Riemann-Roch again that $h^1(D)=0$, and for $n>1$, we have $h^n(D)=0$, due to the vanishing theorem of Grothendieck. From this, $\chi(D)=\deg D$. Now, $\lambda_D$ is the map $a\mapsto \tau_a^*D-D$. We can write $D=[P_1]+...+[P_n]$, and $\lambda_D(a)=[P_1-a]+...+[P_n-a]-[P_1]-...-[P_n]$. Via the isomorphism between $Pic^0(E)$ and $E$, $\lambda_D$ can be identified with the map $[n]$, the multiplication by $n$ map. And our previous result in Section 1 shows that $\deg(\lambda_D)=\deg [n]=n^2=\deg(D)^2=\chi(D)^2$.

3. Riemann-Roch's theorem.

We now turn to the Riemann-Roch's theorem for abelian varieties. It can be seen that for abelian varieties $A$ of dimension $g$, the canonical sheaf on $A$ is trivial, and the tangent bundle of $A$ is also trivial. This yields the $Td(A)=1$, where $Td(A)$ is the Todd's genus of the tangent sheaf of $A$. Let $L$ be any line bundle on $A$, this yields by the theorem of Hizerbruch-Riemann-Roch that

$$\chi(L)=\int_X ch(L)= \frac{\deg c_1(L)^g}{g!}$$

And this is the theorem of Riemann-Roch for abelian varieties. We can use it to deduce the following

Corollary 40. Let $f:A\to B$ be an isogeny of abelian varieties, then for any line bundle on $B$, $\chi(f^*L)=\deg f\chi(L)$.

Proof. Pull-back of a line bundle is a line bundle, so $\chi(f^*L)=\deg (c_1(f^*L)^g)/g!$. Using axioms (C2) in our previous post for Chern's classes, we have $f^*(c_1(L))=c_1(f^*(L))$. And use Proposition 10.2.6 (iv) (Gathmann's note), that states, for $f: X\to Y$ is a morphism, such that $f^*: A_*(X)\to A_*(Y)$ exists. Let $F$ be a vector bundle on $Y$, and $\alpha\in A_*(Y)$, we have

$$c_i(f^*F).f^*\alpha=f^*(c_i(F).\alpha)$$

Now, let $F\equiv L$, and $\alpha=c_1(L)$, we have

$$c_1(f^*L)^2=c_1(f^*L).c_1(f^*L)=c_1(f^*L)f^*(c_1(L))=f^*(c_1(L)^2)$$

Using induction, we get $c_1(f^*L^g)=f^*(c_1(L)^g)$. And by Riemann-Roch's theorem, it is sufficient to check $\deg f^*[P]=\deg f$. But it is clear, since if $f$ is separable, then the last identity holds. Otherwise, $f$ is a composition between a separable isogeny $h$ and a purely inseparable isogeny $g$. If we pull-back $P$ via $h$, we will get exactly $\deg(h)$ points, and for each $\deg h$ points, if we pull them back via $g$, we have exactly $\deg h$ points, with multiplicity $\deg g$ for each. And our desired identity follows from the fact that $\deg f=\deg g\deg h$.

(Q.E.D)

[Abelian Varieties V.2] Cohomology of Line Bundles on Abelian Varieties

We will begin our study about line bundles on abelian varieties. The main result of this note would be following theorem: if $L$ is an ample line bundle on an abelian variety $A$, then the map $\lambda_L$ is surjective. By keeping in mind the circle of ideas we have illustrated in the previous note, it is easier to keep track on the proofs.

1. Kunneth's formula and applications.

Theorem 1.1 (Kunneth's formula). Let $X,Y$ be projective varieties and $F,G$ coherent sheaves on $X,Y$, respectively. Denote $p_1: X\times Y\to X, p_2:X\times Y\to Y$ the projections, and $F\boxtimes G:=p_1^*F\otimes_{\mathscr{O}_{X\times Y}}p_2^*G$. Then 

$$H^n(X\times Y, F\boxtimes G)=\bigoplus_{p+q=n}H^p(X,F)\otimes H^q(Y,G)$$

Using Kunneth's formula, one can deduce the vanishing of cohomology group for non-trivial line bundles on $Pic^0(A)$, where $A$ is an abelian variety.

Proposition 1.2. Let $L\in Pic^0(A)$ be a non-trivial line bundle, then $H^n(A,L)=0$, for all $n\ge 0$, and $H^n(A\times A, L)=0$.

Proof. For $n=0$, we assume that $H^0(A,L)\ne 0$, this yields there exists an effective divisor $D\in Div(A)$, such that $L\cong \mathscr{O}_A(D)$. But then, since $L$ is anti-symmetric, we have $(-1)^*L\cong L^{-1}$, and hence $\mathscr{O}_A((-1)^*D)\cong \mathscr{O}_A(-D)$, and from this, $D+(-1)^*D=0$, and $D=0$, since $D$ is effective. This yields a contradiction, since $L\ne \mathscr{O}_A$.

We now look at the higher cohomology groups of $L$, let $i$ be the smallest positive integer such that $H^i(A,L)\ne 0$. Because $m^*L\cong p_1^*L\otimes p_2^*L$, we have by Kunneth's formula

$$H^n(A\times A, m^*L)\cong H^n(A\times A, p_1^*L\otimes p_2^*L)\cong \bigoplus_{p+q=n}H^p(A, L)\otimes H^q(A,L)$$

This yields by the smallest of $i$, and the zero cohomology group of $L$ vanishes, that $H^n(A\times A, m^*L)=0$. The diagram

$$A\xrightarrow{id\times \{0\}}A\times A\xrightarrow{m}A$$

has the composition of two arrows the identity map. This yields $$H^n(A)\to H^n(A\times A)\to H^n(A)$$ has the composition of two arrows again the identity map. From this, one has $H^n(A)=0$, for all $n$. (Q.E.D)

Let us take a look on the case of elliptic curves $E$, $Pic^0(E)$ consists of divisors of the form $D:=[P]-[O]$, and if $P\ne O$, it is of degree zero but not principal, and hence $h^0(E,D)=0$. And by Riemann-Roch, $h^1(E,D)=0$ again. Because $\dim E=1$, $h^n(E,D)=0$ for all $n\ge 2$.

The result of the proposition above is important and will be used very frequently throughout the note.

2. Two important corollaries from Mumford's book.

During these notes, we often use some techniques: pullback line bundles on an abelian variety $A$ to line bundles on $A\times A$ via the projection maps, or multiplication map, as well as restrict line bundles on $A\times A$ to $A\times \{a\}$, for some $a\in A$. In algebraic geometry language, it is the $A\times \{a\}$ is the fiber of $a$ via the projection map $p_2$, and $L|_{A\times \{a\}}$ is again the fiber of $a$ with respect to the invertible sheaf $L$ via the map $p_2$. The two theorems below gives us the relation between the cohomology groups of the fiber and the higher direct image via proper map between projective varieties. Note that we need the information about higher direct image to exact more information about higher cohomology groups via Leray spectral sequence.

Theorem 2.1 (Corollary 2, Page 50, Mumford's book). Let $f:X\to Y$ be a proper morphism between noetherian schemes, and $F$ is a coherent sheaf on $X$, and $Y$ is reduced and connected, then for all $p$, the following are equivalent:

(i) $y\mapsto \dim_{k(y)} H^p(X_y,F_y)$ is a constant function, where $X_y, F_y$ are fibers of $y$ over $f$.

(ii) $R^pf_*F$ is a locally free sheaf on $Y$, and for all $y\in Y$, $R^pf_*F\otimes_{\mathscr{O}_Y}k(y)\cong H^p(X_y,f_y)$.

If these conditions are satisfied, we have further that $R^{p-1}f_*F\otimes_{\mathscr{O}_Y} k(y)\cong H^{p-1}(X_y, f_y)$.

Now, we will use this fact to deduce some useful information about line bundles on abelian varieties.

Corollary/Lemma 2.2. Let $L$ be a line bundle on $A\times A$, such that $L_{A\times \{a\}}Pic^0(A)$ is non-trivial on $A\times A$, then $H^n(A\times A, L)$ vanishes for all $n$.

Proof. Due to Proposition 1.2, we have $H^n(A\times \{a\}, L|_{A\times \{a\}})=0$, for all $n$, this yields the map $a\mapsto h^p(A\times \{a\}, L|_{A\times \{a\}})$ is a constant function. Hence, by Theorem 2.1, we have $R^qp_{2,*}L$ is trivial. This yields $H^p(A, R^qp_{2,*}L)$ is trivial for all $n$. By Leray spectral sequence, there exists a spectral sequence such that

$$E_2^{p,q}=H^p(A,R^qp_{2,*}L)\Rightarrow H^{p+q}(A\times A, L)$$

And this yields $H^{n}(A\times A, L)$ is zero. (Q.E.D)

Now, if $L$ is an ample line bundle on $A$, and $M$ is any line bundle on $A$, let $N:=\Lambda(L)\otimes p_2^*M=m^*L\otimes p_1^*L^{-1}\otimes p_2^*L^{-1}\otimes p_2^*M$, where $\Lambda(L)$ is the Mumford line bundle on $A\times A$. Once we restrict $N$ to $A\times \{a\}$, we can see $N|_{A\times \{a\}}=\lambda_L(a)\in Pic^0(A)$, and $N|_{A\times \{a\}}$ is trivial iff $a\in K(L)$, which is finite. This yields by Proposition 1.2 and Theorem 2.1 that for any open subset $U\subset A\setminus K(L)$, $R^qp_{2,*}N|_U$ vanishes. Hence $Supp(R^qp_{2,*}N)\subseteq K(L)$, and it has dimension zero. This implies $H^p(A,R^np_{2,*}N)$ vanishes whenever $p\ge 1$. Now, making use of Leray's spectral sequence, we have

$$E_2^{p,q}=H^p(A,R^qp_{2,*}N)\Rightarrow H^{p+q}(A\times A, N)$$

When $p=0$, $E_2^{0,q}=H^0(A,R^qp_{2,*}N)\Rightarrow H^{q}(A\times A, N)$. Using complex sequence in the second page of the spectral sequence, we have

$$E^{1,q-2}=H^1(A,R^{q-2}p_{2,*}N)=0\to E_2^{0,q}\to 0$$

And hence, $E_{\infty}^{0,q}$ is actually stable at the second page. This yields $R^qp_{2,*}N\cong H^{q}(A\times A, N)$. And we have proved

Corollary/Lemma 2.3. Let $L$ be an ample line bundle on $A$, and $M$ is any line bundle on $A$. Let $N:=\Lambda(L)\otimes p_2^*M$ be a line bundle on $A\times A$, then $R^qp_{2,*}N\cong H^q(A\times A, N)$ for all $q$.
 
The following theorem also reflects the relation between higher direct image sheaves and the cohomology groups of fibers.

Theorem 2.4 (Corollary 4, page 53, Mumford's book). Let $X,Y,F$ be defined as in Theorem 2.1, if $R^kf_*F=0$, for $k\ge k_0$, then $H^k(X_y, F_y)=0$ for all $y\in Y$.


Combining all together, we are now ready for the proof of the main theorem.

Theorem 2.5. Let $L$ be an ample line bundle on $A$, then $\lambda(L)$ is a surjective map.

Proof. Assume that $\lambda(L)$ is not surjective, i.e. there exists $M\in Pic^0(A)$, such that $\lambda_L(a)\ne M$ for all $a\in A$, this yields $K_a:=\lambda_L(a)\otimes M^{-1}$ is a non-trivial line bundle on $Pic^0(A)$.

Let $N=\Lambda(L)\otimes p_2^*M^{-1}$. Once we restrict $N$ on $\{a\}\times A$, we get $K_a$, which is nontrivial in $Pic^0(A)$. By Corollary 2.2, $H^n(A\times A, N)$ vanishes. Again, if we restrict $N$ on $A\times \{a\}$, we get $\lambda_L(a)$, by Corollary 2.3, we have $R^qp_{2,*}N\cong H^q(A\times A, N)=0$, for all $q$. Using Theorem 2.4, we have $H^0(A\times \{0\}, N|_{A\times \{0\}})=0$. But it is a contradiction, since $\mathscr{O}_A(A)\ne 0$. Hence, $\lambda(L)$ is surjective. (Q.E.D)

[Abelian Varieties V.1] Leray Spectral Sequence

This is a very quick look on the theory of spectral sequence. It is very useful for our later development for studying cohomology groups of line bundles on abelian varieties. Our main results would be the Riemann-Roch and vanishing theorem. Since we cannot draw diagrams here, I will attach my hand-writing notes.

[Abelian Varieties IV] Dual Isogenies, Polarization and Weil's Pairing

1. Dual Isogenies. Let $f: A\to B$ be a homomorphism between abelian variety, then one has $(f,id): A\times \hat{B}\to B\times \hat{B}$ is a morphism. Let $P_B$ be the Poincare's bundle on $B\times \hat{B}$, and $M:=(f,id)^*P_B$ the pullback of $P_B$. One has
$$M|_{A\times \{ \hat{b}\}}=(f,id)^*P_B|_{A\times \{ \hat{b}\}}\cong f^*P_{ \hat{b}}\in Pic^0(A)$$
due to the following

Lemma 1.1. Let $f: A\to B$ be a homomorphism between abelian varieties, then $f^*: Pic^0(B)\to Pic^0(A)$ is well-defined, and it is a group homomorphism.

Proof. Let $L\in Pic^0(B)$, and for all $a\in A$, we have $\phi_{f^*L}(a)=\tau_a^*f^*L\otimes f^*L^{-1}$. But then, $\tau_a^*f^*=(f\circ \tau_a)^*=(\tau_{f(a)}\circ f)^*=f^*\tau_{f(a)}^*$. This yields $\phi_{f^*L}(a)=f^*(\tau_{f(a)}^*L\otimes L^{-1})$, and it is trivial, since $L\in Pic^0(A)$. Hence, $f^*L\in Pic^0(A)$. (Q.E.D)

From the definition of $M$, one can also see $M|_{0\times \hat{B}}$ is trivial. And due to the universal property of Poincare's bundle, there exists a unique map $ \hat{f}: \hat{B}\to \hat{A}$ such that $M=(id, \hat{f})^*P_A$, where $P_A$ is the Poincare's bundle on $A\times \hat{A}$. In particular, $\hat{f}$ is the unique map from $ \hat{B}$ to $ \hat{A}$ such that $(f,id)^*P_B\cong (id, \hat{f})^*P_A$. The map $ \hat{f}$ is called the dual of $f$. At the level of line bundles on $Pic^0$, one can see $f^*$ is actually the map sends $L\in Pic^0(B)$ to $f^*L\in Pic^0(A)$.

Now, let $L,M$ be ample line bundles on $A,B$, respectively, and $\lambda_L:A\to \hat{A}$, and $\lambda_M: B\to \hat{B}$ the corresponding maps. Then one can hope $\lambda_L= \hat{f}\circ \lambda_M\circ f$, but this does not hold in general. In fact, if we change $L$ by $f^*L$ then the last identity holds.

Lemma 1.2. Let $M$ be any line bundle on $B$, then $\lambda_{f^*M}= \hat{f}\circ \lambda_M\circ f$.

Proof. The proof is very similar to parts of the proof of Lemma 1.1. (Q.E.D)

Now, in case $f$ is an isogeny, i.e. $f$ is surjective and $\dim A=\dim B$, $\hat{f}$ is also an isogeny from $\hat{B}$ to $\hat{A}$, and even more interestingly, we have $\deg f=\deg \hat{f}$. And if $M$ is an ample line bundle on $B$, then $f^*M$ is an ample line bundle on $A$ (We will prove this fact later). This yields $K(\lambda_{f^*M})$ is finite, and it will be killed by some $n\in \mathbb{Z}$, i.e. $K(\lambda_{f^*M})\subseteq A[n]$. From this, one can define the map $\pi: \hat{A}\to A$ sending $\phi_{f^*M}(a)$ to $na$. It is an isogeny, and $\pi\circ \lambda_{f^*M}=[n_A]$. By the previous lemma, we have $[n_A]=\pi\circ \hat{f}\circ \lambda_M\circ f$, i.e. there exists an isogeny $g:B\to A$ such that $g\circ f=[n_A]$. For another version of this observation, we want $n=\deg f$.

Theorem 1.3. Let $f: A\to B$ be an isogeny between abelian varieties, with $n=\deg f$. Then there exists $g: B\to A$ such that $g\circ f=[n_A], f\circ g=[n_B]$.

Proof. First, it can be seen that any element in $\ker f$ is killed by $n$. This can be deduce easily if $f$ is an separable isogeny. Otherwise, one can factor $f$ as $h\circ g$ for some $g: A\to C$ is purely inseparable, and $h: C\to B$ is separable, and from this, $\ker f$ is one-to-one correspondent with $\ker h$. And hence, any element in $\ker f$ is killed by $\deg h$. This also implies any element in $\ker f$ is killed by $\deg f$, since $\deg f=\deg g\deg h$.

One can construct from this the map $g: B\to A$ that sends $f(a)$ to $na$. The composition of $g\circ f$ is exactly $n_A$, and $n_B\circ f=f\circ n_A=f\circ g\circ f$. Now, $f$ is an epimorphism of schemes implies that $n_B=f\circ g$. (Q.E.D)

2. Polarization. Let $A$ be an abelian variety, and $f: A\to \hat{A}$ is a homomorphism. We call $f$ a polarization if $f$ has the form $\phi_L$ for some ample line bundle $L$ on $A$, and $f$ is called principal polarization if $f$ is a polarization, and $\deg f=1$, i.e. $f$ is an isomorphism.

When $A$ is an elliptic curve, one has $P\mapsto [P]-[\infty]$ is the principal polarization. But this does not hold in general that any abelian variety has a principal polarization (See ???). But it is true that any abelian variety is isogeneous with a principal polarized abelian variety. This can be proved with the help of Weil's pairing. The rest of this note is devoted for main ideas of the proof.

Let $[m_A]:A\to A$ the multiplication by $m$ map, where $(m,char(k))=1$. It then induces the dual map $[m_{\hat{A}}]:\hat{A}\to \hat{A}$, and actually, $\hat{A}[m]\cong Hom(A[m], k^\times)=Hom(A[m], \mu_m)$, where $\mu_m$ is the group of $m$-th root of unity in $k^\times$. This yields the pairing

$$e_m: A[m]\times \hat{A}[m]\to \mu_m$$

Similar to the case of elliptic curves, the general Weil's pairing also has a nice interpretation. Let $L\in Pic^0(A)[m]$ (we now identify $\hat{A}$ and $Pic^0(A)$). Then because $L$ is anti-symmetric, we have $[m]^*L\cong L^{\otimes m}\cong \mathscr{O}_A$, i.e. there exists two rational functions $f_L,g_L$ on $A$ such that  $div(f_L)=L^{\otimes m}, div(g_L)=[m]^*L$. From this,

$$div(f_L\circ [m])=[m]^*L^{\otimes m}=([m]^*L)^{\otimes m}=div(g_L^m)$$

This implies there exists a constant $c\in k^\times$ such that $g_L^m(x)=c f_L(mx)$. And then, for all $a\in A[m]$, we have

$$g_L(x+a)^m=c f_L(mx+ma)=c f_L(mx)=g_L(x)^m$$

This implies $\frac{g_L}{g_L\circ\tau_a}$ is a $m$-th root of unity. We now define $e_m(a,L)=\frac{g_L}{g_L\circ\tau_a}$. By using this, we first prove

Lemma 2.1. Let $a\in A[mn], L\in Pic^0(mn)$, then $e_{mn}(a,L)^n=e_m(na,L^{\otimes n})$

Proof. Assume that $(mn)^*L=div(g)$, and $m^*(L^{\otimes n})=div(g')$, then we can see

$$div(g'\circ n)=n^*m^*(L^{\otimes n})=((mn)^*L)^{\otimes n}=div(g^n)$$

This yields there exists $c\in k^\times$ such that $g'(nx)=cg^n(x)$. And from this,
$$e_{mn}(a,L)^n=\frac{g^n(x)}{g^n(x+a)}=\frac{g'(nx)}{g'(nx+na)}=e_m(na,nL)$$
(Q.E.D)

By very similar proofs in Silverman, the general Weil pairing we have just defined is also bilinear, non-degenerate and compatible with the Galois action. Note that because an elliptic curve $E$ has principal polarization, that mean we can identify $Pic^0(E)$ and $E$, and the Weil pairing now can be defined as $e_m: E[m]\times E[m]\to \mu_m$, and it is skew-symmetric, i.e. $e_m(a,b)=e_m(a,b)^{-1}$.

Now, let $\lambda: A\to \hat{A}$ be any homomorphism, then we can define the pairing
$$e_m^\lambda:A[m]\times A[m]\to \mu_m$$
by sending $(a,a')$ to $e_m(a,\lambda(a'))$. Recall that for any $L\in Pic(A)$, one can construct $\lambda_L: A\to Pic^0(A)$ by sending $a$ to $\tau_a^*L\otimes L^{-1}$, and if $\lambda_L\equiv 0$ iff $L\in Pic^0(A)$. This yields if $L,L'\in Pic(A)$, and $L\equiv L'\mod Pic^0(A)$, then $\lambda_L,\lambda_{L'}$ are just the same. This observation gives right to define the Neron-Severi's group of $A$, $NS(A):=Pic(A)/Pic^0(A)$. Now, an important fact mentioned in Milne's note is that if $\lambda$ comes from $\lambda_L$, for some $L\in NS(A)$, then $e_m^\lambda$ is skew-symmetric.

We now reduce to the case $\lambda$ is a polarization, i.e. $\lambda$ comes from $\lambda_L$ for some ample line bundle $L\in Pic(A)$. We can make change a little bit on the definition of the Weil's pairing, by defining it on the kernel of $\lambda$. Assume that $\ker\lambda$ is killed by some integer $m$, then for all $a,a'\in \ker\lambda$, and $b\in A$ such that $mb=a'$, then we define

$$e^\lambda(a,a'):=e_m(a,\lambda(b))$$

We will prove that this definition makes sense, by proving that it is independent on $m,b$. First, $m\lambda(b)=\lambda(mb)=\lambda(a)=0$, so $\lambda(b)\in Pic^0(A)[m]$, so $e_m(a,\lambda(b))$ is well-defined. Second, if we replace $m$ by $mn$, and replace $b$ by some $b'$ such that $mnb'=a$. Let $a_1\in A$ such that $na_1=a$, then it follows by Lemma 2.1 that

$$e_{mn}(a,\lambda(b'))=e_{mn}(na_1,\lambda(b'))=e_{mn}(a_1,\lambda(b'))^n=e_{m}(a,\lambda(nb'))$$

And then by the skew-symmetric property, we have,

$$e_m(a,\lambda(nb'))/e_m(a,\lambda(b))=e_m(a,\lambda(nb'-b))=e_m^\lambda(a,nb'-b)$$
$$=e_m^\lambda(nb'-b,a)^{-1}=e_m(nb'-b,\lambda(a))=1$$

An important application of $e^\lambda$ is that

Theorem 2.2. Let $\lambda: A\to \hat{A}$ be a polarization, and $f: A\to B$ an isogeny, then $\lambda=\lambda_{f^*L}$, for some ample line bundle $L$ on $B$ if and only if $\ker f\subseteq \ker \lambda$, and $e^\lambda$ is trivial on $\ker f\times \ker f$.

Now, with specific $a,a'\in \ker\lambda$, we can even choose some $m$ such that $m$ kills both $a,a'$, then the definition above still works, and we also obtain $e^\lambda(a,a')$ in this case. In the case $a\equiv a_m,a'\equiv a'_m$ is killed by $l^m$, for some prime $l\ne char(k)$, we have

$$e^\lambda(a_m,a'_m)=e_{l^m}(a_m,l^m\lambda(a'_{2m}))=e_{l^m}(l^ma_{2m}, l^m\lambda(a'_{2m}))=e_{l^{2m}}^\lambda(a_{2m},a'_{2m})$$

And this then yields $e^\lambda$ is skew-symmetric on $A[l^m]$, for all prime $l\ne char(k)$. Using this, we obtain our main result of this note

Theorem 2.3. Any abelian variety is isogeneous with a polarized abelian variety.

Proof. Let $\lambda: A\to \hat{A}$ be a polarization with $\deg \lambda$ is co-prime to $char(k)$. If $\lambda$ is principal, then there is nothing to prove. Otherwise, there exists a prime $l|\deg\lambda$, and let $K$ be the subgroup of $\ker \lambda$, whose element is $l$. Then because $e^\lambda$ is skew-symmtric on $N$, it is trivial on $N\times N$. Let $B:=A/N$ be another abelian variety, and $f: A\to B$ is the projection map. Then $N=\ker f\subseteq \ker\lambda$, and $e^\lambda$ is trivial on $N\times N$. This follows from Theorem 2.1 that $\lambda=\lambda_{f^*L}$ for some ample line bundle $L$ on $B$. This follows $\lambda_{f^*L}=\hat{f}\circ \lambda_L\circ f$, i.e. $\deg \lambda=\deg \lambda_L.(\deg f)^2$, and now, this implies $B$ has a polarization of degree $\deg \lambda/l^2<\deg \lambda$. Continue this process, we obtain an abelian $C$, which has a principal polarization. (Q.E.D)

Some questions up to this:

1.For an abelian variety $A$, and a finite subgroup $N$ of $A$, how can we produce an abelian variety $B\cong A/N$.

2. Can we produce an algorithm for Theorem 2.3?

3. What about abelian varieties with complex multiplciations? They are all principally polarized?

[Abelian Varieties III] Dual Varieties and Poincare's Bundles

The notions of dual variety and Poincare's bundle of an abelian variety will be the main theme in this series. Based on this, we can study vector bundles on abelian varieties, with an important tool, so called Fourier-Mukai's transform.

1. More about $Pic^0$. In this section, we will study more on $Pic^0(A)$. We first prove the following

Theorem 1.1. Let $A$ be an abelian variety then the following holds:

1. For all line bundle $L\in Pic(A)$, $\tau_a^*L\otimes L^{-1}$ (in this previous note, it is $\lambda_L(a)$) is in $Pic^0(A)$.

2. Any line bundle $L\in Pic^0(A)$ is anti-symmetric, and in particular, $n^*L\cong L^n$, where $n$ is the multiplication by $n$ map on $A$.

3. For any variety $S$, and $L$ is a line bundle on $A\times S$, such that $L|_{A\times \{s_0\}}\in Pic^0(A)$ for some point $s_0\in S$, then for all $s\in S$, we have $L_s:=L|_{A\times \{s\}}\in Pic^0(A)$.

4. If $L\in Pic(A)$ with $K(L)$ is finite (or equivalently $L$ is an ample divisor), then $\lambda_L$ is surjective.

Proof. 

1. In term of divisors, we will prove $D':=\tau_a^*D-D\in Pic^0(A)$. Based on Proposition 3.3 of the previous note, we need to show $\tau_b^*D'\sim D'$, i.e. $\tau_b^*\tau_a^*D-\tau_b^*D\sim \tau_a^*D-D$. This is equivalent to say $\tau_{a+b}^*+D\sim \tau_a^*D+\tau_b^*D$. But this follows directly from theorem of the square.

2. It just follows from the final part of the proof of Theorem 2.2 in our previous note, where we prove that $L\otimes (-1)^*L$ is trivial for $L\in Pic^0(A)$.

3. It is an application of the theorem of the cube. Consider three maps $(a,b,s)\mapsto (a+b,s)$, $(a,b,s)\mapsto (a,s)$ and $(a,b,s)\mapsto (b,s)$ from $A\times A\times S$ to $A\times S$, which are denoted $m, p, q$ respectively. Let $M=m^*L\otimes p^*L^{-1}\otimes q^*L^{-1}$ be a line bundle on $A\times A\times S$. Then it can be seen $M|_{\{0\}\times A\times S}$ and $M|_{A\times \{0\}\times S}$ are trivial.

Also, for all $s\in S$, we have $m^*L|_{A\times A\times \{s\}}=\{(a,b,s_0,l)|(a+b,s_0)=\pi(l)\}$, where $\pi: L\to A\times S$ is the line bundle map. If we also denote $m:A\times A\to A$ the multiplication map, then $m^*L_s=\{(a,b,l)|a+b=\pi_l|_{A\times \{s\}}\}=m^*L|_{A\times A\times \{s\}}$. We similarly obtain $p^*L|_{A\times A\times \{s\}}=p^*L_s$, and $q^*L|{A\times A\times \{s\}}=q^*L_s$

Also, by assumption, $M|_{A\times A\times \{s_0\}}$ is also trivial (by Proposition 3.3 in our previous note). This yields $M$ is a trivial line bundle by the theorem of the cube. Now, if we restrict it to $A\times A\times \{s\}$ for any $s\in S$, it is also trivial. Hence $L|_{A\times \{s\}}\in Pic^0(A)$ for all $s\in S$.

4. It is a key fact for our applications from later on. The proof of this can be found in Mumford's book on abelian varieties.

(Q.E.D)

Recall that in the last proposition of the previous note, we state that $L\in Pic^0(A)$ iff $m^*L\cong p^*A\otimes q^*A$ on $A\times A$. In general, for any $L\in Pic(A)$, let us denote $\Lambda(L):=m^*L\otimes p^*L^{-1}\otimes q^*L^{-1}$. It can be seen that $Pic^0(A)$ is also the isomorphism classes of line bundles $L$ such that $\Lambda(L)$ is trivial. From this definition, let $M\in Pic^0(A)$, we have 

$$(\Lambda(L)\otimes p^*M)|_{A\times \{a\}}=\tau_a^*L\otimes L^{-1}\otimes M=\lambda_L(a)\otimes M$$

If we choose $L$ an ample divisor, so that $\lambda_L$ is surjective (Theorem 1.1(4)), we can choose $a\in A$ such that $\lambda_l(a)=M^{-1}$. This yields $(\Lambda(L)\otimes p^*M)|_{A\times \{a\}}=\mathcal{O}_A$, and if we choose $a=0$, then $(\Lambda(L)\otimes p^*M)|_{A\times \{0\}}=M$.

2. An application to algebraically equivalent of divisors. Two divisors $L,M\in A$ are called algebraically equivalent if there exists a variety $S$ with two points $s_1,s_2\in S$ and $\mathcal{L}$ on $A\times S$ such that $\mathcal{L}|_{A\times\{s_1\}}=L$, and $\mathcal{L}|_{A\times \{s_2\}}=M$. Via this definition, we can see any divisor in $Pic^0(A)$ is algebraically equivalent to zero. 

Lemma 2.1. With the assumption above, if $L\otimes M^{-1}$ is algebraically equivalent with the trivial sheaf, then $L$ is algebraically equivalent to $M$.

Proof. By definition, there exists a variety $S$ together with two points $s_1,s_2\in S$ and a line bundle $\mathcal{L}$ on $A\times A$ such that $\mathcal{L}|_{A\times \{s_1\}}=L\otimes M^{-1}$, and $\mathcal{L}|_{A\times \{s_1\}}=\mathcal{O}_A$. This yields $\mathcal{L}|_{A\times \{s_1\}}\otimes M=(\mathcal{L}\otimes p^*M)|_{A\times \{s_1\}}=L$, and similarly, $(\mathcal{L}\otimes p^*M)|_{A\times \{s_2\}}=M$. And by definition, $L$ and $M$ are algebraically equivalent (Q.E.D)

Combining these things, one can prove

Theorem 2.2. Let $L,M$ be line bundles on an abelian varieties $A$, then the following are equivalent:

1. $L, M$ are algebraically equivalent.

2. $L\otimes M^{-1}\in Pic^0(A)$.

3. $\lambda_L=\lambda_M$.

Proof.  The equivalence between 2. and 3. easy follows from the definition of $Pic^0$. Assume 1, by definition, there exists a variety $S$ with two points $s_1, s_2\in S$, and a line bundle $L$ on $A\times S$ such that $L_{s_1}=L, L_{s_2}=M$. This yields $L_{s_2}\otimes M^{-1}$ is the trivial bundle, which lies in $Pic^0(A)$. And also, $L_{s_2}\otimes M^{-1}$ is in fact $(L\otimes p^*M^{-1})_{s_2}$. By Theorem 1.1 (3), we have $(L\otimes p^*M^{-1})_{s_1}\in Pic^0(A)$, or equivalently, $L\otimes M^{-1}\in Pic^0(A)$. This yields 1 implies 3.

Now, assume 3, we can see $L\otimes M^{-1}\in Pic^0(A)$. But then, any line bundle on $Pic^0(A)$ is algebraically equivalent to zero. This follows from Lemma 2.1 that $L$ is algebraically equivalent to $M$. (Q.E.D)

3. Dual variety and Poincare's bundle. 

We require the characteristic of our base field is zero (for positive characteristic, the construction is basically the same, but the techniques become much more complicated). Now, if $L$ is an ample line bundle on an abelian variety $A$, we know from Theorem 1.1 that the map $\lambda_L: A\to Pic^0(A)$ is surjective. This then yields $A/\ker\lambda_L\cong Pic^0(A)$. We now define $A/\ker\lambda_L$ the dual variety of $A$, which is denoted by $\hat{A}$.

Denote $\pi: A\to \hat{A}$ the canonical projection map. This then yields the map $(1,\pi):A\times A\to A\times \hat{A}$. Let $\Lambda(L):=m^*L\otimes p^*L^{-1}\otimes q^*L^{-1}$ defined as above, which is a line bundle on $A\times A$, then there exists a line bundle $P$ on $A\times \hat{A}$ such that $(1,\pi)^*P=\Lambda(L)$ (see the book of Mumford for the proof of this fact). The line bundle $P$ is called the Poincare's bundle with $\tau: P\to A\times \hat{A}$ the line bundle map.

For all $a\in A$, we have $(id,\pi)^*P|_{A\times \{a\}}\cong \{(x,p)|\tau(p)=(x,\pi(a))\}=P|_{A\times \{\pi(a)\}}$. On the other hand, one has $(id,\pi)^*P|_{A\times \{a\}}=\Lambda(L)|_{A\times \{a\}}=\tau_a^*L\otimes L^{-1}=\lambda_L(a)\in Pic^0(A)$. So

(1)   $P_{\pi(a)}:=P|_{A\times \pi(a)}\cong \lambda_L(a)\in Pic^0(A)$

Because $L$ is ample, $\lambda_L$ is surjective. This yields $P$ actually parametrizes all line bundles on $Pic^0(A)$ via the isomorphism $Pic^0(A)\cong \hat{A}$ by its restriction to $P_{A\times \{\pi(a)\}}$ for all $a\in A$. This is an important property of Poincare's bundle. Furthermore,

(2)   $P|_{\{0\}\times A}$ is trivial.

By See-saw principle, the properties (1) and (2) uniquely determine Poincare's bundle on $A\times \hat{A}$. Also, we have the universal property for Poincare's bundle as follows. If $S$ is any variety and $L$ is a line bundle on $S$ such that $L|_{A\times \{s\}}\in Pic^0(A)$ for one (and hence, all, by Theorem 1.1) $s\in S$, and $L|_{\{0\}\times S}$ is trivial, then there exists a unique map $\phi: S\to \hat{A}$ such that $L=(id,\phi)^*P$.

[Abelian Varieties II] $Pic^0$

In this post, we will construct the group $Pic^0$ of an abelian variety, that extends the case of elliptic curves. This extension will lead to some important constructions: dual varieties and Weil's pairings.

1. Ample divisors. 

We recall some basic facts about divisors on curves. Let $D$ be a divisor on a non-singular complete curve $X$. Let $n+1:=l(D)$ be the dimension of the Riemann-Roch's space $L(D)$. We know from our previous notes that there exists a morphism $\phi_D$ from $X$ to $\mathbb{P}^{n-1}$ in the case $D$ is base-point free, defined by $x\mapsto (f_0(x):...:f_{n-1}(x))$, where $(f_0,...,f_{n-1})$ is the basis of $L(D)$. The same things hold for higher dimensional non-singular complete varieties. And $D$ is called very ample divisor if $\phi_D$ is a closed immersion, i.e. $\phi_D(X)$ is a closed subvariety of $\mathbb{P}^n$ and it is isomorphic to $X$. For curves, $\phi_D$ is a closed immersion iff $l(D-P-Q)=l(D)-2$, for all points $P,Q\in X$ (Note that: $l(D-2P)=l(D)-2$ implies that $\phi_D$ separates tangents, and $l(D-P-Q)=l(D)-2$, for $P\ne Q$ implies that $\phi_D$ separates points).

Definition 1.1. A divisor on a non-singular complete variety $X$ is ample if $nD$ is very ample for some $n> 0$.

We will prove some statements related to ample divisors.

Proposition 1.2. Let $X$ be a non-singular complete variety and $D,D'$ are ample divisors in $X$, the the following holds

1. $nD$ is ample for all $n> 0$.

2. $D+D'$ is ample.

3. Assume that $X$ is an abelian variety, $D+(-1)^*D$ is also ample.

4. If $D$ is trivial, then $D$ is an ample divisor if $X$ is a point.

5. If $X$ is a curve, then $D$ is ample iff $\deg D>0$

Proof. The first statement follows easily from the fact that $L(D)\subset L(nD)$, for all $n>0$. For the second, there exists some $n>0$ such that both $nD$ and $nD'$ are very ample. This yields $l(nD')\ge 1$, and therefore, there exists an effective divisor $D''$ such that $nD'\sim D''$. Now, $L(nD)\subset L(nD+D'')$, and so, $nD+D''$ is also very ample. This yields $nD+nD'$ is also very ample (since $nD+D''\sim nD+nD'$). By definition, $D+D'$ is an ample divisor.

For the third statement, because $(-1)$ is an isomorphism, $(-1)^*D$ is also an ample divisor on $X$. From the second statement, $D+(-1)^*D$ is an ample divisor. Next, we can see $D\sim 0$, and hence $nD\sim 0$, for all $n>0$, and in particular, $l(nD)=1, \forall n>0$. We choose $n>0$ such that $nD$ is very ample. The definition implies that $X\to \{pt\}\equiv \mathbb{P}^0$ is a closed immersion. Hence, $X$ is just a point.

Finally, assume that $X$ is a curve and $D$ is ample, if $\deg D<0$, then $\deg nD<0$, for all $n\ge 0$, and $L(nD)=\{0\}$ in this case. So, $D$ cannot be ample. Assume that $\deg D=0$, then $\deg nD=0$, and $l(nD)\ge 1$, for all $n>0$. If $D$ is ample, then we choose $n$ such that $nD$ is very ample. And in this case, $X\cong \{pt\}$, a contradiction. Hence, if $D$ is an ample divisor, $\deg D>0$. Conversely, if $\deg D>0$, Riemann-Roch theorem implies that for sufficient large $n$, $l(nD-P-Q)=l(nD)-2,\forall P,Q\in X$. This yields $nD$ is very ample.

(Q.E.D)

2. $Pic^0$ of an abelian variety from a special point of view.

We now turn to the construction of $Pic^0$, and give full proofs for the case of elliptic curves. The general case will be revisited in the next section. We first need the theorem of square.

Theorem 2.1 (Theorem of square). Let $A$ be an abelian variety, $\tau_a$ the translation by $a$, for $a\in A$, and $L$ a line bundle on $A$. Then for all $a,b\in A$, $\tau_{a+b}^*L\otimes L\cong \tau_a^* L\otimes \tau_b^*L$.

Proof. It is just an application of the theorem of the cube. Now, using Corollary 2.3 in the previous note for $f=id, g=(x\mapsto a), h=(x\mapsto b)$. This will yield $f+g+h=\tau_{a+b}$, $f+g=\tau_a, f+h=\tau_b$, and $g,h,g+h$ are constant map. This implies $g^*L, h^*L, (g+h)^*L$ are trivial line bundles. By Corollary 2.3, the line bundle $\tau_{a+b}^*L\otimes \tau_a^*L^{-1}\otimes \tau_b^*L^{-1}\otimes L$ is trivial. (Q.E.D)

Now, $\tau_{a+b}^*L\otimes L\cong \tau_a^*L\otimes \tau_b^*L$ implies that $\tau_{a+b}^*L\otimes L^{-1}\cong (\tau_a^*L\otimes L^{-1}) \otimes (\tau_b^*L\otimes L^{-1})$. We can see from this that the map $\lambda_L: A\to Pic(A)$ defined by $a\mapsto \tau_a^*L\otimes L^{-1}$ is a homomorphism (in term of divisor, this is the map $\lambda_D: a\mapsto \tau_a^*D-D$). An important theorem is

Theorem 2.2. Let $D$ be a divisor on $A$ such that $L(D)\ne \{0\}$. Then $D$ is ample iff $\ker\lambda_D$ is finite.

Proof. We will prove this theorem in the case of elliptic curves. We know from Proposition 1.2 (5) that $D$ is an ample divisor iff $\deg D>0$. We assume that $\deg D=n\ge 1$. Then Riemann-Roch theorem implies that $l(D)=\deg D>0$, i.e. there exists an effective divisor $D'$ such that $D\sim D'$. And we can write $D=[P_1]+...+[P_n]$. Now, $\tau_a^*D=[P_1-a]+...+[P_n-a]$, and hence $\tau_a^*D\sim D$ iff $na=O$, i.e. $a\in E[n]$. And in this case $\ker\lambda_D=E[n]$. But we have known in our previous note that $E[n]$ is finite. So, $D$ is ample implies $\ker\lambda_D$ is finite.

Conversely, assume that $\ker\lambda_D$ is finite and $L(D)\ne \{0\}$, we will prove that $\deg D>0$. Because $L(D)=\{0\}$ for all $D$ such that $\deg D<0$, and $D$ is not principal, we need to prove the converse for $D\sim 0$. But this follows directly from a simple observation: when we pull back a principal divisor, we will get a principal divisor, so $\tau_a^*D-D$ is principal for all $a\in A$. This is a contradiction to the finiteness of $D$. Hence, $\deg D>0$. (Q.E.D)

This theorem gives us a criterion to know when $D$ is an ample divisor on an abelian variety. In fact, it is proved that if $D$ is an ample divisor on $A$, then $3D$ is very ample (easy to check for elliptic curves, by R-R). Using this, one can see that $A$ can be embedded into projective space, i.e. abelian varieties are projective.

We now care about for which $D$, the map $\lambda_D$ is zero, i.e. the kernel in this case is not finite anymore. Again, we can take a look on elliptic curves. First, if $\deg D>0$, the previous theorem implies that $\ker\lambda_D$ is always finite. Next, if $\deg D<0$, then we can write $D=[P_1]+...+[P_n]-[Q_1]-...-[Q_m]$, where $m>n$, and $\tau_a^*D=[P_1-a]+...+[P_n-a]-[Q_1-a]-...-[Q_m-a]$. This yields $\tau_a^*D\sim D$ iff $(m-n)a=0$, i.e. $a\in E[m-n]$. And in this case $\tau_D$ is also non-zero. In the case $\deg D=0$, one can write $D=[P_1]+...+[P_n]-[Q_1]-...-[Q_n]$, and by similar argument, we can see $\tau_a^*D=[P_1-a]+...+[P_n-a]-[Q_1-a]-...-[Q_n-a]$. And $sum(\tau_a^*D-D)=\infty$, and $\deg(\tau_a^*D-D)=0$. This yields $\tau_a^*D-D$ is principal for all $a\in A$. And hence, $\lambda_D$ is the zero map iff $\deg D=0$ in the case of elliptic curves.

Now, $Pic^0(A)$ is defined to be the isomorphism classes of line bundle $L$ such that $\lambda_L$ is the zero map. In the case of elliptic curve, any divisor of degree zero is linearly equivalent to a unique divisor of the form $[P]-[\infty]$. So, in this case, $Pic^0(A)=\{[P]-[\infty]|P\in A\}$ and this gives a bijective map between $Pic^0(A)$ and $A$.

3. Proof of Theorem 2.2 in the general case.

We will prove the general version of Theorem 2.2 in the general case. Recall that we already know about theorem of the square and theorem of the cube. We now turn to another theorem, which also play an important role in studying abelian varieties, so called the see-saw principle. We first begin with the following

Theorem 3.1. Let $V,T$ be varieties, where $V$ is complete, and $L$ is a line bundle on $V\times T$ such that $L_t:=L|_{V\times \{t\}}$ is trivial for all $t\in T$. Then there exists a line bundle $N$ on $T$ such that $L=q^*T$, where $q:V\times T\to T$ the projection on the second coordinate.

Proof. See Milne's note (Theorem 5.16)

Now, assume that $L,M$ are line bundles on $V\times T$ such that $L_t=M_t$ for all $t\in T$. Then it can be seen $(L\otimes M^{-1})_t$ is trivial. This yields by the previous theorem that $L\otimes M^{-1}=q^*N$, for some line bundle $N$ on $T$, and $L=M\otimes q^*N$. Now, if we assume further that there exists a point $v\in V$ such that $L|_v=M|_v$, this yields $(q^*N)_v$ is trivial. But then, $(q^*N)_v=\{(v,t,n)\in \{v\}\times T\times N|\pi(n)=q(v,t)=t\}\cong \{(\pi(n),n)\in T\times N\}\cong N$. So, $N$ is trivial, and hence, $q^*N$ is also trivial. We have proved the see-saw principle.

Corollary 3.2 (See-Saw Principle). Let $V,T$ be varieties, where $V$ is complete, and $L, M$ are line bundles on $V\times T$ such that $L_t=M_t$ for all $t\in T$. Furthermore, if there exists a point $v\in V$, such that $L_v=M_v$, then $L=M$.

We now use it to study more about the kernel of $\lambda_L$ in the second section. First, let $m:A\times A\to A$ the multiplication map, $p: A\times A\to A$ the projection onto the first coordinate, and $L$ is a  line bundle on $A$. We can look at the line bundle $m^*L\otimes p^*L^{-1}$ on $A\times A$, and define $K(L):=\{a\in A|(m^*L\otimes p^*L^{-1})_{A\times \{a\}} \text{ is trivial}\}$. But then, when we restrict $m^*L$ on $A\times \{a\}$, it is $\tau_a^*L$-the pullback of the translation map by $a$, and when we restrict $p^*L^{-1}$ on $A\times \{a\}$, it is exactly $L^{-1}$. So, in fact, $K(L)=\{a\in A| \tau_a^*L\otimes L^{-1} \text{ is trivial}\}=\ker \lambda_L$. And hence, $Pic^0(A)$ can be defined as isomorphism classes of line bundles $L$, such that $K(L)$ is trivial.

Proposition 3.3. The following are equivalent:

1. $K(L)=A$.

2. $\tau_a^*L \cong L$ for all $a\in A$.

3. $m^*L\cong p^*L\otimes q^*L$.

Proof. The equivalence between $(1)$ and $(2)$ follows directly from our earlier discussion. Assume (1), we can see $(m^*L\otimes p^*L^{-1})|_{V\times \{a\}}$ is trivial for all $a\in A$. Also, $(q^*L)|_{V\times \{a\}}$ is trivial for all $a\in A$. Furthermore, $m^*L|_{\{0\}\times V}$ is $L$, and $p^*L^{-1}|_{\{0\}\times A}$ is trivial. And $q^*N|_{\{0\}\times A}$ is $L$. So, due to the see-saw principle, we have $m^*L\otimes p^*L^{-1}\cong q^*L$. Conversely, assume (3), for all $a\in A$, the restriction $q^*L|{A\times \{a\}}$ is just trivial. So, $m^*L\otimes p^*L^{-1}$ is trivial on $A\times \{a\}$ for all $a\in A$. This yields $K(L)=A$. (Q.E.D)

We are now ready for half of the proof of Theorem 2.2. Assume that $L$ is ample, we will prove that $K(L)$ is finite. Let $B$ be a connected component containing $0$ in $K(L)$. It is also an abelian variety, which we will call it $B$. We now have $L_B$ is also ample, and $K(L_B)=B$. From Proposition 3.3, we have $m^*L_B\otimes p^*L_B^{-1}\otimes q^*L_B^{-1}$ is trivial on $B\times B$.

But then, consider the map $(1,-1): B\to B\times B$ that sends $b\mapsto (b,-b)$. The pullback $(1,-1)^*m^*L_B^*$ is trivial, $(1,-1)^*p^*L_B=L_B$, and $(1,-1)^*q^*L_B^{-1}=(-1)^*L_B^{-1}$. And so, $L_B\otimes (-1)^*L_B^{-1}$ is trivial on $B$. And it is also ample (Proposition 1.2 (3)). This yields $B$ is just a point (Proposition 1.2 (4)). This yields the dimension of $B$ is just 0, and it is finite.

[Abelian Varieties I] Rigidity Theorem and Theorem of the Cube

We now turn to discuss about the higher dimensional analogue of elliptic curves: abelian varieties. Most of results from elliptic curves presented in the book of Silverman can be generalized to abelian varieties, with more abstract proofs. In this post, we will prove some fundamental results in the theory of elliptic curves based on the algebro-geometric techniques.

Theorem 0. Let $E_1, E_2$ be elliptic curves defined over a field $k$, and $\alpha: E_1\to E_2$ is a regular map, then $\alpha$ is a composition of a homomorphism and a translation map.

Theorem 1. Let $E$ be an elliptic curve defined over an algebraically closed field $k$. Denote $n: E\to E$ the multiplication by $n$ map, then $\deg n=n^2$, and $n$ is an unramified map iff $(n,char(k))=1$. And in this case, $E[n]$-the $n$-torsion subgroup of $E$ has $n^2$ elements and $E[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$.

1. Rigidity theorem. First, let us define what an abelian variety is. One can see for elliptic curves, the map $m: E\times E\to E$ sends $(p,q)$ to $p+q$ and $i: E\to E$ send $p\to -p$ are given by polynomials, and they are actually regular maps, and it satisfies the group law on $E$ with an identity element $0\in E$. Now, an abelian variety is defined as a complete connected variety $A$ such that there exists regular maps $m: A\times A\to A$, $i: A\to A$ together with an element $0\in A$ such that $(m,i,0)$ satisfies the group law in $A$.

From the definition, an abelian variety is automatically smooth, for the set of non-singular points on $A$ is an open dense subset. If we choose a point $p\in A$ is smooth, then for all point $q\in A$, the translation map $\tau_{-p+q}$, is an automorphism of $A$. This yields $q$ is also a smooth point of $A$. Hence, $A$ is a smooth variety. An abelian variety is defined to be a complete variety, but it can be proved that $A$ is projective. This result is due to Weil.

We should recall the definition of an elliptic curve $E$ over a field $k$: it is a non-singular projective curve of genus 1 with at least one rational point, denoted $O$. The Riemann-Roch's theorem implies that there exists a bijective map $E\to Pic^0(E)$ by sending $P\to [P]-[O]$, and the group structure on $E$ is induced naturally from the group structure on $Pic^0(X)$. It is abelian. However, it is difficult to prove the converse.

Theorem 1.1. A projective curve with group law defined by regular maps is actually an elliptic curve.

Proof. The proof is presented on Milne's note on abelian varieties, introduction pages. It is an amazing proof, where one has to use Lefschetz's fixed point formula, for etale cohomology. It is a very motivated proof for those who wants to study etale cohomology (including me). (Q.E.D)

As we can see, the group law on an abelian variety of dimension 1 (elliptic curve) turns out to be commutative. This also holds for higher dimension abelian varieties.

Theorem 1.2 (Rigidity Theorem). Let $U,V,W$ be varieties, with $V$ is complete, $V\times W$ is connected and $\alpha: V\times W\to U$ is a regular map. Assume that there exists $u_0\in U, v_0\in V,w_0\in W$ such that $\alpha(\{v_0\}\times W)=\{u_0\}=\alpha(V\times \{w_0\})$, then $\alpha(V\times W)=\{u_0\}$.

Proof. (See Milne's note on Abelian Varieties-Theorem 1.1)

By using Rigidity Theorem, we will prove the general version of Theorem 0.

Corollary 1.3. Let $\alpha: A\to B$ be a regular map between abelian variety. Then $\alpha$ is a composition of a translation and a homomorphism.

Proof. By a suitable translation, we can assume that $\alpha(0)=0$. Consider the map $h: A\times A\to B$ sending $(a_1,a_2)$ to $\alpha(a_1+a_2)-\alpha(a_1)-\alpha(a_2)$. The restriction of $h$ to $\{0\}\times A$ and $A\times \{0\}$ is just $0$. This yields $h(A\times A)=\{0\}$ by rigidity theorem, and $\alpha$ is a homomorphism (Q.E.D)

Using this, we can prove that the group law on an abelian variety.

Corollary 1.4. The group law on an abelian variety $A$ is commutative.

Proof. We now use a trick: a group $A$ is abelian iff the map $-1: a\mapsto -a$ is a homomorphism. Using Theorem 1.3, -1 is a regular map (by definition), which sends $0$ to $0$, and hence, it is a homomorphism. This yields $A$ is an abelian group. (Q.E.D)

2. Theorem of the cube. We now turn to the main theorem in our notes. The theorem of the cube will be one of the main tools for studying line bundles on abelian varieties.

Theorem 2.1 (Theorem of the cube). Let $U, V, W$ be complete irreducible varieties, and $L$ is a line bundle on $U\times V\times W$. Assume that there exists $u_0\in U,v_0\in V,w_0\in W$ such that $L$ restricts on $\{u_0\}\times V\times W, U\times \{v_0\}\times W, U\times V\times \{w_0\}$ are trivial. Then $L$ is trivial.

Now, it has the following important corollary

Corollary 2.2. Let $A$ be an abelian variety and $L$ is a line bundle on $A$. Let us denote $p_{123}:A\times A\times A\to A$ the map sends $(p,q,r)$ to $p+q+r$, $p_{12}: A\times A\times A\to A$ sends $(p,q,-)$ to $p+q$ (similarly for $p_{23}, p_{13}$), and $p_1:A\times A\times A\to A$ sends $(p,-,-)$ to $p$ (similarly for $p_2,p_3$). Then
$$L':=p_{123}^*L\otimes p_{12}^*L^{-1}\otimes p_{23}^*L^{-1}\otimes p_{13}^*L^{-1}\otimes p_1^*L \otimes p_2^*L \otimes p_3^*L$$
is a trivial line bundle.

Proof. We recall that if $\pi: L\to V$ is a line bundle and $f: W\to V$ is a regular map, then the pullback bundle of $L$ is defined $f^*L:=\{(w,l)\in W\times L| f(w)=\pi(l)\}$, and $f^*L$ is a line bundle on $W$ together with $\pi': f^*L\to W$ the projection map into the first coordinate.

Now, one can restrict $L'$ to $B:=\{0\}\times A\times A$, this can be seen

$$p_{123}^*L=\{(a,b,c,l)\in A\times A\times A\times L|a+b+c=\pi(l)\}$$

And $p_{123}^*L|_B=\{(0,b,c,l)\in \{0\}\times A\times A\times L|b+c=\pi(l)\}=p_{23}^*L$. This yields $p_123^*L\otimes p_{23}^*L^{-1}$ is trivial on $B$.

Also

$$p_{12}^*L=\{(a,b,-,l)\in A\times A\times A\times L|a+b=\pi(l)\}$$

And $p_{12}^*L|_B=\{(0,b,-,l)\in A\times A\times A\times L|b=\pi(l)\}=p_2^*L$. This yields $p_{12}^*L^{-1}\otimes p_2^*L$ is trivial on $B$. The similar result holds for $p_{13}L^{-1}\otimes p_3^*L$.

Finally, it can be seen that $p_1^*L|_B=\{(0,-,-,l)\in A\times A\times A\times L|\pi(l)=0\}$ is also a trivial bundle on $B$. This shows $L'$ is a trivial bundle on $B$. Due to the symmetry of the index, $L'$ is also trivial on $A\times \{0\}\times A$ and $A\times A\times \{0\}$. By theorem of the cube, $L'$ is trivial. (Q.E.D)

Corollary 2.3. Let $f,g,h$ be regular maps from a variety $V$ to an abelian variety $A$, and $L$ is a line bundle on $A$. Then
$$L':=(f+g+h)^*L\otimes (f+g)^*L^{-1}\otimes (g+h)^*L^{-1}\otimes (h+f)^*L^{-1}\otimes f^*L\otimes g^*L\otimes h^*L$$
is a trivial bundle.

Proof. Let us define the map $V\xrightarrow{(f,g,h)} A\times A\times A$. Then $L'$ is just the pullback bundle of the line bundle on the previous corollary. And the pull back of a trivial bundle is a trivial bundle. (Q.E.D)

Using this, we can prove the important

Corollary 2.4. Let $A$ be an abelian variety and $n: A\to A$ the multiplication by $n$ map. Then
$$n^*L\cong L^{(n^2+n)/2}\otimes (-1)^*L^{(n^2-n)/2}$$
In particular, if $L$ is a symmetric line bundle, i.e. $L\cong (-1)^*L$, then $n^*L\cong L^{n^2}$, and if $L$ is anti-symmetric, i.e. $L\cong (-1)^*L^{-1}$, then $n^*L\cong L^n$.

Proof. We just use the previous corollary for $(f,g,h)=(n,1,-1)$ and use induction on $n$. (Q.E.D)

We are now ready for the proof of Theorem 1.

Theorem 1. Let $E$ be an elliptic curve defined over an algebraically closed field $k$. Denote $n: E\to E$ the multiplication by $n$ map, then $\deg n=n^2$, and $n$ is an unramified map iff $(n,char(k))=1$. And in this case, $E[n]$-the $n$-torsion subgroup of $E$ has $n^2$ elements and $E[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$.

Proof. We recall that if $f: X\to Y$ is a non-constant regular map between smooth projective curves, then for all $Q\in Y$, we have pull-back of divisor $f^*[Q]=\sum_{P\in Y, f(P)=Q}e_{f,P}[P]$, where $e_{f,P}$ is the ramification index at $P$ with respect to $f$. Let $D$ be a symmetric divisor on $E$, i.e. $(-1)^*D=D$ (for example, one can choose $D=[O]$). Then it can be seen by the previous corollary that $n^*[O]\sim n^2[O]$.

Also, $n^*[O]=\sum_{P\in E[n]}e_{n,P}[P]$, and $(-n)^*[O]=\sum_{P\in E[n]}e_{-n,P}[P]$. Because $-1$ is an automorphism of $E$, $e_{n,P}=e_{-n,P}$, i.e. $n^*[O]=(-n)^*[O]$. But then, $(-n)^*=(-1)^*n^*$. And this yields

$$(-1)^*(n^*[O])=(-1)^*(\sum_{P\in E[n]}e_{n,P}[P])=\sum_{P\in E[n]}e_{n,P}[-P]$$

From this, one can see $e_{n,P}=e_{n,-P}$, and hence $\sum_{P\in E[n]}e_{n,P}P=O$. And we know that $\deg(n)=\sum_{P\in E[n]}e_{n,P}$. This implies $n^*[O]-\deg(n)[O]$ is a principal divisor, i.e. $n^*[O]\sim \deg(n)[O]$. Now, we can see $n^2[O]\sum \deg(n)[O]$. This yields $\deg n=n^2$. This proves our first statement.

For the second statement, the map $\alpha: A\to B$ between abelian varieties will induce the map $d\alpha: T_{A,0}\to T_{B,0}$, where $T_{A,0}$ is the tangent space of $A$ at $0$. And it can be seen that $\alpha\mapsto d\alpha$ is a homomorphism. This yields $dn$ is actually the multiplication by $n$ map on the tangent space of $A$ at $0$. It will be $0$ iff $char(k)$ divides $n$. Otherwise, it is an isomorphism. Now, we use the fact that if $f: X\to Y$ is a finite map between varieties, and $f(x)=y$, then $f$ is unramified at $x$ iff the induced map on tangent space $T_{X,x}\to T_{Y,y}$ is injective (Stack Project, Lemma 32.16.8). This yields $n$ is unramified at 0 iff $(n,char(k))=1$. Because the translation map is an automorphism (of curves), this will induce an isomorphism on tangent spaces. Hence, $n$ is unramified iff $(n,char(k))=1$.

With the assumption that $(n,char(k))=1$, the multiplication map by $n$ is unramified, i.e. $e_{n,P}=1$ for all $P\in E[n]$. This yields $\#E[n]=n^2$. By using the fundamental theorem of finite abelian group, we will obtain $E[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$ (Q.E.D)

Remark. The more general version of Theorem 1 for abelian varieties can be found in Milne's note (Theorem 7.2).

[Elliptic Curves II] A Note on Weil Pairing

$E$ in this note is an elliptic curve defined over an algebraically closed field $k$.

1. The definition of Weil's pairing. From the previous note, we know that there exists an isomorphism $\phi$ between $E$ and $Pic^0(E)$, defined by $P \mapsto [P]-[\infty]$. And we know that any principal divisor on $E$ has degree 0. It is natural to ask, when is a divisor of degree 0 is a principal divisor?

Assume that $D:=\sum_{i=1}^n [P_i]-\sum_{i=1}^n [Q_i]\in Div(E)$, where $P_i, Q_i$ are not necessarily distinct. Then $D$ is a principal divisor iff $D\equiv 0$ in $Pic^0(E)$. Via the inverse map $\phi^{-1}$, we get $D\mapsto \sum_{i} P_i - \sum_{i} Q_i$. So $D=0$ in $Pic^0(E)$ iff $\sum_{i} P_i - \sum_{i}Q_i =0$ in $E$.

Now, if we write $D=\sum_{P\in E}n_P[P]\in Div(E)$, then we define $sum(D):=\sum_{P\in E}n_PP$. Note that the later sum is a point in $E$. By our earlier discussion, we get an important

Theorem 1.1. If $D\in Div(E)$ is a divisor, then $D$ is principal iff $\deg(D)=0$ and $sum(D)=\infty$.

Now, let $m$ be a positive integer such that $(m,char(k))=1$, we define $E[m]:=\{P\in E|mP=\infty\}$. It follows from Chapter III of Washington's book that $E[m]\cong \mathbb{Z}_m\oplus \mathbb{Z}_m$. Then for any $T\in E[m]$, we define the divisor $D_T$ as follows

$$D_T:=\sum_{Q\in E, mQ=T}([Q])-\sum_{P\in E[m]}[P]$$

Remark. It can be seen that $[m]: E\to E$ defined by $P\mapsto mP$ is a separable isogeny of $E$, so by our Remark 2, Section 4 of the previous note, $\deg [m]=\#[m]^{-1}(\infty)=E[m]=\#[m]^{-1}(P)=m^2$, for any point $P\in E$. That means the ramification index at all point respected to $[m]$ is just 1. So in this case, the pull-back map of divisors is defined as $[m]^*[P]=\sum_{Q\in E, mQ=P}[Q]$. And it can be seen that in fact, $D_T$ defined above is $[m]^*([T]-[\infty])$ because

$$[m]^*([T]-[\infty])=\sum_{Q\in E, mQ=T}[Q]-\sum_{P\in E, mP=\infty}[P]=\sum_{Q\in E, mQ=T}([Q])-\sum_{P\in E[m]}[P]$$

We will prove that

Lemma 1.1. $D_T$ defined above is a principal divisor.

Proof. Due to the remark, we can see, $\#[m]^{-1}(\infty)=E[m]=\#[m]^{-1}(T)$. This implies there exists a point $Q$ such that $mQ=T$ (because $\#E[m]=m^2$). Then it can be seen for any point $P\in E[m]$, $m(P+Q)=\infty+T=T$. Also, for any $P_1,P_2\in E[m], P_1\ne P_2$, then $P_1+Q\ne P_2+Q$. This yields $[m]^{-1}(T)=\{P+Q|P\in E[m]\}$. And we can now rewrite $D_T$ as

$$D_T=\sum_{P\in E[m]}([P+Q])-[P])$$

It can be seen that $\deg D_T=0$, and $\sum(D_T)=\sum_{P\in E[m]}(P+Q-P)= m^2Q= mT = \infty$. By Theorem 1.1, $D_T$ is a principal divisor. (Q.E.D)

Now, it follows that there exists a rational function $g_T$ in $k(E)$ such that $div(g_T)=D_T$. For $S\in E[m]$, the Weil's pairing is defined as follows

$$e_m(S,T):=\frac{g_T(S+X)}{g_T(X)}$$

for any point $X\in E$ such that $g_T(S+X), g_T(X)\in k^*$. We will prove that

Lemma 1.2. $e_m(S,T)$ is well-defined, i.e. it does not depend on the choice of $X$.

Proof. We actually want to show that $\frac{g_T(S+X)}{g_T(X)}$ is a constant function. That is equivalent to say $div(g(X+S))-div(g(X))=0$. Let $h_T(X):=g_T(S+X)$, it is equivalent to prove that $div(h_T)=div(g_T)$. It can be seen that $g_T$ has zeros at $P+Q$, for some $Q\in E$ such that $mQ=T$, and $P\in E[m]$. This yields the zeros of $h_T$ are $P+Q-S$. Also, poles of $g_T$ are of the form $P$, where $P\in E[m]$. This yields poles of $h_T$ are actually $P-S$. So we get

$$div(h_T)-div(g_T)=\sum_{P\in E[m]}([P-S+Q]-[P-S])-\sum_{P\in E[m]}([P+Q]-[P])$$

But then, we can see that $S\in E[m]$, so $\{P-S|P\in E[m]\}=E[m]$. This yields $div(h_T)-div(g_T)=0$. This yields $h_T/g_T$ is a constant. (Q.E.D)

We will show next that in fact, $e_m(S,T)\in \mu_m$, where $\mu_m$ is the group of $m$-th root of unity in $k^\times$.

Lemma 1.3. $e_m(S,T)\in \mu_m$.

Proof. It can be seen that $m([T]-[\infty])$ is a principal divisor by Theorem 1.1. This implies there exists a function $f_T\in k(E)$ such that $div(f_T) = m([T]-[\infty])$. We already show that there exists a point $Q\in E$, such that $mQ=P$. And we can see

$$div(f_T\circ [m])=m\sum_{P\in E[m]}([P+Q]-[P])$$

since $f_T$ has zeros at $T$ with order $m$, so $f_T\circ [m]$ has zeros at $[m]^{-1}(T)=\{P+Q|P\in E[m]\}$, and each is of order $m$. Similarly for the poles. But then, the right hand side is just $mD_T=mdiv(g_T)=div(g_T^m)$. This yields $g_T^m=\lambda f_T\circ[m]$, for some $\lambda\in k^\times$. By possibly change $f_T$ by $\lambda f_T$, we can assume $g_T^m=f_T\circ [m]$. That means, for all points $X\in E$, we have $g_T^m(X)=f_T(mX)$, and for $S\in E[m]$, we have $g_T(X+S)^m=f_T(mX+mS)=f_T(mX)$. Hence, $\frac{g_T^m(X+S)}{g_T^m(X)}=1$. This implies $e_m(S,T)$ is a $m$-th root of unity. (Q.E.D)

So, we have proved if $m$ is any positive integer such that $(m,char(k))=1$, then there exists a well-defined map

$$e_m: E[m]\times E[m]\to \mu_m$$

where $\mu_m$ is the multiplicative group of $k^\times$ consisting of all $m$-th root of unity. The next section will be devoted to prove the properties of Weil's pairing.

2. Properties of Weil's pairing.

Proposition 2.1 (Bilinear). The Weil's pairing defined above is bilinear, i.e. for all $S_1, S_2, T_1, T_2\in E[m]$, we have $e_m(S_1+S_2,T)=e_m(S_1,T)e_m(S_2,T)$ and $e_m(S,T_1+T_2)=e_m(S,T_1)e_m(S,T_2)$.

Proof. If we denote $\tau_S(P):=P+S$, for all $P\in E$, i.e. $\tau_S$ is a translation with respect to $S$. Then it can be seen that $e_m(S_1+S_2, T)=\frac{g_T(S_1+S_2+X)}{g_T(X)}$ and $e_m(S_1, T)e_m(S_2,T)=\frac{g_T(S_1+X)}{g_T(X)}\frac{g_T(S_2+X)}{g_T(X)}$. Then $e_m(S_1+S_2,T)=e_m(S_1,T)e_m(S_2,T)$ iff $\frac{g_T(S_1+S_2+X)}{g_T(S_1+X)}=\frac{g_T(S_2+X)}{g_T(X)}=e_m(S_2,T)$. But this follows since $e_m(S_2,T)$ does not depend on the choice of $X$. If we replace $X$ by $S_1+X$, we get $e_m(S_2,T)=\frac{g_T(S_2+S_1+X)}{g_T(S_1+X)}$. And the conclusion follows.

The second property is more difficult. If we let $T_3:=T_1+T_2$, then by Theorem 1.1, we get $[T_1]+[T_2]-[T_3]-[\infty]=div(h)$ for some function $h\in k(E)$. We also have by definition

$$div(g_{T_1})=[m]^*([T_1]-[\infty]), div(g_{T_2})=[m]^*([T_2]-[\infty])$$
$$div(g_{T_3})=[m]^*([T_3]-[\infty])=[m]^*([T_1]+[T_2]-div(h)-2[\infty])=$$
$$=[m]^*([T_1]-[\infty]+[T_2]-[\infty]+div(h))=div(g_1)+div(g_2)+div(h\circ [m])=$$
$$=div(g_1.g_2.(h\circ[m])$$

So, in particular, we get $g_{T_3}=g_1.g_2.(h\circ[m])$. Hence,

$$e_m(S,T_1+T_2)=e_m(S,T_3)=\frac{g_{T_3}(S+X)}{g_{T_3}(X)}=$$
$$=\frac{g_1(S+X)g_2(S+X)h(mS+mX)}{g_1(X)g_2(X)h(mX)}$$

But then, since $mS=\infty$, we have $\frac{h(mS+mX)}{h(mX)}=1$. This yields by our previous calculation $e_m(S, T_1+T_2)=e_m(S,T_1)e_m(S,T_2)$. (Q.E.D)

Proposition 2.2 (Alternating). For any $T\in E[m]$, we have $e_m(T,T)=1$. And one can easily deduce $e_m(S,T)=e_m(T,S)^{-1}$ for all $S,T\in E[m]$.

Proof. Because $E[m]\cong \mathbb{Z}_m\oplus \mathbb{Z}_m$, and the bilinear property, it is sufficient for us to prove $e_m(T,T)=1$, for $T$ is a point with the order exactly $m$. Let $h_i(X):=g_T(X+iT')$, because $div(g_T)=\sum_{P\in E[m]}[P+Q]-[P]$, where $Q\in E, mQ=T$. Let $h_i(X)=g(X+iQ)$, for some $0\le i\le m-1$. Then it can be seen

$$div(h_i)=\sum_{P\in E[m]}([P+Q-iQ]-[P-iQ])=\sum_{P\in E[m]}[P+(1-i)Q]-[P-iQ]$$

Because $mQ=T$, this yields

$$div(\prod_{i=0}^{m-1}h_i)=\sum_{i=0}^{m-1}div(h_i)=\sum_{P\in E[m]}([P-Q]-[P+(m-1)Q])=$$
$$=\sum_{P\in E[m]}[P-Q]-\sum_{P\in E[m]}[P-T-Q]$$

But then, the later divisor is just 0, because $T\in E[m]$. This yields $\prod_{i=0}^{m-1}h_i=1$, i.e, it is a constant. This yields
$$1=g_T(X)\prod_{i=1}^{m-1}g_T(X+iQ)=\prod_{i=0}^{m-1}g_T(X+iQ)=\prod_{i=1}^{m}g_T(X+iT)=\prod_{i=1}^{m-1}g_T(X+iQ)g_T(X+T)$$

Note that the third identity follows when we replace $X$ by $X+Q$, and the last identity follows since $mQ=T$. In particular, we get $g_T(X+T)=g_T(X)$. This implies $e_m(T,T)=1$. The second statement easily follows from Proposition 2.1 and the first statement. (Q.E.D)

Proposition 2.3 (Non-degenerate). The Weil's pairing is non-degenerate, i.e. if $e_m(S,T)=1$ for all points $S\in E[m]$, then $T=\infty$.

Proof.

Remark 2.4. We will contemporarily accept this fact, which will be proved later: if $g\in k(E)$ such that $g\circ \tau_S=g$ for all $S\in E[m]$, i.e. $g$ is invariant under the translation then $g=h\circ [m]$, for some $h\in k(E)$.

Due to the assumption, $g_T$ now is invariant under the translation map $\tau_S$, for all $S\in E[m]$. This yields $g_T$ is of the form $h\circ [m]$, for some $h\in k(E)$. Due to the proof of Proposition 2.1, we get $g_T^m=f_T\circ [m]$, i.e. $(h\circ[m])^n=f_T\circ [m]$. This yields for all point $P\in E$, $h(mP)...h(mP)=f_T(mP)$. Since $[m]$ is surjective, we have $h^m=f_T$. But then, the divisor of $f_T$ is $m([T]-[\infty])$. This yields $div(h)=[T]-[\infty]$. Due to Theorem 1.1, we have $T-\infty=\infty$, i.e. $T=\infty$.

So, it remains to prove Remark 2.4. We first begin with an important lemma

Lemma 2.5. If $\alpha$ is a separable isogeny from $E_1$ to $E_2$, then the map $\ker (\alpha) \to Aut(k(E_1)/\phi^*(k(E_2)))$ define by $T\mapsto \tau_T^*$ is a group isomorphism.

Proof. It can be seen first that for any $T\in E_1$, the translation $\tau_T^*: k(E_1)\to k(E_1)$ define by $g\mapsto g\circ \tau_T$ defines an automorphism of field, since its inverse is $\tau_{-T}^*$.

For any $T\in \ker\alpha$, and for all $f\in k(E_2)$, we have $\tau_T^*(\alpha^* f)=f\circ \alpha\circ \tau_T$. So for any point $P\in E_1$, we have $f\circ \alpha\circ \tau_T(P)=f\circ\alpha(P+T)=f\circ\alpha(P)$, since $\alpha(P+T)=\alpha(P)+\alpha(T)=\alpha(P)$. This yields $\tau_T^*(\alpha^* f)=\alpha^* f$, i.e. $\tau_T^*$ fixes the subfield $\phi^*(k(E_2))$ of $k(E_1)$.

Now, it is a group homomorphism since for $T,S\in \ker\alpha$, we have $T+S\mapsto \tau_{T+S}^*=\tau_T^*\circ \tau_S^*$. Now, it follows from Galois theory and our Remark 2 in the previous note that $\#Aut(k(E_1)/\phi^*(k(E_2)))=\deg \alpha =\#\ker \alpha<\infty$. So, we just need to prove the map is injective. But it is obvious, since if $\tau_T^*=\tau_\infty^*$, then for sure $T=\infty$ (Q.E.D)

Now, if we let $\alpha=[m]$ and $E_1=E_2=E$, then $\ker\alpha = E[m]$. Now, $g$ is fixed under the translation $\tau_S$, for all $S\in E[m]$ yields $g\in [m]^*(k(E))$, i.e. there exists $h\in k(E)$ such that $g=[m]^*h=h\circ [m]$. So, the proof of Remark 2.4 follows. (Q.E.D)

Proposition 2.6 (Compatible). Let $\varphi$ be any endomorphism of $E$, then for any $S, T\in E[m]$, we have $e_m(\varphi(S),\varphi(T))=e_m(S,T)^{\deg\varphi}$.

Proof. Let $\deg\varphi =d$. We want to prove for all point $P\in E$,

$$\frac{g_{\varphi(T)}(\varphi(P)+\varphi(S))}{g_{\varphi(T)}(\varphi(S))}=\bigg(\frac{g_T(P+S)}{g_T(P)}\bigg)^d$$

If we denote $\tau_S$ the translation by $S$ (i.e. $P\mapsto P+S$), the statement is equivalent to
\begin{align}
\frac{g_{\varphi(T)}\circ \tau_{\varphi(S)}\circ\varphi}{g_{\varphi(T)}\circ\varphi}=\bigg(\frac{g_T\circ\tau_S}{g_T}\bigg)^d (1)
\end{align}

One can see $\tau_{\varphi(S)}\circ\varphi(P)=\varphi(P+S)=\varphi\circ\tau_S(P)$. Therefore
$$\frac{g_{T\varphi(T)}\circ\tau_{\varphi(S)}\circ\varphi}{g_{\varphi(T)}\circ\varphi}=\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_{\varphi(T)}\circ\varphi}$$
And for the RHS of (1), because $(g_T\circ\tau_S)^d=g_T^d\circ\tau_S$, we have (1) is equivalent to
$$\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_{\varphi(T)}\circ\varphi}=\frac{g_T^d\circ\tau_S}{g_T^d}\Leftrightarrow\frac{g_{\varphi(T)}\circ\varphi\circ\tau_S}{g_T^d\circ\tau_S}=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}\circ\tau_S=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}$$

That means, we have to prove that $\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}$ is invariant under the translation of any $S\in E[m]$. We need the following

Lemma 2.7. If $r=s\circ[m]$ (i.e. $r=[m]^*s$), where $s\in k(E)$, then $r$ is invariant under the translation of any $S\in E[m]$.

Proof. For any $S\in E[m]$, we have $r(P+S)=s\circ [m](P+S)=s(mP+mS)=s\circ[m](P)=r(P)$, for all $P\in E$. (Q.E.D)

Hence, it is sufficient for us to prove that $F:=\frac{g_{\varphi(T)}\circ\varphi}{g_T^d}=[m]^*[s]$ for some $s\in k(E)$. Let us compute the divisor of $F$, where $div(g_T)=[m]^*([T]-[\infty])$, we have
$$div(F)=(\varphi^*[m]^*)([\varphi(T)]-[\infty])-d[m]^*([T]-[\infty])=[m]^*(\varphi^*([\varphi(T)]-[\infty])-d([T]-[\infty]))$$
And hence, it is sufficient for us to prove that $D:=(\varphi^*([\varphi(T)]-[\infty])-d([T]-[\infty]))$ is a principal divisor. By definition, we have
$$\varphi^*([\varphi(T)]-[\infty])=\sum_{R\in \ker\varphi}e_R([T+R]-[R])$$
where $e_R$ is the ramification index at $R$. Hence, $\deg D=0$, because $\sum_{R\in \ker\varphi}e_R=\sum_{R\in\varphi^{-1}(\infty)}e_R=d$. And by the similar reason, the sum of $D$ is
$$\sum_{R\in \ker\varphi}e_R(T+R-R)-d(T-\infty)=\infty$$
That means, $D$ is a principal divisor, by Theorem 1.1. This implies $F=[m]^*s$, for some $s\in k(E)$. And hence, the statement now follows. (Q.E.D)

Proposition 2.8 (Galois invariant). If $E$ is defined over $k$ (may not be algebraically closed), then for all $\sigma\in Gal(\bar{k}/k)$, we have $e_m(S^\sigma,T^\sigma)=e_m(S,T)^\sigma$.

Proof. It just follows if we consider the action of $\sigma$ on $g$. (Q.E.D)

[Elliptic Curves I] A Note on Isogenies

This short note is taken from the book of Washington "Elliptic Curves: Number Theory and Cryptography", and Milne "Elliptic Curves".

1. Brief review on Weierstrass $\wp$-function. We first begin with a lattice $L:=\mathbb{Z}\omega_1+\mathbb{z}\omega_2$, where $\omega_1,\omega_2\in \mathbb{C}$ are linearly independent over $\mathbb{R}$. The fundamental domain $\mathcal{F}$ of $L$ is the set $\mathcal{F}:=\{a\omega_1+b\omega_2|a,b\in [0,1)\} = \mathbb{C}/L$. One can consider $\mathcal{F}$ as a group by the following addition law: $z_1+z_2 = z_1+z_2\mod L$. The fundamental domain $\mathcal{F}$ is called a torus associated to $L$. The Weierstrass $\wp$-function associated to $L$ is defined as follow:

$$\wp(z) := \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}[\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}]$$

It can be seen that $\wp(z)$ is a non-constant meromorphic even function on $\mathbb{C}$. More clearly, poles of $\wp(z)$ is exactly the lattice points of $L$, and $\wp(z)=\wp(-z)$. If we take the derivative of $\wp(z)$, we get 

$$\wp'(z) = -2\sum_{\omega\in L}\frac{1}{(z-\omega)^3}$$

We can see that $\wp'(z)$ is an odd function, and both $\wp(z)$ and $\wp'(z)$ are doubly periodic, i.e. for all $\omega\in L$, we have $\wp(z+\omega)=\wp(z)$, and $\wp'(z+\omega) = \wp'(z)$. We next define the $k$-th Eisentein's series associated to $L$ as follows.

$$G_k=\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^k}$$

Around a point $\omega$ in $L$, we can can see

$$\wp(z) = \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\omega^{-2}[\frac{1}{(z/\omega-1)^2}-1]$$

Using the geometric series, we get

$$\wp(z) = \frac{1}{z^2} + \sum_{\omega\in L\setminus \{0\}}\omega^{-2}(\sum_{n\ge 1}(z/\omega)^n)^2=\frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\sum_{n\ge 1}(n+1)\frac{z^n}{\omega^{n+2}}=$$

$$=\frac{1}{z^2}+\sum_{n\ge 1}(n+1)z^n\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^{n+2}}=\frac{1}{z^2}+\sum_{n\ge 1}G_{n+2}(n+1)z^n$$

Because $\wp(z)$ is an even function, we get

$$\wp(z) = \frac{1}{z^2}+\sum_{n\ge 1}G_{2n+2}(2n+1)z^{2n}$$

Using this identity, we can obtain the power series expansion for $\wp(z)$ as follows

$$\wp'(z) = \frac{-2}{z^3}+\sum_{n\ge 1}G_{2n+2}(2n+1)(2n)z^{2n-1}$$

and by direct computation, we have

$$f(z):=\wp'(z)^2 - 4\wp(z)^3 +  60G_4\wp(z) + 140G_6 = c_1z+c_2z^2 + ...$$

That means $f(z)$ is a holomorphic function, this yields $f(z)$ is bounded in the fundamental domain $\mathcal{F}$,  and it is doubly periodic (since $\wp(z)$ and $\wp'(z)$ are). Now, it follows that $f(z)$ is an entire funcion and bounded. By Louville's theorem, $f(z)=f(0)$ is a constant function. Because $f(0)=0$, we have $f(z)\equiv  0$. This yields a following important identity:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2\wp(z) - g_3$$

where $g_2 = 60G_4, g_3 = 140 g_6$. That means, $(\wp(z),\wp'(z))$ is a point on a curve $ (E): y^2= x^3 - g_2x - g_3$. This curve is non-singular, and for any $(x,y)$ on the curve, there exists $z$ such that $(x,y)=(\wp(z),\wp'(z))$. Moreover, one obtain the following important bijection $\mathbb{C}/L \ to (E)$ by sending $z\ne 0 \mapsto (\wp(z):\wp'(z):1)$ and $0\mapsto (0:1:0)$. This gives a natural group law on $(E)$ induced from the group law on $\mathbb{C}/L$. For any $z\ne 0 \in \mathbb{C}/L$, because $z+0=0+z=z$, we can define 

$$(\wp(z):\wp'(z):1) + (0:1:0) = (\wp(z):\wp'(z):1)$$

If $z=0$, we have $(0:1:0)+ (0:1:0) = (0:1:0)$. If $z_1,z_2\in \mathbb{C}/L\setminus\{0\}$, and $z_1\ne z_2$, then $z_1+z_2$ correspond to the point $(\wp(z_1+z_2):\wp'(z_1+z_2):1)$ on the curve. So, we can define

$$(\wp(z_1):\wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1):=(\wp(z_1+z_2):\wp'(z_1+z_2):1)$$

Similarly, when $z_1 = -z_2 in \mathbb{C}/L$, and both are non-zero, because $z_1+z_2=0$, we can define

$$(\wp(z_1), \wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1)=(0:1:0)$$

This gives $(E)$ the natural group structure, and $(E)\cong \mathbb{C}/L$ as abelian groups.

NOTE. We are familiar with the fact that the group law on an elliptic curve comes from the intersection of curves with lines. Yes, it is true also in this context. Let us consider the line goes through two (affine) points $(\wp(z_1), \wp'(z_1))$ and $(\wp(z_2),\wp'(z_2))$ with equation: $y-ax-b=0$, it will cut the curve at the third point: $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$.

To prove this fact, let $f(z) = \wp'(z) - a\wp(z) - b$ (correspond to our line: $y-ax-b$), then we know that $f(z)=0$ at two points $z_1,z_2 in \mathbb{C}/L$, and it has unique pole in this domain (at 0) with multiplicity 3 (since $\wp'(z)$ has). We now use a theorem from complex analysis, if $f$ is a meromorphic map from $\mathbb{C}/L$ to $\mathbb{C}$, and it has zero and poles at $z_i$ with multiplicity $n_i$, then $\sum_{i}n_i=0$, and $\sum_{i}n_iz_i=0$. From this, we can see $f$ has another zero $z$ such that $z_1+z_2+z+0=0$, i.e. $z=-z_1-z_2$. 

This will yields the third intersection point between the line and our curve is $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$. So, taking the inverse point, we get $(\wp(-z_1-z_2), -\wp'(-z_1-z_2))$. Because $\wp(z)$ is an even function, and $\wp'(z)$ is an odd function, we get $((\wp(-z_1-z_2), -\wp'(-z_1-z_2))=(\wp(z_1+z_2),\wp'(z_1+z_2))$. And this is identical with our familiar definition on group law.

2. Complex theory of isogenies. Let $L_1, L_2$ be lattices in $\mathbb{C}$ with fundamental domain $E_1:=\mathbb{C}/L_2,E_2:=\mathbb{C}/L_2$, which are considered as elliptic curves defined over $\mathbb{C}$. Let $\alpha\in \mathbb{C}$ such that $\alpha L_1\subset L_2$, we define the isogeny $[\alpha]: E_1\to E_2$ by sending $z\mod L_1$ to $\alpha z\mod L_2$. This is a well-defined map, and is a homomorphism of groups.

If $\alpha\ne 0$, it can be seen that $\alpha L_1$ is a sublattice of $L_2$, i.e. a subgroup of rank 2 of $L_2$. This will yields the index $[L_2:\alpha L_1]$ is finite. In the case $\alpha\ne 0$, the degree of $[\alpha]$ is defined by the index $[L_2:\alpha L_1]$. If $\alpha=0$, we define the degree of $[\alpha]$ as 0.

Assume that $\alpha\ne 0$ and $n$ is the degree of the map $\alpha$, we have $n L_2\subset \alpha L_1$. This implies $\hat{\alpha} L_2\subset L_1$, where $\hat{\alpha} := (n/\alpha)$ and it will yields the isogeny $[\hat{\alpha}]$ from $E_2$ to $E_1$, which is called the dual of $[\alpha]$. We will prove that $\deg [\alpha] = \deg [\hat{\alpha}]$.

In fact, let $\{\omega_1,\omega_2\}$ is the basis for $L_1$, and $\{\omega_3,\omega_4\}$ is the basis for $L_2$. Because $\alpha L_1\subset L_2$, one can represent $\alpha \omega_1 = a\omega_3 + b\omega_4$, $\alpha \omega_2=c\omega_3 + d\omega_4$, with $a,b,c,d\in \mathbb{Z}$. Let us denote $A$ the $2\times 2$ matrix with the first row is $a,b$ and second row is $c,d$. Then the degree of $[\alpha]$ is $|\det A| = n$ Also, since $n/\alpha L_2\subset L_1$, we can also represent $n \omega_3 = e (\alpha\omega_1) + f (\alpha\omega_2) = (ea+fc)\omega_3 + (eb+fd)\omega_4$, and $n \omega_3 = g(\alpha \omega_1) + g(\alpha\omega_2) = (ga+hc)\omega_3 + (gb+hd)\omega_4$. If we denote $B$ the matrix with the first row $e,f$ and the second row $g, h$, then $\deg([\hat{\alpha}])=|\det B|$. Furthermore, it can be seen that $BA$ is the matrix send $(\omega_3, \omega_4)$ to $(n\omega_3, n\omega_4)$, i.e. $|\det B||\det A|=n^2$. This yields $\deg [\hat{\alpha}]=\deg \alpha$.

One can also easily deduce that $[\hat{\hat{\alpha}}]=\alpha$, and $[\alpha]\circ [\hat{\alpha}] = [\deg \alpha]$ is an isogeny from $E_1$ to $E_1$. Also, $[\hat{\alpha}]\circ [\alpha]=[\deg \alpha]=[\deg \hat{\alpha}]$ is an isogeny from $E_2$ to $E_2$. The kernel of $[\alpha]$, it is exactly $z\in E_1$ such that $\alpha z\in L_2$. 

Furthermore, we can deduce the complex version of Velu's formula.

Proposition 2.1. Let $G\subset E_1$ is a finite subgroup. Then there exists a lattice $L_2$ and an isogeny from $E_1\to E_2$ such that its kernel is $G$.

Proof. Due to the correspondence theorem of groups, there exists $L_2\subset \mathbb{C}$ such that $G=L_2/L_1$. If we denote $\#G= n$, then it can be seen that $L_1\subset L_2\subset (1/n)L_1$, i.e. $L_2$ is also a lattice. The map $E_1\to E_2$ sending $z\mod L_1$ to $z\mod L_2$ has the kernel $G$. (Q.E.D)

We can also obtain the complex version of the following statement

Proposition 2.2. Let $f$ be a holomorphic map between $E_1$ and $E_2$, then $f(z)$ is of the form $\alpha z+\beta$ for some $\alpha,\beta \in \mathbb{C}$, i.e. $f$ is a composition of an isogeny and a translation map.

Proof. We will use a bit theory of Riemann surfaces, because $\mathbb{C}$ is a universal covering of $E_1, E_2$, with projection maps $\pi_1, \pi_2$. The holomorphic map between $E_1$ and $E_2$ will induce the holomorphic map from $\tilde{f}: \mathbb{C}\to \mathbb{C}$ such that $\tilde{f}(z\mod L_1)=\tilde{f}(z)\mod L_2$, i.e. it makes the diagram commutes (the reader should draw it, I cannot draw it here). 

From this, for any $\omega\in L_1$, we have 

$$\tilde{f}(z+\omega) \equiv \tilde{f}(z+\omega)=\tilde{f}(z)\equiv \tilde{f}(z)\mod L_2$$

If we let $g(z) = \tilde{f}(z+\omega) - \tilde{f}(z)$, then $g(z)$ is a holomorphic function, and $g(z)$ takes values only on $L_2$, which is a discrete subset of $\mathbb{C}$. Hence, $g(z)$ is a constant function, then $0=g'(z) = \tilde{f}'(z +\omega)-\tilde{f}'(z)$, i.e. $\tilde{f}'(z) = \tilde{f}'(z+\omega)$, i.e. $\tilde{f}'$ is a bounded entire function. By Louville's theorem again, $\tilde{f}'(z)$ is a constant function. This yields, $\tilde{f}(z) = \alpha z + \beta$, for some $\alpha,\beta \in \mathbb{C}$. (Q.E.D)

3. Brief review on divisors. We will recall something about intersection numbers and divisors. Let $E$ be any projective curve in $\mathbb{P}^2$, the formal sum $D:=\sum_{P\in E}n_P [P]$, where $n_P\in \mathbb{Z}$ and $n_P\ne 0$ at finitely many points $P\in E$, is called a divisor on $E$. The set of all divisors on $E$ is denoted $Div(E)$. It can be seen that $Div(E)$ has the abelian group structure induced from $\mathbb{Z}$. The degree of $D$, denoted by $\deg D$ is defined as $\sum_{P\in E}n_P$.

If $E, F$ are projective curves in $\mathbb{P}^2$, without common irreducible components. it can be seen that $E\cap F$ is a finite set (since $E,F$ is of dimension 1, their intersection is of dimension 0, and hence, discrete and finite). At a point $P\in E$, we can define the intersection multiplicity (with $F$) at $P$, which is denoted $I_P(E,F)$, it is a non-negative integer, and is positive if $P\in E\cap F$. Furthermore, it has the following properties: $I_P(E,F)=I_P(F,E)$ and $I_P(E,FG) = I_P(E,F)+I_P(E,G)$, where we identify $FG$ with the zeros of $fg$, with $F, G$ are given by $f=0, g = 0$ resp.  From this, one obtains the Bezout's theorem:

$$\deg E \deg F = \sum_{P\in E\cap F} I_P(E,F)$$

If $E$ is a curve, and $f:=f_1/f_2$ is a rational function, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$. We define

$$div(f) := \sum_{P\in E\cap F_1}I_P(E,F_1)[P] - \sum_{Q\in E\cap F_2}I_P(E,F_2)[Q]$$

Because $\deg F_1=\deg F_2$, due to Bezout's theorem, we get $\deg (div(f))=0$. If $D\in Div(E)$, and there exists $f=f_1/f_2$, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$, such that $D = div(f)$, then $D$ is called a principal divisor. Due to the properties of the intersection numbers, we have $div(fg)=div(f)+div(g)$, i.e. the set of all principal divisors form a subgroup of $Div(E)$. We denote this group $Prin(E)$. If $D_1, D_2\in Div(E)$, and $D_1-D_2\in Prin(E)$, we denote $D_1\sim D_2$. It is obvious to see that $\sim$ is an equivalent relation.

It follows from above that any principal divisor has degree 0. Hence, $Prin(E)\subset Div^0(E)$, which is the subgroup of $Div(E)$ containing all degree zero divisors.The quotient group $Div^0(E)/Prin(E)$ is denoted $Pic^0(E)$.

If $E$ is an elliptic curve (we now view $E$ a projective curve), and $L$ is a line. There is 3 possibilities:

1. $L$ cuts $E$ at three distinct points
2. $L$ cuts $E$ at two distinct points, and the intersection multiplicity at a point is $2$, i.e. $L$ is a tangent line of $E$ at this point.
3. $L$ cuts $E$ only at 1 point, and the intersection multiplicity at this point is $3$. In this case, $L$ is also a tangent line of the curve at this point. For example, if we consider the Weierstrass form of $(E): y^2z=x^3+Axz^2+Bz^3$, then the line at infinity $(L): z=0$ cuts the curve only at $(0:1:0)$. Hence, $I_{(0:1:0)}((E), L)=3$. Another example is $(E): y^2z+yz^2=x^3$. The line $(L): y=0$ cuts the curve only at $(0:0:1)$, and $I_{(0:0:1)}(E, L)=3$.

Let $P,Q$ are two distinct points on $E$, the line $L_1$ given by equation $f_1=0$ connecting $P,Q$ will cut the curve at the third point $R$. If this line is the tangent line at $P$ (or $Q$), then $R$ can be considered as $P$ (or $Q$, resp.). If we take the line $L_2$ connecting $R$ and $-R$, given by equation $f_2=0$ then it will cut the curve at the third point $\infty$. So, in this case, by definition,

$$div(f_1/f_2)=[P]+[Q]+[R]-([R]+[-R]+[\infty])= [P]+[Q] - [-R] - [\infty]$$

This implies $[P]+[Q]\sim [-R]+[\infty]$. If $P\equiv Q$, we take the tangent line $L_1$ of $E$ at $P$, this will cut the curve at $R$ ($R\equiv P$ in the case the tangent line cuts the curve only at 1 point), we again take the line connecting $R$ and $-R$, this will cut the curve at $\infty$. We again get $div(f_1/f_2)=2[P] - [-R] - [\infty]$, i.e. $2[P]\sim [-R]+[\infty]$.

What we can see from this is that we can define the group law on $E$ as $P + Q = -R$. It is identical with our familiar definition for the group law on an elliptic curve, and it is more accurate, because in the case of tangent line of order 3, we are tricked by our geometric intuition. And we do not have to prove the associativity law, because it is natural induced from the group law on $Pic^0(E)$. Furthermore, one obtains the bijection (and hence, a group isomorphism) between $E$ and $Pic^0(E)$ given by $P\mapsto [P]-[\infty]$.

4. A quick look on algebraic theory of isogenies. Let $E_1,E_2$ are two elliptic curves over an algebraically closed field $k$. An isogeny is both a non-constant morphism and a group homomorphism from $E_1$ to $E_2$, i.e. if we look at affine parts of $E_1$ and $E_2$, an isogeny a rational map between $E_1$ and $E_2$ that sends $\infty$ to $\infty$. Let $\alpha: E_1\to E_2$ be an isogeny, it will induce the injective field homomorphism $\alpha^*: k(E_2)\to k(E_1)$. The field extension $k(E_1)$ of $\alpha^*(k(E_2))$ is finite, and the degree of $\alpha$ is defined as the degree of the field extension. If the field extension is separable, $\alpha$ is called separable isogeny. Otherwise, $\alpha$ is called inseparable.

 It follows from Chapter II of Washington's book that:

1. An isogeny is always surjective (We can use a little bit of algebraic geometry to deduce this fact: $E_1$ is projective variety, and hence, is complete, and $\alpha(E_1)$ is closed and irreducible in $E_2$ (since $E_1$ is irreducible), i.e. $\alpha(E_1)=E_2$, since $E_2$ is also irreducible and of dimension 1, and $\alpha$ is non-constant).

2. The kernel of an isogeny $\alpha$ is always finite, and $\# \ker \alpha =\deg \alpha$ if $\alpha$ is separable, i.e. $\deg \alpha$ is actually the number of points in a fiber $\alpha^{-1}(\infty)$. It can be proved that in the case $\alpha$ is an separable isogeny, the fiber of $\alpha$ at every point is equal, and it is $\deg \alpha$.

3. If $\alpha$ is non-separable, then $\deg \alpha >\#ker \alpha$. An example is the Frobenius endomorphism, its degree is $q$ (in the case $E$ are defined over $\mathbb{F}_q$), and it is bijection from $E(\overline{\mathbb{F}_q})$ to $E(\overline{\mathbb{F}_q})$, hence, the its kernel is just $\infty$.

So from 2 and 3, we can see that the kernel of an isogeny is always finite.

4. If $\alpha: E_1\to E_2$ is a morphism, then it induces the push-forward map $\alpha_*$ from $Div(E_1)\to Div(E_2)$ defined by $\alpha_*(\sum_{P\in E} n_P [P])=\sum_{P\in E} n_P [\alpha(P)]$. This can be seen that $\alpha_*$ is a group homomorphism. If furthermore, $\alpha(\infty)=\infty$, then it can be proved $\alpha_*$ maps principal divisors to principal divisors, i.e. the induced map $\alpha_*: Pic^0(E)\to Pic^0(E)$ is well-defined.

Using the isomorphism between $E$ and $Pic^0(E)$, we can prove a beautiful

Proposition 4.1. Let $E_1,E_2$ as above, and $\alpha:E_1\to E_2$ is a non-constant morphism, and $\alpha(\infty)=\infty$, then $\alpha$ is an isogeny.

Proof. It is sufficient to prove that $\alpha$ is a group homomorphism. From the isomorphism $\phi_i$ between $E_i$ and $Pic^0(E_i)$, which sends $P\to [P]-[\infty]$, and Remark 4 above, we have

$$\phi_2^{-1}\circ \alpha_*\circ \phi_1(P) = \phi_2^{-1}\circ \alpha_* ([P]-[\infty])=\phi_2^{-1}([\alpha(P)]-[\infty])=\alpha(P)$$

Because $\phi_i$ and $\alpha_*$ are group homomorphism, we have $\alpha$ is also a group homomorphism. (Q.E.D)

In section 3, we give the proof of the Velu's theorem for complex elliptic curves. It is the original version of Velu's theorem.

Proposition 4.2. Let $E_1$ be an elliptic curve defined over $k$, and $G\subset E_1$ a finite subgroup. Then there exists the curve $E_2$ and an isogeny $\alpha: E_1\to E_2$ such that $\ker(\alpha)=G$ and $E_2, \alpha$ can be computed explicitly via $G$ and the equation of $E_1$.

The existence of dual isogenies (for the separable case) can be deduced by the proposition above and the following suprising

Lemma 4.3. Let $E_1, E_2, E_3$ be three elliptic curves, with $\alpha_2: E_1\to E_2$ and $\alpha_3: E_1\to E_3$ are separable isogenies, and $\ker \alpha_2=\ker \alpha_3$. Then $E_2\cong E_3$, via the isomorphism $\beta$ such that $\alpha_3=\beta\circ\alpha_2$.

We now deduce the existence of dual isogeny for separable case (note that this also holds for the non-separable case).

Proposition 4.4. Let $\alpha: E_1\to E_2$ be a separable isogeny, then there exists an isogeny $\hat{\alpha}:E_2\to E_1$ such that $\hat{\alpha}\circ \alpha= [\deg \alpha]$.

Proof. Let $n:=\deg \alpha$, we will prove this proposition in the case $char(k)$ does not divide $n$.  Since $\alpha$ is separable, we have $\deg\alpha=\#\ker\alpha$. It follows $n=\#\ker\alpha$, and hence, $\ker\alpha\subset E_1[n]$, where $E_1[n]$ is the $n$-torsion subgroup of $E_1$. It can be seen then $\alpha(E_1[n])\cong E_1[n]/\ker\alpha$, i.e. $\#\alpha(E_1[n])=n$, since $E_1[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$ in the case $char(k)\nmid n$.

Due to Velu's formula, there exists a curve $E_3$ with an isogeny $\alpha_1$ from $E_2$ to $E_3$ such that $\ker\alpha_1 = \alpha(E_1[n])$. Hence, the composition map $\alpha_1\circ \alpha$ form $E_1$ to $E_3$ has the kernel $E_1[n]$. This is also the kernel of the map $[n]: E_1\to E_1$ sending $P$ to $nP$. Due to Lemma 4.3, we have an isomorphism $\beta$ between $E_3$ and $E_1$, such that $\beta\circ \alpha_1\circ \alpha=[n]$. Now, we just let $\hat{\alpha}:=\beta\circ \alpha_1$ (Q.E.D)