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.
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