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.
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)
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 $(\Lambda(L)\otimes p^*M)|_{A\times \{0\}}=M$.
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 $L=(id,\phi)^*P$.
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 $L=(id,\phi)^*P$.
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