Divisors in an abelian surface

How to compute the Néron-Severi group of the abelian surface $Y = \mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i]$. More generally, are there any result that compute the Néron-Severi group of product of curves?

Suppose that surface $Y$ has a divisor class $[F]$ such that $[F]^{2} > 0$. Why is $[F]$ or $-[F]$ ample?

Thanks.

Solution 1:

For your first question, the answer is based on your comments.

By Lefschetz (1,1) theorem, $rk(NS(Y))=\dim_\mathbb{Q}H^2(Y,\mathbb{Q})\cap H^{1,1}(Y)$, so for any abelian surface $Y$, we have $rk(NS(Y))\leq h^{1,1}(Y)=4$.

Claim: Suppose $Y=\mathbb{C}/\mathbb{Z}[i]\times \mathbb{C}/\mathbb{Z}[i]$, then $rk(NS(Y))=4$.

Proof: Let $(z,w)$ be the complex coordinate of $\mathbb{C}^2$, the universal cover of $Y$. Here $Y=\mathbb{C}^2/\Lambda$, $\Lambda$ is a free abelian group generated by $(1,0),(i,0),(0,1),(0,i)$.

Consider $\alpha=idz\wedge d\bar z$, $\beta=idw\wedge d\bar w$, $\gamma=i(dz\wedge d\bar w+dw\wedge d\bar z)$, $\delta=dz\wedge d\bar w-dw\wedge d\bar z$, so $\{\alpha,\beta,\gamma,\delta\}$ is a basis for the complex vector space $H^{1,1}(Y)$.

Moreover, it follows by simple calculation that $[\mathbb{C}/\mathbb{Z}[i]\times\{\ast\}]=\beta/2$, $[\{\ast\}\times\mathbb{C}/\mathbb{Z}[i]]=\alpha/2$, $[\{(z,z)\in Y\}]=-\gamma/2$, $[\{(z,iz)\in Y\}]=\delta/2$, so $\alpha,\beta,\gamma,\delta\in H^2(Y,\mathbb{Q})\cap H^{1,1}(Y)$ and they are linearly independent over $\mathbb{Q}$. QED

For your second question, since $K_Y=0$, by Riemann-Roch theorem $$ h^0(nF)+h^0(-nF)=h^0(nF)+h^2(nF) \geq \frac{(F^2)}{2}n^2 +\chi(\mathcal{O}_Y). $$

Therefore, for $n>>1$, either $nF$ or $-nF$ is effective. WLOG, we assume $D=nF$ is effective. Since $Y$ is abelian surface, every curve $C$ in $Y$ is movable (in the sense of algebraic equivalence), so for any two (possibly reducible) curves $C_1$ and $C_2$ we have $(C_1\cdot C_2)\geq 0$. In particular, for any curve $C$ we have $(D\cdot C)\geq 0$.

If $(D\cdot C)=0$ for some curve $C$, together with $(D^2)>0$, the Hodge index theorem implies $(C^2)<0$, we get a contradiction! Thus for any curve $C$, we have $(D\cdot C)>0$, together with $(D^2)>0$, the Nakai-Moishezon criterion tells us that $D=nF$ is ample, so $F$ is ample.

Remark: This argument for your second question applies for any abelian surface $Y$.

Solution 2:

In general, for two curves $C$ and $C'$ you have that $\mbox{Pic}(C\times C')\simeq\mbox{Pic}(C)\times\mbox{Pic}(C')\times\mbox{Hom}(JC,JC')$, where $JC$ denotes the Jacobian of $C$, and I'm sure there must be a similar situation with the Néron-Severi group for the product of curves. Now, if $(A,\lambda)$ is a polarized abelian variety, you have that $\mbox{NS}(A)\simeq\mbox{End}_\lambda(A)$, that is, the endomorphisms of $A$ that are symmetric (invariant under the Rosati involution defined by $\lambda$). In the case of the product of elliptic curves, it is easy to see that we have an isomorphism $$\mbox{End}_{\lambda_1}(E_1)\times\mbox{End}_{\lambda_2}(E_2)\times\mbox{Hom}(E_1,E_2)\simeq\mathbb{Z}^2\times\mbox{Hom}(E_1,E_2)\to\mbox{End}_{\lambda_1\otimes\lambda_2}(E_1\times E_2),$$ given by $$(\alpha,\beta,f)\mapsto\left(\begin{array}{cc}\alpha &\lambda_2^{-1}\hat f\lambda_1\\f&\beta\end{array}\right)$$ where $\lambda_1\otimes\lambda_2$ denotes the product polarization and $\lambda_i$ corresponds to the polarization on $E_i$ given by the divisor of a single point. This then induces an isomorphism $\mathbb{Z}^2\times\mbox{Hom}(E_1,E_2)\to\mbox{NS}(A)$ given by $$(a,b,f)\mapsto (a-\deg f)[\{0\}\times E_2]+(b-1)[E_1\times\{0\}]+[\Gamma_{-f}],$$ where $\Gamma_{-f}$ is the divisor corresponding to the graph of $-f$. This answer is independent of the ground field.

To answer your second question, at least for abelian varieties over $\mathbb{C}$, it is interesting to note that if $(A,\Theta)$ is a polarized abelian variety of dimension $n$, then a divisor $D$ is nef if and only if $(D^{k}\cdot\Theta^{n-k})\geq 0$ for all $k$ (see Lemma 1.1 in http://arxiv.org/pdf/alg-geom/9712019v1.pdf), and it is ample if it is a strict inequality. So for dimension 2, we have that if $(F^2)>0$, then either $(F\cdot\Theta)>0$ (and in this case $F$ is ample) or $(F\cdot\Theta)\leq 0$ and so $-F$ is at least nef. if $(F\cdot\Theta)=0$, then by the same Lemma, $F$ is linearly equivalent to an effective divisor, and since the intersection of an ample divisor and an effective divisor is positive (if the effective divisor is not numerically trivial), we must have that $F\equiv 0$, and so $(F^2)=0$, a contradiction.