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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.

rla
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    If I can find where I've seen this I'll flesh it out to an answer. Here's my guess though. One way to think of the NS group is as the image under the first chern class map $c_1: Pic(Y)\to H^2(Y, \mathbb{Z})$. For any abelian surface, this is free of rank $6$. But better, the Lefschetz 1-1 theorem says the image actually lands in $H^1(X, \Omega^1)$ under the Hodge decomposition. But this has dimension $4$ (and intersected back with $H^2(X, \mathbb{Z})$ is a rank $4$ free abelian group). This gives the classical result that the $rk(NS(Y))\leq 4$. – Matt May 31 '13 at 01:13
  • Here's what I don't remember. I think there is a theorem that says it equals $4$ if and only if you are a product of two supersingular elliptic curves. So my guess is that in this case $NS(Y)\simeq \mathbb{Z}^4$. – Matt May 31 '13 at 01:14
  • @Matt you are right. The generators are $[\mathbb{C}/\mathbb{Z}[i] \times { \ast }], [{ \ast } \times [\mathbb{C}/\mathbb{Z}[i]], [{ (z, z) \in Y }]$ and $[{ (z, iz) \in Y }]$. Maybe that is the way to describe the NS group: find the rank and try the trivials curves in that product. – rla May 31 '13 at 01:48

2 Answers2

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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 $Y$.

Yuchen Liu
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    Dear jerry, Regarding "This argument applies for any abelian surface $Y$", I'm not sure exactly what you have in mind, but for a typical abelian surface the NS group will have rank $1$, not $4$. For a product of elliptic curves, the rank of NS is typically $2$, or $3$ if they are isogenous, or $4$ if they are isogenous CM curves (as in this particular example). Regards, – Matt E May 31 '13 at 05:06
  • Yes, you are right, but my remark here is only related with the second question, not the first question. I will edit it to be more precise. Thanks! – Yuchen Liu May 31 '13 at 05:23
  • Dear jerry, Ah, that makes sense! Cheers, – Matt E May 31 '13 at 05:50
  • @jerrysciencemath are you applying Hodge Index Theorem correctly? I saw in Hartshorne "Algebraic Geometry" that you should have $D$ ample. – rla Jun 01 '13 at 01:35
  • @rla The Hodge index theorem says that the intersection form on $NS(X)\otimes \mathbb{R}$ has signature $(1,\rho-1)$ where $\rho=rk(NS(X))$. Hartshorne seems give a different version of the Hodge index theorem, but actually they are equivalent to each other: If $D$ is ample and $(D\cdot C)=0$ implies $(C^2)<0$, then of course the intersection form has signature $(1,\rho-1)$. Conversely, if the intersection form has signature $(1,\rho-1)$, then for any divisor $D$ such that $(D^2)>0$ (in particular $D$ is ample) and $(D\cdot C)=0$, we have $(C^2)<0$, this is a linear algebra exercise. – Yuchen Liu Jun 01 '13 at 01:56
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    Dear @jerrysciencemath, once again I'm impressed by the maturity you display at 22 years of age. I think you can be proud of yourself and I would be very happy to see you sign your posts with your real name (unless you have a strong reason not to do so). Anyway, +1. – Georges Elencwajg Jun 02 '13 at 07:34
  • @GeorgesElencwajg Thanks for your encouragement! Would you like to chat with me online? I found the chat room interesting and convenient, I'm at the room "Mathematics" now. – Yuchen Liu Jun 02 '13 at 07:46
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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.

rfauffar
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