# Effective semi-group of a singular abelian surface

Let $$A$$ be a singular abelian surface over $$\mathbb{C}$$; that is, an abelian surface of maximal Picard rank $$\rho(A)=4$$. By Shioda-Mitani we know $$A \cong E \times E'$$ where $$E,E'$$ are isogenous elliptic curves with CM in an imaginary quadratic field $$\mathbb{Q}(\sqrt{-d})$$. I'm not sure if this is standard terminology, but by the effective semi-group, I mean the semi-group $$\text{NS}^{+}(A) \subset \text{NS}(A)$$ of integral points in the effective cone of $$A$$.

We can take as a basis of $$\text{NS}(A)$$ the four classes $$v, h, \Gamma, \Gamma_{\text{CM}}$$, where $$v,h$$ are the vertical and horizontal classes in $$E \times E'$$, $$\Gamma$$ is the graph of an isogeny between $$E, E'$$, and $$\Gamma_{\text{CM}}$$ is the graph of the CM map. Obviously we get effective classes by taking non-negative integer linear combinations of these basis elements. However, $$\text{NS}^{+}(A)$$ is not finitely generated (see, page 1 of https://arxiv.org/pdf/alg-geom/9712019.pdf). So my questions are:

1. Do we have any understanding of the lattice points in $$\text{NS}^{+}(A)$$ which are not non-negative linear combinations of $$v, h, \Gamma, \Gamma_{\text{CM}}$$? Has this been studied anywhere? There are infinitely many such points, but I'm really lacking intuition for these.

2. Given an explicit class in $$\text{NS}(A)$$, is there any useful way of determining when it is effective? Other than the fact that it must intersect positively with an ample class. I haven't heard of such a condition in general, but I'm hoping maybe this particular case is easier.

• Related: the situation for $\operatorname{NS}(E \times E)$ for a non-CM elliptic curve $E$ is worked out in Lazarsfeld's Positivity in algebraic geomtery I, Lemma 1.5.4. This should at least provide some intuition, even if the CM case might be a bit harder. May 11, 2020 at 22:08

Below is a summary of the discussion in Lazarsfeld's Positivity in algebraic geometry I, Ex. 1.4.7, Lem. 1.5.4, and Rmk. 1.5.6.

Lemma. Let $$D$$ be an $$\mathbf R$$-divisor on an abelian surface $$A$$. Then the following are equivalent:

1. $$D$$ is nef;
2. $$D$$ is pseudo-effective;
3. $$D^2 \geq 0$$ and $$D \cdot H \geq 0$$ for any ample divisor $$H$$.

Proof. Implication 1 $$\Rightarrow$$ 3 is clear. For 2 $$\Rightarrow$$ 1, it suffices to treat the case where $$D$$ is effective and irreducible. Any translate $$D + a$$ for $$a \in A$$ is algebraically equivalent to $$D$$, so $$D^2 = D(D + a) \geq 0$$ as $$D \neq D + a$$ for $$a \in A$$ general. Finally, for 3 $$\Rightarrow$$ 2 it suffices to show that if $$D$$ is an integral divisor with $$D^2 > 0$$ and $$D \cdot H > 0$$, then some multiple of $$D$$ is linearly equivalent to an effective divisor. This follows from Riemann–Roch for abelian surfaces. (In fact $$D$$ is ample; see e.g. this post, or Prop. 1.5.17 in Lazarsfeld.) $$\square$$

Example. For example, if $$E$$ is an elliptic curve with CM in $$\mathbf Z[\sqrt{-n}]$$ for $$n > 0$$, and $$\Delta, \Gamma \subseteq E \times E$$ are the diagonal and the graph of "multiplication by $$\sqrt{-n}$$" respectively, then the matrix of the intersection form with respect to the basis $$(h,v,\Delta,\Gamma)$$ is $$\begin{pmatrix}0 & 1 & 1 & n \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1+n \\ n & 1 & 1+n & 0 \end{pmatrix}.$$ A better basis is $$(h+v,h-v,\Delta-h-v,\Gamma-h-nv)$$, which gives the matrix $$\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2n \end{pmatrix}.$$ With respect to this basis, the equations become \begin{align*} a^2 \geq b^2 + c^2 + nd^2, & & a \geq 0 \end{align*} for a divisor $$D = a(h+v) + b(h-v) + c(\Delta-h-v) + d(\Gamma-h-nv)$$. These describe a circular cone in $$\operatorname{NS}(A)_{\mathbf R} \cong \mathbf R^4$$, so you can produce many effective classes close to the boundary with negative $$\Delta$$ or $$\Gamma$$ components. For example $$(a,b,c,d) = (2m^2+1,2m^2,-2m,0)$$ gives a divisor $$D$$ with $$D^2 = 2\Big((2m^2+1)^2 - (2m^2)^2 - (-2m)^2\Big) = 2\Big( 4m^4 + 4m^2 + 1 - 4m^4 - 4m^2 \Big) = 2,$$ so $$D$$ is effective (even ample). Its coefficient in $$\Delta$$ is $$-2m$$.

Remark. I don't know if every pseudo-effective class is algebraically equivalent to an effective one. (This is certainly false for "linearly equivalent", as can be seen with $$p \times E - q \times E$$ for different points $$p, q \in E$$.) On a general abelian surface I don't expect this to be true, because every effective class is ample if $$A$$ is simple, but I imagine there might be classes on the boundary of the nef cone (if $$A$$ has complex multiplication).

For a product of isogenous CM elliptic curves, there is a little more hope.

• Thanks for your wonderful answer! It clears part of my confusion up. I suppose another complication is that certain classes on the boundary of the circular nef cone might actually be effective right? For example, the classes $v, h, \Delta, \Gamma$ are all on the boundary. And presumably any effective curve of square zero lies on the boundary. May 12, 2020 at 4:59
• The boundary is exactly given by $D^2 = 0$, since $D^2 > 0$ and $D \cdot H \geq 0$ force $D \cdot H > 0$. So effective classes on the boundary correspond to square $0$ curves, which have arithmetic genus $1$ by Riemann–Roch/adjunction, hence are elliptic since $A$ has no rational curves. Every such elliptic curve is isogenous to $E$ because it is an isogeny factor of $A = E \times E$. I don't know a way to predict whether a given class on the boundary comes from an elliptic curve, but you should be able to write down many examples by thinking of étale correspondences. May 12, 2020 at 17:54