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.

  • 1
    $\begingroup$ 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. $\endgroup$ May 11, 2020 at 22:08

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Benighted
    May 12, 2020 at 4:59
  • 1
    $\begingroup$ 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. $\endgroup$ May 12, 2020 at 17:54

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