Let $C_2$ be a smooth genus $2$ curve and $J(C_2)$ its Jacobian. It is well known that the blow-up of $J(C_2)$ at the origin $o$ is isomorphic to the second symmetric product $\textrm{Sym}^2(C_2)$, and that the blow-up morphism $$\pi \colon \textrm{Sym}^2(C_2) \to J(C_2)$$ coincides with the Abel-Jacobi map.

The exceptional divisor $E \subset \textrm{Sym}^2(C_2)$ corresponds to the unique $g^1_2$ of $C_2$, so it intersects transversally the diagonal $\Delta \subset \textrm{Sym}^2(C_2)$ at six points, corresponding to the six Weierstrass points of $C_2$. Hence the image $\pi(\Delta) \subset J(C_2)$ is a curve with an ordinary singular point of multiplicity six at $o$.

Finally, the curve $\Delta$ is the branch locus of the double cover $C_2 \times C_2 \to \textrm{Sym}^2(C_2)$. This implies that its class is $2$-divisible in the Néron-Severi group $\textrm{NS}(\textrm{Sym}^2(C_2))$, and so the same holds for the class of $\pi(\Delta)$ in $\textrm{NS}(J(C_2))$.

More precisely, it is possible to show that $\pi(\Delta)$ is algebraically equivalent to $4 \Theta$, where $\Theta$ is the theta divisor, namely the principal polarization of the Jacobian (see for instance this MO question).

In particular, we have $$\pi(\Delta) \cdot \pi(\Delta) = 16 \Theta^2 = 32.$$

I would like to know whether this is the only possible occurrence for this situation, so let me ask the following

Question. Let $A$ be an abelian surface and $D$ be an effective divisor on it such that

  • $D$ has an ordinary point of multiplicity $6$ at the origin $o \in A$ and no other singularities;
  • $D^2=32;$
  • the class of $D$ is $2$-divisible in $\textrm{NS}(A)$.

Is the pair $(A, \, D)$ necessarily of the form $(J(C_2), \, \pi(\Delta))$ for some genus $2$ curve $C_2$?

  • 1
    $\begingroup$ This seems very likely. A possible approach might be to first compute the geometric genus of $D$ using the first two points. Assuming one gets it to be $2$, let $C$ be the normalization of $D$ and consider the map $J(C) \to A$ induced by the map $C \to D$. Again using the first two points one should probably get that this is essentially the multiplication by $2$ map on $J(C)$. This should imply what you want (so the last condition might be superfluous). $\endgroup$
    – naf
    Jun 18, 2016 at 10:23
  • $\begingroup$ @ulrich: thanks for the suggestion. How do you know that $C$ is irreducible? $\endgroup$ Jun 18, 2016 at 12:01
  • 1
    $\begingroup$ Sorry, I was assuming that $D$ was irreducible though you didn't say that. If this is not assumed I guess one might have to consider some cases, but first one should check that the argument does indeed work if $D$ is assumed to be irreducible. $\endgroup$
    – naf
    Jun 19, 2016 at 10:00


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