I am reading Fulton's but I cannot find a useful result for my problem that seems to be something well known.

Let $\mathcal{J}$ be the Jacobian of a hyperelliptic curve of genus $2$. Consider $D_1,D_2\in \text{Div}(\mathcal{J})$ (two curves inside $\mathcal{J}$) and let $D_2'$ be a translation by a $2$-Torsion point of $\mathcal{J}$.

Q: Is it true that $D_1\cdot D_2=D_1\cdot D_2'$ ?

I cannot see directly that $D_2\sim D_2'$ always. I will appreciate a source in case of being true, or a condition for $D_2$ for this to happen.

  • $\begingroup$ A divisor is always algebraically equivalent to its translates. No ampleness is needed $\endgroup$
    – user45150
    Dec 11, 2017 at 20:47

1 Answer 1


The answer is yes if $D_2$ is an ample curve. More generally, the following result holds:

Two ample line bundles on an abelian variety are algebraically equivalent if and only if they differ by a translation.

This is essentially Corollary 2.5.4 in Birkhenake-Lange book Complex Abelian Varieties (it is at p. 40 in the second edition), where it is stated for analytically equivalent line bundles on (non necessarily projective) complex tori.

  • $\begingroup$ Dear Francesco, Thanks for the quick answer, ampleness is very special, in fact I am intersecting the $\Theta$ divisor (which is ample) by the image of the $\Theta$ divisor under an endomorphism of the Jacobian. The translation I am doing is with the second divisor which I think it is always ample. But I was wondering if it is not ample, the Two torsion preserves symmetry, therefore maybe the intersection numbers will be the same. $\endgroup$ Dec 11, 2017 at 19:58
  • $\begingroup$ If the self-intersection of the divisor is $>0$, then it is ample (again, this is Corollary 2.5.4 in BL). It seems to me that this is the case, or not? $\endgroup$ Dec 11, 2017 at 20:02
  • $\begingroup$ At any rate, the general Jacobian has the Néron-Severi group generated by (the class of) the Theta divisor, so every effective curve will be ample in that case. $\endgroup$ Dec 11, 2017 at 20:03
  • 1
    $\begingroup$ The hard part of the theorem, is definitely the other direction. A divisor is always algebraically equivalent to one if its translates. No ampleness is needed $\endgroup$
    – user45150
    Dec 11, 2017 at 20:49

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