Let $X$ be the Jacobian of a genus 2 curve over $\mathbb{C}$. Let $L=\mathcal{O}(nC)$, where n is an even number. Is it possible to find a smooth curve from $|L|$ which is fixed by the involution $x\mapsto -x$ and which passes through the sixteen 2-torsion points? I have the following ideas:

1) if we take $n$ to be sufficiently large, $L$ will be very ample. Consider $\hat{X}$, the blow up of $X$ at the sixteen 2-torsion points, $b:\hat{X}\longrightarrow X$. Let $E_1,...,E_{16}$ be the exceptional divisors, let $E=\Sigma E_i$. Then the curves I am interested in $|b^*L-E|$. This has no basepoints. So by Bertini there is a open dense subset of the complete linear system consisting of smooth curves. Now the global sections $H^0(b^*L-E)$ breaks up into $+$ and $-$ eigen spaces because the involution acts on it. Curves coming from either the $+$ space or the $-$ space passes through all 16 points and is fixed by the involution. But will they be smooth? (Thanks @Francesco Polizzi and @abx for the answers to two related questions that I asked from which this idea is entirely based on : Curve through the 16 singular points of a Kummer surface and A curve in an abelian surface and its image in the Kummer surface).

2) The other idea is this. If $X=J(C)$, we can make sure that the involution $i$ on $X$ restricts to the hyperelliptic involution on $C$. We can also make $C$ pass through 0. So $i^*\mathcal{O}(C)=\mathcal{O}(C)$. Now we have $[2]:X\longrightarrow X$, multiplication by 2. Choose $L=\mathcal{O}(nC)$ where $4|n$. Under $[2]$, $\mathcal{O}(\frac{n}{4}C)$ pulls back to $L$. So $nC$ maps to some curve $C'\in|\mathcal{O}(\frac{n}{4}C)|$ which will contain 0? And so $nC$ contains all 2-torsion points?

I am not quite confident about these arguments. But I am required to use these in the course of work I do. I have asked other similar questions to which I have got some very enlightening answers. I would be grateful for help in this direction too!

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    $\begingroup$ In idea (2), why not just take the preimage of $C$ through $[2]$? $\endgroup$ – Daniel Litt Nov 6 '15 at 2:55
  • $\begingroup$ Will it be smooth? @Daniel Litt $\endgroup$ – gradstudent Nov 6 '15 at 3:09
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    $\begingroup$ Yes, the map $[2]$ is a local analytic isomorphism, so the preimage of anything smooth is smooth. $\endgroup$ – Daniel Litt Nov 6 '15 at 3:10
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    $\begingroup$ I assume you mean "preserved by the involution" rather than "fixed by the involution." Yes. Suppose that $x$ is an element of the preimage, i.e. $[2](x)\in C$; then you wish to show that $-x$ is as well. But $[2](-x)=-[2](x)$ which is in $C$ because $C$ is preserved by the involution. $\endgroup$ – Daniel Litt Nov 6 '15 at 4:03
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    $\begingroup$ Oh, I'm sorry, I didn't see that you wanted this to belong to $|L|$, this actually requires some argument. See Corollary 3 on page 59 of Mumford's book Abelian Varieties, and use the fact that $[-1]^*L=L$ (as $C$ is preserved by $[-1]$) to conclude that $[2]^*L=L^{\otimes 4}$. Hence $[2]^{-1}(C)\in |O(4C)|$. $\endgroup$ – Daniel Litt Nov 6 '15 at 4:15

We have $X=\text{Jac}(C)$, for $C$ some hyperelliptic curve. By appropriate choice of basepoint, we can arrange that the embedding $C\to X$ passes through $0\in X$ and is that $[-1]$ on $X$ restricts to the hyperelliptic involution on $C$.

Let $C'=[2]^{-1}(C)$. I claim that this curve satisfies all of the desired properties. It is (a) smooth, as $[2]$ is a local analytic isomorphism and $C$ is smooth, (b) preserved by $[-1]$, by group theory, (c) passes through the $2$-torsion of $X$ (as $0\in C$), and (d) is in $|\mathscr{O}(4C)|$, as $[2]^*\mathscr{O}(C)=\mathscr{O}(4C)$.

To see this last, we use Corollary 3 on page 59 of Mumford's book Abelian varieties, which implies that if $L$ is a line bundle with $[-1]^*L\simeq L$ (i.e. $L$ is symmetric), $[n]^*L=L^{\otimes n^2}$. $\mathscr{O}(C)$ is symmetric because $C$ itself is preserved by $[-1]$.

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  • $\begingroup$ Probably you should mention the small argument to justify that $C'$ is also connected. $\endgroup$ – nfdc23 Nov 28 '15 at 21:58
  • $\begingroup$ Sure; this follows from the surjectivity of the map $\pi_1(C)\to \pi_1(\text{Jac}(C))$, for example. $\endgroup$ – Daniel Litt Nov 28 '15 at 22:06

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