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SOrry for the very specific question, but curiosity bites....

So here's the story: an idecomposable principally polarized abelian surface is embedded in $P^8=|3\Theta |^* $ as a deg 18 surface A. Moreover $|3\Theta|^*$ decomposes into two eigenspaces w.r.t. the canonical involution: one $P^3$ and one $P^4$. The $P^3$ is a sublinear system with base points the 10 even 2-torsion points, whereas the $P^4$ has the 6 odd 2-torsion points as base points. On the other hand $P^3\cap A \subset P^8=$ 6 odd 2-tors points and $P^4\cap A \subset P^8=$ 10 even 2-torsion pts. It is known that the projections of A on the $P^3$ and $P^4$ are, respectively, a quartic 6-nodal K3 surface, and a sextic 10-nodal K3 in $P^4$.

Now from a trivial degree computation I see that the degree of both projections is two, can you see an easy direct way to show that it is indeed two?

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I'd speculate that these are the linear systems on the Kummer surface given by twice the curves described in Hudson's "Kummer's quartic surface" sections 90 and 91: identifying the Kummer surface with it's image in $\mathbb{P}^3$ (as a quartic surface with 16 nodes at the images of the 2-torsion points), and choosing a 6-tuple of nodes which sit on a plane, there is exactly one reduced sextic on the Kummer surface which passes only on this 6-tuple of nodes out of the 16, and exactly one sextic on the Kummer surfaces which pass only on the residual 10 nodes out of the 16. The numerology (degrees of these divisors, their uniqueness, the nodes they pass through) suggests that these are halves of the linear systems in question.

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I went back to Hudson and read the actual section (instead of writing from memory - bad habit): there is a 5-dimensional family of sextics through these 6 nodes, and a 4-dimensional family of sextics through the residual ten. The corresponding linear systems are 4 and 3 dimensional respectively, they are both invariant under the kummer involution, and are non intersecting subspaces $|3\Theta|\cong\mathbb{P}^8$ (and therefor span it). It is also easy to check that they are base point free. I think this is closer to being a complete answer to the question

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  • $\begingroup$ yes I do agree with you, this should be so. I wondered if there was any modular explaination to the the 2:1 cover, but maybe it is just that they are the halves, as you point out. thank you for your answer. $\endgroup$
    – IMeasy
    Commented Nov 18, 2012 at 9:03
  • $\begingroup$ If there was a "classically known" modular explanation, it should have been in Bruce Hunt's "The geometry of some special arithmetic quotients" in chapter 4 or 5. I don't have it around, but I don't remember it being there - as far as I recall he just performs the computation you presented above. I agree that your question is natural, and deserves a modular answer. Weird part is - it also deserves the answer I've given - and I've never seen it either. $\endgroup$
    – David Lehavi
    Commented Nov 19, 2012 at 2:11
  • $\begingroup$ the only thing I can say is that both the weddle quartic and the sextic are birational models of the kummer so it is natural that there is a (ramified) deg 2 map from the ab surface embedded in $P^8$, but that's all I can say. $\endgroup$
    – IMeasy
    Commented Nov 19, 2012 at 9:49

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