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This question arose while I was studying some finite covers of abelian surfaces.

Let $(A, \mathscr{L})$ be a $(1,3)$-polarized abelian surface over the complex numbers and consider the vector space $H^0(A, \mathscr{L})\cong \mathbb{C}^3$. If $A$ is simple, then the linear system $|\mathscr{L}|$ is base-point-free, hence its general element is smooth by Bertini theorem.

Let us consider now the Heisenberg group $$\mathscr{H}_3:=\{(k, \, t, \, l) \mid k \in \mathbb{C}^* , \, t \in \mathbb{Z}/3 , \, l \in \widehat{\mathbb{Z}/3} \}$$ whose group law is $$(k, t, l)\cdot (k', t', l')=(kk'l'(t), t+t', l+l').$$ By [Birkenhake-Lange, Complex Abelian Varieties, Chapter 6] there exists a canonical representation, known as the Schrodinger representation, of $\mathscr{H}_3$ on $H^0(A, \mathscr{L})$, where the latter space is identified with the vector space $V:=\mathbb{C}(\mathbb{Z}/3)$ of all complex valued functions $f \colon \mathbb{Z}/3 \longrightarrow \mathbb{C}$.

Such an action is given by $$ (k, t, l)f(x)=kl(x)f(t+x).$$

Finally, let $X, Y, Z$ be the elements in $H^0(A, \mathscr{L})$ corresponding to the characteristic functions of $0, 1, 2$ in $V$, respectively.

Question. Is it true that if $(A, \mathscr{L})$ is general then the three distinguished sections $X$, $Y$, $Z \in H^0(A, \mathscr{L})$ correspond to smooth curves in $|\mathscr{L}|$ which intersect transversally?

This seems to me reasonable, but so far I could neither find any reference nor prove it.

If possible, I would also be pleased to see any explicit example where $X$, $Y$, $Z$ are smooth and intersect transversally.

This question can be generalized in a straightforward way to the case where $\mathscr{L}$ is a polarization of type $(1,d)$, but let us consider only $d=3$ for simplicity.

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For $d=5$ you get Horrocks-Mumford surfaces, which (unlike the case of $d=3$) are known quite explicitly; did you try that case? – Noam D. Elkies Nov 22 '11 at 18:37
@Noam: The problem I have in mind concerns the case $d=3$, so I did not try with Horrocks-Mumford surfaces. I will, anyway. Thank you for your suggestion. – Francesco Polizzi Nov 22 '11 at 21:01
A question and a remark - A) why do you say that the Schrodinger representation is unique, I see # SP_2(Z/3) of them ? B) If you want to work with explicit models then all (1,3) polarized Kummer varieties are quartic surfaces with 16 lines, after you blowdown the lines ( contains more ref's than you'll need); note that the images of 1,2 are interchagned by the Kummer involution, and that the Kummer involution restrcits to an involution on the image of 0. – David Lehavi Nov 22 '11 at 21:08
@david: I did not say that the representation is unique, but the one I have given should be canonical in some sense. Furthermore, it is the one I'm using for my computations. If you have time, could you please expand your nice observation in a proper answer, adding some detail on the action of the Kummer involution on the three sections? I will be glad to upvote it. Thanks! – Francesco Polizzi Nov 23 '11 at 13:36

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