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 $(A, \mathscr{L})$ isgeneralthen the three distinguished sections $X$, $Y$, $Z \in H^0(A, \mathscr{L})$ correspond tosmoothcurves in $|\mathscr{L}|$ which intersecttransversally?

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.