Let $E$ be a smooth elliptic curve over an algebraically closed field of characteristic zero. Let $\mathcal{L}$ be a line bundle of degree $3$. Heisenberg group $H\_3$ acts on global sections of $\mathcal{L}$. Let me choose basis $z\_1, z\_2, z\_3$ in $H^0(E, \mathcal{L})$ such that generators of of order three points acts in a standard way 
$$
z\_i \mapsto \rho^i z\_i \\
z\_i \mapsto z\_{i-1},
$$
where $\rho$ is a primitive cubic root. Is it true that embedding of $E$ into projective plane given by these sections produce a cubic curve in a Hesse form? Is it possible to determine parameter in Hesse pencil in terms of moduli of $\mathcal{L}$?



Hesse pencil and Schrodinger representation of Heisenberg group
The answers are **yes** and **no**.


**Yes.** Denote by $g$ the first of mentioned generators of order $3$. Consider the embedding $E\subset{\mathbb P}\_k^2$ given by $\mathcal L$, pick as $0\in E$ one of the inflection points, and denote by $\oplus$ the operation on $E$. Let $p\_1+p\_2+p\_3$ be the divisor of the intersection of $E$ with the line $z\_i=0$. Then $p\_1\oplus p\_2\oplus p\_3=0$. By construction,
$p\_1+p\_2+p\_3=(p\_1\oplus g)+(p\_2\oplus g)+(p\_3\oplus g)$, implying that the $p\_i$'s are of order $3$. Hence, all $9$ inflection points lie on the "parallel" lines $z\_i=0$ and, therefore, provide a Hesse pencil (see <http://en.wikipedia.org/wiki/Hesse_configuration> and <http://en.wikipedia.org/wiki/Hesse_pencil> for details).


**No.** As line bundles of degree $3$ up to an isomorphism are divisors of degree $3$ up to linear equivalence, the moduli space of such line bundles is isomorphic to the proper elliptic curve $E$. At the level of divisors, every divisor $D$ on $E$ of degree $3$ is equivalent to $2\cdot0+p$, $p\in E$, so we map $[D]\mapsto p$. On the other hand, there is a map
$j:{\mathbb P}\_k^1\to{\mathbb P}\_k^1$ of degree $9$ (first ${\mathbb P}\_k^1$ is the Hesse pencil) that calculates the $j$-invariant of $E$ (see <http://arxiv.org/pdf/math/0611590v3.pdf>). Which of the $9$ possible points of the Hesse pencil you get depends on the choices you made while constructing the pencil and not on the choice of ${\mathcal L}$"$\in$"$E$.


