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 $\mathcal{L}$?
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$\begingroup$ Didn't Hesse write Steppenwolf and Siddharta? $\endgroup$– Will JagyCommented Jan 11, 2014 at 1:57
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$\begingroup$ en.wikipedia.org/wiki/Hermann_Hesse and en.wikipedia.org/wiki/Helmut_Hasse $\endgroup$– Will JagyCommented Jan 11, 2014 at 2:13
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1$\begingroup$ en.wikipedia.org/wiki/Otto_Hesse $\endgroup$– Sasha PavlovCommented Jan 11, 2014 at 2:34
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$\begingroup$ Amazing. Never heard of him before this. Maybe that's just me. Will people (well, people who might be able to answer your question) know what a Hesse pencil is? $\endgroup$– Will JagyCommented Jan 11, 2014 at 2:47
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$\begingroup$ Hessian matrix (second order terms in Taylor expansion) is named after him. As for Hessian pencil I believe it is standard terminology, although I was not able to find any reasonable sources to read about it. $\endgroup$– Sasha PavlovCommented Jan 11, 2014 at 2:58
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2 Answers
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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 ${\mathcal L}$"$\in$"$E$.
answered Jan 11, 2014 at 9:57
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Yes, the equation of the curve will be in Hesse form (that is, $z_0^3+z_1^3+z_2^3=3\mu z_0z_1z_2$). Indeed it must be invariant under the Heisenberg representation; it is an easy exercise to see that the space of invariant cubic forms is spanned by $z_0^3+z_1^3+z_2^3$ and $z_0z_1z_2$.
The $j$-invariant of this elliptic curve is easily calculated (see for instance here, p. 4-5): it is equal to $-2^{-8}\Bigl(\dfrac{\mu (\mu ^3+8)}{\mu ^3+1}\Bigr)^3$ (you better check the constant!).
