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Suppose I have an entire function $f : \mathbb{C} \longrightarrow \mathbb{C}^n$ for $n \geq 1$.

Let $C$ be the curve $f(\mathbb{C})$ in $\mathbb{C}^n$.

Let $\Lambda$ be a lattice in $\mathbb{C}^n$ (free $\mathbb{Z}$-submodule of rank $2n$) and assume that $\mathbb{C}^n/\Lambda$ is algebraic - it is an abelian variety.

Suppose that $C$ is stable by $\Lambda$ i.e. $C + \lambda = C$ for all $\lambda$ in $\Lambda$.

What conditions that imposes on $f$? I think it should impose that $f$ is a linear function but I have no proof.

This question is probably trivial to an expert in complex analysis..

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  • $\begingroup$ If $n \geq 2$, there is in general no such a function. Indeed, the $\Lambda$-invariant curve $C$ would give rise to a compact $1$-dimensional submanifold in the complex torus $T = \mathbb{C}^n / \Lambda$, but it is known that for a general choice of the lattice such a torus contains no $1$-dimensional complex submanifolds at all. So, your lattice must be special. For instance, you can require that the torus $T$ is algebraic (i.e, an abelian variety), which means that $\Lambda$ satisfies the Riemann bilinear conditions. $\endgroup$ Mar 26, 2015 at 10:58
  • $\begingroup$ Thanks for this reply! Yes, in fact I forgot to mention - my torus IS algebraic, it is an abelian variety $\endgroup$
    – user42721
    Mar 26, 2015 at 11:36
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    $\begingroup$ For $n=1$, all your condition $C + \lambda = C$ does is tell you that $f:\mathbb{C}\to\mathbb{C}$ is surjective. (If it weren't surjective, then it would have to miss a lattice of values, and so, by big Picard, it would have to be constant, which is clearly incompatible with $C + \lambda = C$.) There are many, many such surjective holomorphic maps (for example, all of the non constant polynomials); they need not be linear. $\endgroup$ Mar 26, 2015 at 11:54
  • $\begingroup$ Thanks for this; what about n>1? $\endgroup$
    – user42721
    Mar 26, 2015 at 11:59
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    $\begingroup$ Well, $n>1$ is still going to have this problem because if $f:\mathbb{C}\to\mathbb{C}^n$ does actually satisfy your condition, then $f\circ h:\mathbb{C}\to\mathbb{C}^n$ will also satisfy your condition for any $h:\mathbb{C}\to\mathbb{C}$ that is surjective. By the way, for $n>1$, there is no linear $f$ that satisfies your assumptions because you can always arrange that $f(0)=0$, and then your condition would imply that the complex line $f(\mathbb{C})$ contains the generators of the lattice $\Lambda$, which is impossible for $n>1$. $\endgroup$ Mar 26, 2015 at 12:09

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