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Let $v_1 =\lambda_1 \zeta_1$ and $v_2 = \lambda_2 \zeta_2$ with $\zeta_1 = \frac{4\pi i\omega}{3}$ and $\zeta_2 = \frac{4\pi i\omega^2}{3}$ where $\omega = e^{2\pi i/3}$ is the third root of unity and $\lambda_1,\lambda_2$ some positive integers.

I would like to ask if there is an entire function $f$ such that $$ f(z+ v_1) = f(z) e^{2\pi i \lambda_1 Cz/3 }$$ and $$ f(z+ v_2) = f(z) e^{2\pi i \lambda_2 Cz/3 }$$ where $C \in \mathbb R$ is a constant such that $\frac{\lambda_1 \lambda_2 C (\zeta_2 -\zeta_1)}{3}=1$. This looks a bit like periodic boundary conditions, but since $z \in \mathbb C$ the modulus of these boundary conditions has of course a growing/decaying direction.

This looks pretty similar to something related to theta functions, but I don't quite get it together, as the underlying lattice looks rather different.

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    $\begingroup$ Presumably second equation should have $f(z+v_2)$, not $f(z+v_1)$. $\endgroup$ Commented Nov 10, 2022 at 20:57
  • $\begingroup$ @NoamD.Elkies absolutely, thank you! $\endgroup$
    – Guido Li
    Commented Nov 10, 2022 at 21:06

1 Answer 1

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The answer seems to be negative. Suppose that an entire function $f$ satisfies $f(z+v_i)=e^{A_iz}f(z)$, where $v_1$ and $v_2$ generate a lattice. Let $\Pi$ be the fundamental parallelogram of this lattice and integrate $f'/f$ over $\partial \Pi$. You obtain the ``Legendre's relation'': $$v_2A_1-v_1A_2=2\pi in,$$ where $n$ is the number of zeros of $f$ in $\Pi$.

Substituting your values, we see that $n=1$. Now $(f'/f)'$ is doubly periodic with respect to our lattice, having a single double pole per parallelogram. So we may assume (by shifting a pole to the origin) that $(f'/f)'=\wp+c,$ and two integrations integrations give $$f(z)=e^{P(z)}\sigma(z),$$ where $\sigma$ is the Weierstrass sigma function and $P$ is a polynomial of degree at most $2$. This is the general form of your $f$ (modulo a shift of the origin), if it exists.

Now let us try to find $P$. Sigma satisfies $$\sigma(z+v_j)=-e^{\eta_j(z+v_j)}\sigma(z),$$ where $\eta_j=\zeta(\omega_j)$, and $\zeta$ is the Weierstrass zeta function ($\zeta'=-\wp$), which gives $$P(z+v_j)=P(z)-\eta_j(z+v_j)+\pi i, \quad j=1,2.$$ Trying to find such a polynomial with your data, we just set $P(z)=az^2+bz+c$, and try to find $a,b,c$. Equation for $a$ is satisfied in view of Legendre's relation, and we have $$a=(A_j-\eta_j)/(2v_j), \quad j=1,2,$$ but for $b$ we obtain $$av_j^2+bv_j+\eta_jv_j=\pi i,\quad i=1,2.$$ These two equations with one variable must be consistent, which is very unlikely, certainly not for all $\lambda_1,\lambda_2$. To check this one has to compute $\eta_j$ for your lattices.

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