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Sorry lads, but my analysis is long in the past...

Given a lattice $\Gamma$ in the complex plane (or more generally: in a complex vector space $V$), why does there not exist a holomorphic function $f$ on the whole plane (or $V$) which satisfies:

$f(z+\gamma)=f(z)-\bar{\gamma}$

for all $z \in \mathbb C$ (or $V$) and $\gamma \in \Gamma$ ?

(here $\bar{\gamma}$ denotes the complex conjugate).

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How about $f(z) = \sin z - z$, with $\gamma=2\pi$? –  Michael Hardy Sep 29 '11 at 20:43
    
Oh..... "lattice" must have meant having two periods that are linearly independent over $\mathbb{R}$. –  Michael Hardy Sep 29 '11 at 20:44
    
Yes, that's a lattice. –  Veen Sep 29 '11 at 20:47
    
What do you take a lattice in a general complex vector space to mean here? Would $\{(2\pi a, 2\pi b)\mid a,b\in\mathbb Z\}\subseteq \mathbb C^2$ qualify? –  Henning Makholm Sep 29 '11 at 21:54
    
A subgroup generated by a real basis of $V$. Just look at a complex tours e.g. –  Veen Sep 30 '11 at 7:41

2 Answers 2

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Assume there exists such a function $f.$

Recall that any entire function $g:\mathbb{C} \rightarrow \mathbb{C}$ satisfying the equation $g(z + \omega) = g(z)$ for all $\omega \in \Gamma$ is constant. And observe that the derivative of $f$ is of this form. It follows $f(z) = az +b \text{ for }a,b \in \mathbb{C}.$

But then

$$a(z + \gamma) + b = az + b - \overline{\gamma}$$ for all $\gamma \in \Gamma.$

Hence, $-a\gamma = \overline{\gamma}$ for all $\gamma \in \Gamma.$ As the $\mathbb{R}$-span of $\Gamma$ is equal to $\mathbb{C}$ and complex conjugation and complex multiplication are both $\mathbb{R}$-linear, it follows that multiplication by $-a$ and complex conjugation are equal as automorphisms of $\mathbb{C}.$ But the former is holomorphic and the latter is not. This is a contradiction. It follows no such function $f$ exists.

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Assume without loss of generality that $1\in\Gamma$. (Otherwise just replace $f$ with $g(z)=f(\gamma z)/\bar\gamma$ for some nonzero $\gamma\in\Gamma$).

Let $\gamma$ be any lattice point, and integrate $f$ around the parallelogram $0\to\gamma\to1+\gamma\to1\to0$. The terms that involve $f$ cancel due to the periodicity, and the result ends up being just $\gamma-\bar\gamma$. This is twice the imaginary part of $\gamma$, which is nonzero for at least some lattice points -- but if the integral around a closed curve is nonzero, then $f$ cannot be entire.

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