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Does the Gamma function $\Gamma: \mathbb{C} \to \mathbb{C}$ preserve the Kummer ring $\mathbb{Z}[\exp(2\pi\imath/m)]$? And if not, then what about the Gaussian integers $\mathbb{Z}[\imath]$ or the Eisenstein integers $\mathbb{Z}[\exp(2\pi\imath/3)]$?

Is it possible to characterize holomorphic function which do preserve these lattices?

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    $\begingroup$ The gamma function has rapid decay in vertical strips, so I don't think it can preserve any lattices. I assume you are asking if it's possible that $\Gamma(L) \subset L$ where $L$ is a lattice. $\endgroup$
    – Matt Young
    Commented Oct 15, 2014 at 20:15
  • $\begingroup$ Yes, that's what I'm asking. I know there is a lot to know about $\Gamma$ but I never learned about it's behaviour on the pure imaginary numbers. $\endgroup$
    – Vít Tuček
    Commented Oct 15, 2014 at 21:59

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Note that

$$\frac{\Gamma(z) \Gamma(1-z)}{\Gamma(2z) \Gamma(1 - 2z)} = 2 \cos(\pi z),$$

and the RHS is transcendental for any non-rational algebraic number $z$ (by the Gelfond–Schneider theorem). So $\Gamma$ certainly won't preserve any number field $K$. It's most likely true that $\Gamma(z)$ is transcendental for algebraic $z \notin \mathbf{Z}$, but I'm not sure if that's known.

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  • $\begingroup$ I think the same argument works with the simpler expression $\Gamma(z)\Gamma(1-z)$. $\endgroup$
    – GH from MO
    Commented Oct 16, 2014 at 15:23
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    $\begingroup$ But then you get a factor with $\pi$. This expression just gets rid of it and it lets you use Gelfond-Schneider theorem directly. $\endgroup$
    – Vít Tuček
    Commented Oct 17, 2014 at 0:29
  • $\begingroup$ @VítTuček: Thank you, that makes sense. $\endgroup$
    – GH from MO
    Commented Oct 20, 2014 at 11:54
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I don't know about characterization, but there are lots of such functions. In fact, for any map $g$ of the lattice $L$ into itself, there are continuum-many entire functions $f$ such that $f(z) = g(z)$ for $z \in L$. This is because you can get entire functions that take prescribed values on any subset of $\mathbb C$ without limit points.

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  • $\begingroup$ Thanks! It seems that this extension problem has been dealt with on MO at least 3 times. :-) $\endgroup$
    – Vít Tuček
    Commented Oct 15, 2014 at 22:49

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