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I have a holomorphic function $G(z_1,z_2)$ in 2 variables, such that $G(z_1 + 1, z_2) = G(z_1,z_2)$, hence for a fixed $z_2$, $G(z_1,z_2)$ has a Laurent expansion in $e^{2\pi i z_1}$. I'm trying to show that this function doesn't have an essential singularity, so the Laurent series in $e^{2\pi i z_1}$ has only a finite negative terms. I would like to ask if anyone knows a theorem that is related to the problem? Something like if we know some properties of $G(z_1,z_2)$ then we know that it cannot have an essential singularity in the variable $z_1$ at $\infty$ or a relationship between singularities of the 2 variables.

Long
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2 Answers2

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Pick any entire function $f$ of one variable which is not a polynomial, say $f=\sum_{k=0}^\infty a_k z^k$. Then $f(e^{-2\pi i z_1})$ is a counterexample, as it has the form $$ f(e^{-2\pi iz_1}) = \sum_{k=0}^\infty a_k (e^{2\pi iz_1})^{-k}$$ considered as a Laurent series in $e^{2\pi iz_1}$.

vujazzman
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  • huh? what are you talking about? OP is talking about a holomorphic function of $2$ variables. – mathworker21 Apr 02 '20 at 23:51
  • My example function is constant in $z_2$ – vujazzman Apr 02 '20 at 23:54
  • okay. but something feels off to me. OP wants to show $G(z_1,z_2)$ doesn't have an essential singularity (tell me if you disagree with this). It seems if you used $f(e^{2\pi iz_1})$ instead of $f(e^{-2\pi i z_1})$, the Laurent series would have only positive terms and thus no essential singularity. But doesn't $f(e^{2\pi i z_1})$ have an essential singularity iff $f(e^{-2\pi iz_1})$ does? I'm very bad with this stuff, so sorry if I'm being stupid. – mathworker21 Apr 03 '20 at 00:02
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    $G$ itself does not have any singularity, much less essential. OP correctly remarks that any such periodic function factors through $F: (z_1,z_2)\mapsto (e^{2\pi iz_1},z_2)$, say, as $G = H\circ F$. The claim seems to be that $H$ must not have an essential singularity, but I provide a counterexample. – vujazzman Apr 03 '20 at 00:07
  • what about the rest of my comment? Isn't it true that $f(e^{-2\pi iz_1})$ has an essential singularity iff $f(e^{2\pi iz_1})$ does? And it seems the latter doesn't, since you can just say it is $\sum_{k=0}^\infty a_k (e^{2\pi i z_1})^k$. – mathworker21 Apr 03 '20 at 00:09
  • Yes that is true, but the question is not of $G$ having a singularity (which you are writing), but the corresponding function $H$. To be clear $G(z_1,z_2) = f(e^{-2\pi iz_1})$ is holomorphic, as asked. But as a function in $e^{2\pi i z_1}$ has an essential singularity. – vujazzman Apr 03 '20 at 00:11
  • That's not what I am writing. I thought your $H$ is $f$. – mathworker21 Apr 03 '20 at 00:15
  • $H$ here is $H(z_1,z_2) = f(1/z_1)$ – vujazzman Apr 03 '20 at 00:16
  • okay. I think I'm being stupid. I'll exit. I'll come back in a few hours – mathworker21 Apr 03 '20 at 00:17
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This answer may be seen a completion of the one of vujazzman: even if $G(z_1,z_2)$ has a Laurent expansion respect to $e^{2\pi i z_1}$ with no negative term (and thus it is a Taylor expansion), it can have an essential singularity on the whole complex line $\{(z_1,z_2)\in \Bbb C^2\,:\, z_2=0\}$ as, for example of the function $$ G(z_1,z_2)=\sum_{k=0}^\infty \frac{e^{2k\pi iz_1}}{k!z_2^{2k}}=\exp\left({\frac{e^{2\pi iz_1}}{z_2^2}}\right) $$ This implies that the behavior of a holomorphic function of several variables cannot be predicted only from its properties as a holomorphic function of a single one of its variables. Also the above examples shows a deep property of holomorphic functions of several variables: they cannot have compact singularities (by Hartogs's Extension theorem).

Daniele Tampieri
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