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It is known that the cylinder $X:=S^1\times \mathbb R$ is a complex manifold. Does $X$ have analytic functions. i.e, if $f:X\to \mathbb C$ is analytic then is $f$ necessarily constant?

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    $\begingroup$ It has a continuum of mutually non-isomorphic complex structures, e.g., annuli with varying ratios of inner radius to outer radius. All of these examples have many non-constant holomorphic functions. Is this not what you are asking about? $\endgroup$ – paul garrett Jan 25 at 1:46
  • $\begingroup$ Can you explain your example to me please? Since I think this is what I am asking about. Is $X$ by the way Stein? $\endgroup$ – Amrat A Jan 25 at 2:07
  • $\begingroup$ $X$ is $\Bbb C/\Bbb Z$. A function $f: X \to \Bbb C$ lifts to the universal cover $\tilde f: \Bbb C \to \Bbb C$ with $\tilde f(z+1) = \tilde f(z)$, and conversely such a function descends to the quotient. So a holomorphic function $f: X \to \Bbb C$ is the same as a periodic holomorphic function on $\Bbb C$. You know many of these. $\sin(z/2\pi), \cos(z/2\pi), e^{z/2\pi}$. $\endgroup$ – user98602 Jan 25 at 3:27
  • $\begingroup$ Many smooth periodic functions ($\sum_n |a_nn^k| < \infty$ for every $k$) are not analytic. Iff $\sum_n |a_n|e^{ n y} < \infty$ for every $y\in (-r,R)$ then $\sum_n a_n e^{i n x}$ extends to an analytic function function $\sum_n a_n e^{i n z}, \Im(z) \in (-r,R)$ or $ \sum_n a_n z^n, |z|\in (e^{-R},e^r)$. Then for any $\lambda \in \mathbb{C}^*$, $\mathbb{C}/\lambda \mathbb{Z}$ is just a change of variable of $\mathbb{C}/\mathbb{Z}$, the two Riemann surfaces are isomorphic (in contrary to complex tori $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ non-isomorphic to $\mathbb{C}/(\mathbb{Z}+2i\mathbb{Z})$) $\endgroup$ – reuns Jan 25 at 4:24
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    $\begingroup$ Such regions are discussed in great detail in Ahlfors, if I recall correctly. (And as far as "Stein-ness" and so on go, these several complex variable notions that have content in several complex variables tend to degenerate in a single variable. E.g., it's easy to construct holomorphic functions with pretty random natural boundary sets.) $\endgroup$ – paul garrett Jan 25 at 13:35

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