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Let $(X, \mathcal{O}_X)$ be a complex analytic space in the sense of Grauert, i.e., a $\mathbb{C}$-analytic ringed space which is locally isomorphic to a local model. We may assume that $X$ is a complex manifold for our purposes here.

Let $\mathcal{O}_X$ denote the sheaf of analytic functions on $X$ and $\mathcal{O}_X^{\ast}$ denote the associated multiplicative sheaf. For each $x \in X$, we define a morphism on stalks $\exp : \mathcal{O}_{X,x} \longrightarrow \mathcal{O}_{X,x}^{\ast}$ which assigns to each analytic function $f \in \mathcal{O}_{X,x}$ the non-zero analytic function $\exp(f)\in \mathcal{O}_{X,x}^{\ast}$.

Claim: $\ker(\exp) = 2\pi i \mathbb{Z}$, where $\mathbb{Z}$ is the stalk of the constant sheaf at $x$. The subscript $x$ is omitted for obvious reasons.

It is obvious that $2\pi i \mathbb{Z} \subset \ker \exp$,but it is not immediate that the reverse inclusion is also true. A standard argument, which can be found in Kaup and Kaup's text on several complex variables is of the following form:

For $f \in \ker(\exp)$, assume that $f(x) =0$. Expanding $\exp(z)$ into a power series centered at $x$, we obtained a stalk $g_x \in \mathcal{O}_{X,x}$ such that $\exp(f) = 1 +f - g f^2$. Since $\exp(f) =1$, it follows by induction that $$f = f^2 g = \cdots = f^{j+1} g^j \in \bigcap_{j=1}^{\infty} \mathfrak{m}_x^j = (0),$$ where the last line follows from elementary properties of analytic algebras and $\mathfrak{m}_x$ denotes the maximal ideal of the local ring $\mathcal{O}_{X,x}$. Explicitly, $\mathfrak{m}_x = \{ \varphi \in \mathcal{O}_{X,x} : f(x) =0 \}$.

I am curious as to whether anyone is aware of other ways of proving this result, perhaps in a more analytic or geometric manner?

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    $\begingroup$ I don't get it. If $\exp(f)=1$ on a connected open subset of $X$, $f$ is constant. $\endgroup$ – abx Jul 10 '18 at 6:19
  • $\begingroup$ @abx Does the identity theorem hold on general complex analytic spaces? (possibly non-reduced). $\endgroup$ – AmorFati Jul 10 '18 at 23:03
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Let $f\in\ker(\exp)$; by definition: \begin{equation} \forall x\in X,\,\exp_x(f_x)=1\in\mathcal{O}_{X,x}^{\times}, \end{equation} in other words: \begin{equation} \exp_x(f_x)=\exp_x(Re(f_x))[\cos(Im(f_x))+i\sin(Im(f_x))]=1\Rightarrow\\ \Rightarrow f_x=2k\pi i\,\text{for some}\,k\in\mathbb{Z}. \end{equation} By definition of stalks: there exists an open neighbourhood $U$ of $x$ such that $f_{|U}=2k\pi i$, that is $f$ is locally constant to value $2k\pi i$ for some $k\in\mathbb{Z}$.

So: \begin{equation} \forall x\in X,\,\ker(\exp)_x\subseteq2\pi i\mathbb{Z}_x, \end{equation} by previous reasoning: $\ker(\exp)=2\pi i\mathbb{Z}$.

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