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Who can tell me why the compactification of RS of Log is just Riemann Sphere, please?

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2 Answers 2

The Riemann surface is just a copy of $\mathbb C$. Compactifying it (by adding one point) gives therefore the usual sphere.

In general, if $f:\mathbb C\to U$ is an surjective holomorphic function with non-zero derivative with image an open subset of $\mathbb C$, the Riemann surface of the inverse function $f^{-1}:U\to\mathbb C$ (which is in general not-single valued, so I am abusing notation a bit here...) is just the graph of $f$, the set $\Gamma=\{(z,f(z)):z\in\mathbb C\}$. The restriction of the first projection $\mathbb C\times\mathbb C\to\mathbb C$ gives a holomorphic function $\Gamma\to\mathbb C$ which is a biholomorphism. Your question involves the case wher $f(z)=\exp(z)$.

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The Riemann surface of $\log z$ is an infinite cylinder since $\log z = \log|z| + \mathrm{arg}(z) + 2 \pi i k$.

To compactify, we're going to add two points at $z = \pm i\infty$, closing the cylinder to make a sphere.

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That surface is diffeomorphic to a plane. Adding two points to it cannot give a sphere. –  Mariano Suárez-Alvarez Oct 11 '12 at 17:23

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