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I am looking for universal covering spaces and I am now wondering if the Riemann surface for the square root $z^{1/2}$ (or even more general for $z^{1/n}$) is simply-connected and therefore a universal covering space for the punctured complex plane?

If so, is (the topology of) the Riemann surface for $log(z)$ homeomorphic to the Riemann surface for $z^{1/n}$? If both Riemann surfaces are simply-connected covering spaces for the punctured complex plane, then they clearly must be homeomorphic, but it doesn't seem obvious just from looking at them.

I am not familiar with complex analysis (I am specializing in geometry), but need an explanation for a result in geometry...so any help is greatly appreciated!

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  • $\begingroup$ The punctured plane is homeomorphic to $\mathbb S^1\times\mathbb R$, and hence its universal covering space is $\mathbb R\times\mathbb R$. Typically, this is represented by the exponential function $z=e^w$. $\endgroup$ – Thomas Andrews Mar 21 '15 at 4:22
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The Riemann surface for $z^\frac{1}{2}$ is a copy of $\mathbb{C}^*$, and the covering map you want is $z \mapsto z^2$.

For $z^\frac{1}{n}$, it's still $\mathbb{C}^*$, and the covering map is $z \mapsto z^n$.

For $\log(z)$, the Riemann surface is $\mathbb{C}$, with covering map $z \mapsto e^z$.

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I'm assuming you're talking about the Riemann surface $\{w^2=z, w\neq0\}$ and the like. Then it is not simply connected. To see this, first note that it is a two sheeted covering of $\mathbb C\backslash\{0\}$, because it is locally a homeomorphism and every point in $\mathbb C\backslash\{0\}$ has two preimages. Then its fundamental group is a subgroup of index 2 of $\pi_1(\mathbb C\backslash\{0\})=\mathbb Z$, which is not trivial. More directly, if you let $z$ wind 2 times around $0$, $w$ will make a full turn and thus you trace out a loop in $\{w^2=z, w\neq0\}$. But if you lift this loop to the universal covering of $\mathbb C\backslash\{0\}$ by the logarithm (see the next paragraph), you will see you have moved $4\pi i$, so it is not a loop in the universal covering. The same argument applies to $\{w^n=z, w\neq0\}$.

In contrast, the Riemann surface $\{e^w=z\}$ is simply connected because it is homeomorphic (in fact biholomorphic) to $\mathbb C$. It is a covering of $\mathbb C\backslash\{0\}$ via the exponential map, and so you can lift any loop in the latter via the logarithm to see if it is nullhomotopic.

Conclusion: the Riemann surface $\{e^w=z\}$, being simply connected, is not homeomorphic to $\{w^n=z, w\neq0\}$ for any $n>1$.

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  • $\begingroup$ This is very helpful! But I have one more issue: You stated: "The Riemann surface {ew=z} is simply connected because it is homeomorphic to the complex plane. 1) How can there be a map beween the infinite-sheeted Riemann surface and the plane, which is a well-defined bijective function and a homeomorphism? I mean, how can it be injective ("one-to-one") if a point from the plane is mapped on infinite many sheets on the Riemann surface? 2) The Riemann surface for the square root is not simply-connected. Is this Riemann surface homeomorphic to the (multiply connected) punctured complex plane? $\endgroup$ – eigenvalue Mar 21 '15 at 21:10
  • $\begingroup$ @eigenvalue That's why I used the set $\{e^w=z\}=\{(w,z)\in\mathbb C^2:e^w=z\}$ to identify the Riemann surface. It is homeomorphic to the complex plane via the map $w\to (w,e^w)$. The infinite-sheeted covering is $(w,z)\to z$, or $(w,e^w)\to e^w$, and it's a covering of the punctured complex plane, not the complex plane itself. In short they're two different maps, and what's on the other side of the map is not the same thing. $\endgroup$ – Fan Zheng Mar 22 '15 at 0:47

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