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!
Best Answer
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 $n>1$.