# How to understand multivalued functions in terms of Riemann surfaces

This note says in the last sentence, "The discontinuities of multivalued functions in the complex plane are commonly handled through the adoption of branch cuts, but use of Riemann surfaces is another possibility."

I have never heard of this before. Can someone explain what Riemann surfaces have to do with multivalued functions? For example, can one explain the meaning and holomorphicity of a function like $f(z) = \sqrt[m]{z}$ in terms of Riemann surfaces instead of branch cuts?

• For every point, consider the germ of a branch of $f$ at that point. That set of germs form a Riemann surface $S$. Then you can define an analytic function $F:S\to\mathbb{C}$ that sends a point in $S$ (a germ of $f$) to the value at the point. There is a projection from $\pi: S\to\mathbb{C}$ sending a germ to its base point. Then $f\circ\pi=F$. – blueInk Mar 18 '18 at 19:01
• For the case of the the $f(\cdot)=\sqrt[m]{\cdot}$, the Riemann surface is going to be the quotient of an helicoid identifying points that project vertically to the same point of the plane, but only after every $m$-th turn. The value of $F$ at a point on the $k$-th sheet of the helicoid, counting the number of sheets by the number of times you go over what was before the branch cut, would be the value of the corresponding branch of $f$ at the projection of the point on the plane. – blueInk Mar 18 '18 at 19:28
• The whole idea is just to drop the multivalued $f$ and just talk about the (single valued) holomorphic function $F$ between Riemann surface ($S$ and $\mathbb{C}$). – blueInk Mar 18 '18 at 19:31

Riemann surfaces become handy for multivalued functions because they help you to "transform" a multivalued function in a univalued function. Usually the most common example is the m-th root of z. The idea is that for the m-th root of z is that there are m points which m power is z. If you take this m points, they form an m-gon (if this is not so clear, do an example with the unity), that satisfies that each vertex has as it's m power z. This implies that $z^m$ is not inyective and this is the reason why $z^{1/m}$ is multivalued.