It seems that there exists ring structure on all multi-valued holomorphic functions on a punctured disc. Can someone explain the formal definition of multi-valued holomorphic function?

I only know some examples which are multi-valued, for example $x^\lambda$, $log x$.

Edit: My motivation to ask this question, is to understand the mutilvalued holomorphic solution of system of analytic differential equations. For example the solution of $x\partial_x f=\lambda f$ is $x^\lambda$


3 Answers 3


If $U$ is an open (connected...) subset in $\mathbb C$, a multivalued holomorphic function on $U$ is an holomorphic function on $\tilde U$, the universal cover of $U$, endowed with its natural structure of complex curve; if $\pi:\tilde U\to U$ is the covering map, then univalued holomorphic functions are, in this picture, those of the form $f\circ\pi$ with $f:U\to\mathbb C$ an holomorphic function.

In particular, such multivalued holomorphic functions more or less obviously form a ring.

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    $\begingroup$ Are two multivalued holomorphic functions which differ by a deck transformation different? $\endgroup$ Commented Apr 4, 2010 at 23:27
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    $\begingroup$ Reid makes an important point. Namely, given an open subset $U\subseteq\mathbb C$, I only know what its universal cover is up to isomorphism. And not even up to canonical isomorphism: its automorphism group as a universal cover is $\pi_1U$. So I don't see how, without choosing once and for all a universal cover, to define the ring structure. $\endgroup$ Commented Apr 4, 2010 at 23:36
  • $\begingroup$ In this picture, yes. I want $\sqrt z$ and $-\sqrt z$ to be two different functions, for example. $\endgroup$ Commented Apr 4, 2010 at 23:43
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    $\begingroup$ Precisely. The problem is that people often say things like "the multivalued holomorphic functions $z^\lambda$, $\log z$" without explicitly describing the relevant choices. It is not meaningful to ask, for example, whether $z^\lambda + \log z$ on some given domain $U$ has a multiplicative inverse, without specifying a particular choice of $\tilde U$ and representatives of the deck transformation equivalence classes $z^\lambda$, $\log z$ on $\tilde U$. $\endgroup$ Commented Apr 4, 2010 at 23:56

If you choose a basepoint of the disc and a local section there, you can employ analytic continuation to interpret a multivalued holomorphic function as a holomorphic function on the universal cover of the disc, i.e., (some open subset of) the complex upper half plane. If you have a formula in terms of the coordinate $x$, you can replace all appearances of $x$ with $e^{2 \pi i z}$. For example, $x^\lambda$ on the disc becomes $e^{2 \pi i \lambda z}$ on the upper half plane. You will end up with an ambiguity of lift given by maps $z \mapsto z+n$ for integers $n$.

Edit: I just realized that I hadn't answered the title question. We define a multivalued holomorphic function on a pointed connected complex manifold $(X,x)$ (such as an open punctured disc with a distinguished point) to be a regular function on the pointed universal covering space $(\widetilde{X},\widetilde{x})$. The set of multivalued functions then forms an augmented commutative ring under pointwise operations. This definition generalizes to pointed connected complex analytic spaces.

You might ask: What if I don't want a basepoint? In that case, "universal covers" are not canonical (as Theo mentioned in the comments to Mariano's answer), and even if you fix a cover $\widetilde{X} \to X$, there is no canonical choice of lift of a locally defined function to $\widetilde{X}$. Often in practice, you will not see a basepoint specified, but you might find people using the equivalent technique of choosing a preferred simply connected neighborhood (e.g., the complement of negative reals in $\mathbb{C} \setminus \{0\}$ when working with logarithms). In this case, it is essential that you specify a distinguished preimage of that neighborhood in $\widetilde{X}$ in order to get a well-defined ring structure on locally defined functions.

  • $\begingroup$ So we can only think of addition and multiplication of multivalued function when they are lifted as functions on the covering space. A little confused by your definition that the analytic subset to $D^\circ$ is a smooth covering, when we take a local section, do you require it is analytic? $\endgroup$
    – JJH
    Commented Apr 4, 2010 at 17:39
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    $\begingroup$ Sorry, that was a poor choice of words. Yes, we require local sections to be analytic, and we need to choose preferred branches before we can describe a ring structure. $\endgroup$
    – S. Carnahan
    Commented Apr 4, 2010 at 17:47

It seems this notion causes some confusion not only on me. Let me give my own definition, by taking into account with yours.

Let $X$ be an analytic manifold, and let $\pi:\tilde{X}\mapsto X$ be some universal covering of $X$. A multivalued holomorphic function on $X$ is a holomorphic function on $U$, where $U$ is some contractible open subset of $X$, if the function $\pi\circ f$ on some component $\mathcal{C}$ of $\pi^{-1}(U)$ can be analytically extended to $\tilde{X}$. Denote this function by $\tilde{f} _\mathcal{C}$

Two multivalued holomorphic functions $(f,U)$ and $(g,V)$ are equivalent, if there exists some component $\mathcal{C}$ of $\pi^{-1}(U)$ and $\mathcal{D}$ of $\pi^{-1}(V)$, such that $\tilde{f}_\mathcal{C} = \tilde{g}_\mathcal{D}$.

This definition doesn't depend on the choice of universal covering.


Here is a more rigorous definition:

Definition. A multivalued holomorphic function on $X$ based at $x$ is a function $f$ in $\mathcal{O}_x$, such that one component of $\pi^* f$ can be analytically extended to $\tilde{X}$, where $\pi: \tilde{X}\mapsto X$ is some universal covering.

Denote the ring of multivalued holomorphic functions on $X$ based at $x$ as $\tilde{\mathcal{O}} _x$, which is a subring of $\mathcal{O}_x$. Any loop based at $x$ gives an action on $\tilde{\mathcal{O}} _x$, which is essentially the monodromy.

Claim. $\tilde{\mathcal{O}} _x$ is locally free $\mathcal{O}(X)$-module with rank the order of fundemental group.

It is essentially equivalent to the defintion that a mutilvalued fundtion is defined as function on covering space. But i feel it is more intuitive and doesen't depend on the choice of covering space.

Under this definition, I can make sense of the solution of mulivalued function of some analytic differential equation, which is actually my original motivation to ask this question.

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    $\begingroup$ Equivalence classes don't form a ring. $\endgroup$
    – S. Carnahan
    Commented Apr 5, 2010 at 17:42
  • $\begingroup$ ...and I am pretty sure this makes $\sqrt z$ and $-\sqrt z$ the same 'function', so that $2\sqrt z=0$. $\endgroup$ Commented Apr 5, 2010 at 17:53
  • $\begingroup$ I didn't claim by this definition, they will form a ring. But it's really a problem to give a ring structure. If we fix a point $x$ in $X$ , and forget equivalent class, we do get ring structure. Maybe rigorously, Let $\mathcal{O}$ be the sheaf of holomorphic function on $X$ , let $\mathcal{O}_x$ be the stalk at $x$. We call the function in $x$ a mutivalued function, if my above definition holds. Any comment on this? $\endgroup$
    – JJH
    Commented Apr 5, 2010 at 18:13
  • $\begingroup$ (Assuming you meant "we call the functions in $\mathcal O_x$ multivalued functions) That definition introduces a synonym for 'germ', so it is not very useful. $\endgroup$ Commented Apr 5, 2010 at 18:29
  • $\begingroup$ A analytic manifold is by definition is a pair $(X, \mathcal{O})$, the notion "germ" isn't natural and immediate? By this definition, mutivalued function gives a subring of $\mathcal{O}_x$. $\endgroup$
    – JJH
    Commented Apr 5, 2010 at 18:35

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