In Brown and Churchill's book, the concept of multivalued functions is not discussed in a very rigorous way (if at all). But I can see that branch cuts have importance in complex analysis, so I want to clarify my understanding of multivalued functions.

Is there a rigorous development of the definition of a multivalued function somewhere, along with branch cuts? Or is the whole idea of a multivalued function just a way of saying, "Hey, there's no unique way of defining the logarithm function here, so we're going to use whatever is convenient at the time"? And if the latter, where does a rigorous understanding of branch cuts fit in? Or are they also more of an intuitive term rather than a real defined mathematical object?

If they are rigorous, would a multivalued function be something like $f: \mathbb{C} \to \mathbb{C}^\infty$? I've never seen an infinite dimensional space before so I don't really know how that is developed.

  • 7
    $\begingroup$ It's rigorous and deep to put it lightly. The way I'd say it is that the 'correct' domain for a complex function like, say, $\sqrt{z(z^2-1)}$ is not the complex plane but an appropriate (Riemann) surface: understood as a function on that surface, that square root is a single-valued function. (But that's a slogan, not an explanation!) $\endgroup$ Jul 29, 2014 at 21:55
  • 1
    $\begingroup$ Also, at the level of set theory, a multivalued function can be thought of as a regular function that takes values that are subsets of the domain. This formalism isn't exactly the same as the Riemann surface way to make functions single valued, but I know very little about that way. $\endgroup$
    – Jeff
    Jul 29, 2014 at 23:40
  • 1
    $\begingroup$ One possible definition is given here, where I discuss them in connection to branch points. $\endgroup$
    – Zhen Lin
    Jul 30, 2014 at 0:23
  • 1
    $\begingroup$ @semiclassical It is a solution to the biharmonic with a jump / discontinuity in the first derivative along the positive x-axis (representing the branch cut). It represents an idealized line defect in a solid. en.wikipedia.org/wiki/Dislocation $\endgroup$ Jul 30, 2014 at 2:23
  • 2
    $\begingroup$ @Jeff Yes, it's possible to talk about multivalued functions that way but there's a reason we usually don't: As soon as we try, we find it's impossible to calculate anything with that definition. The size of the set of the sum of a few such functions explodes quite fast, for example. The Riemann surface viewpoint is better; it shows that what we're really doing is trying to invert a finite map and should look to construct a domain manifold for it to study. $\endgroup$ May 21, 2015 at 13:07

1 Answer 1


Multivalued analytic functions can be made (and have been made) a rigorous notion. This notion is sometimes useful. But modern textbooks prefer not to use it, because it is hard to deal with rigorously. (What is a sum or product of multivalued functions?)

There are several substitutes:

  1. To use only single valued branches. For this you need to restrict the region (usually by making some "branch cuts"). This is the way most elementary textbook take.

  2. To use sheafs (which can be considered a rigorous framework into which multivalued functions fit, though there are alternative approaches). This is the approach used in the standard graduate textbook of Ahlfors.

  3. To translate everything to the language of functions on Riemann surfaces. This is perhaps the most useful approach, at least in one complex variable, which was proposed by Riemann.

  • $\begingroup$ Thank you. I will take a look at Ahlfors when I get a chance. $\endgroup$ Mar 29, 2015 at 18:19
  • $\begingroup$ @AmadeusDrZaius This Book, though old and not as comprehensive as Ahlfors, gives very nice and concise explanations. $\endgroup$ Oct 3, 2015 at 15:47
  • $\begingroup$ Yes, this is a very good book. $\endgroup$ Oct 3, 2015 at 19:29
  • $\begingroup$ I'm not sure what this book by Ahlfors is. Could someone link to it, or give a more complete citation? $\endgroup$
    – Galen
    Feb 10, 2021 at 19:55
  • $\begingroup$ The book of Ahlfors is called Complex Analysis, (McGraw Hill, 1979, 3-d edition) and it is the standard textbook in the US universities. $\endgroup$ Feb 10, 2021 at 21:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .