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Could someone please explain the concept of branch points to me? I have tried searching online and had a read of the textbook Visual Complex Analysis by T. Needham, but I am still not very clear how they work.

An excerpt I found online from Introduction to Complex Analysis by H. Priestly says that

$a$ is a branch point for [$w(z)$] if, for all sufficiently small $r>0$, it is not possible to choose $f(z)\in[w(z)]$ so that $f$ is a continuous function on $\gamma(a;r)^*$.

Firstly, I couldn't find what $\gamma(a;r)^*$ is ... I presume it is an open ball around the point $a$ with radius $r$?

Secondly, I just don't understand what it is saying. Why is there no continuous function? How when asked to find branch points would I know which points in $\mathbb C$ have this property?

Needham's book basically says a branch point is one which if we circle it once we don't get back to the same point... but I still don't get it!

Then I read something about branch cuts and Riemann spheres which really don't help to clarify anything at all!

Thank you for your time.

[Added] For example if I have a map of the form $f(z) =[(z-a)(z-b)...(z-n)]^{1\over m}$ how may branch points are there?

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The square root would be a rather simple example. Seeing the square root's branch point at the origin can be facilitated by looking at its Riemann surface. Note that line where the surface intersects itself; that comprises the branch cut, and its endpoint is the branch point. –  J. M. Feb 2 '12 at 10:56
    
@J.M.: Thanks, would you mind explaining what Priestly's definition is saying? –  Josef Feb 2 '12 at 11:25
    
Take the square root again: you can consider it to have two branches, $\sqrt{z}$ (a.k.a. the "principal branch"), and $-\sqrt{z}$. Consider only the principal branch for the time being. You'll notice that there is no way to draw a circle centered at the origin that isn't cut through by the branch cut of the square root. That's my interpretation of Priestly's words. –  J. M. Feb 2 '12 at 11:52
    
@J.M.: Thanks again! So basically a branch point is where the complex function either equals to $0$ or $\infty$? –  Josef Feb 2 '12 at 12:02
    
No; a branch point is the endpoint of a branch cut. A branch cut is a necessary discontinuity of a complex function; a number of functions are "multivalued", like the square root (two) or the logarithm (infinitely many). So that your function is single-valued, it is necessary to select only one branch, and with the selection of the branch, there is a necessary "cutting" of the corresponding Riemann surface (did you notice the gaping gap in the surface plots of the square root that I showed you?) –  J. M. Feb 2 '12 at 12:08

2 Answers 2

up vote 7 down vote accepted

Given a non-constant morphism of Riemann surfaces $f:X\to Y$, a critical point of $f$ is a point $x\in X$ such that equivalently :

$\bullet$ For every neighbourhood $x\in U\subset X$ the restriction $f|U:U\to Y$ is not injective.
$\bullet$ The differential $df(x)=0$
$\bullet$ In local cooordinates at $x$ and $f(x)$, $f$ can be written as $z\mapsto z^n $ with $n\geq 2$

The critical values of $f$ are the $y\in Y$ which can be written $y=f(x)$ for some critical $x\in X$.
The terminology "branchpoint" is unfortunately ambiguous: some authors (e.g. Griffiths and Forster) use it for critical point and others (e.g. Miranda) for critical values.

For example, given distinct $a_1, a_2,...,a_n\in \mathbb C$, if you consider the Riemann surface $X$ of the "function"$\sqrt {(z-a_1)(z-a_2)...(z-a_n)}$, you will obtain a morphism $f:X\to \mathbb P^1(\mathbb C)$ whose critical values in $\mathbb C$ are $a_1,a_2,..., a_n$ .
Moreover $\infty\in \mathbb P^1(\mathbb C) $ will also be a critical value precisely when $n$ is odd.
In this simple but basic example each critical value has exactly one critical point mapping to it.

Although the above might look a bit abstract, be very wary of the "concrete" approach of this kind of problems by cut and paste techniques with pictures of sheets crossing themselves.
It has elicited some very harsh words from Serge Lang (for example) in his book here , Chapter XI, §1: "It should be emphasised that the picture is totally and irretrievably misleading".
I recommend Forster's Riemann Surfaces for a completely rigorous and definitive treatment.

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Thanks, Georges, this is vey helpful! :) may I ask why $\infty$ is a critical point precisely when $n$ is odd? Also, without knowing how the Riemann surface looks like, is it possible to tell which points are critical? It seems to me that they tend to be the roots of the function concerned, but I don't think that is generally true? Thanks. –  Josef Feb 2 '12 at 19:12
    
Perhaps you wouldn't mind taking a look at this question? –  Josef Feb 2 '12 at 19:21

Answer to added question: if $a,b,\dots,n$ are all distinct, then each one of them is a branch point.

EDIT: As indicated in the comments, I have been careless about the possible branch point at infinity. I think the discussion at this link is pretty good [EDITEDIT although it sticks to the square root case]. In particular, it explains why the even and odd cases differ.

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Thanks, Gerry, jUst one little thing, sometimes, $\infty$ is also said to be a branch point for cases like these, when is that true? –  Josef Feb 2 '12 at 11:57
    
@Josef: Yes, provided you have either one point or more than two points (all distinct). There is only one "square root of $\infty$", so to speak. If there are exactly two points (both distinct), then $\infty$ is not a branch point. –  Zhen Lin Feb 2 '12 at 17:32
    
@ZhenLin: I'm afraid I don't quite follow, would you mind elaborating a little bit? Thanks. –  Josef Feb 2 '12 at 19:14
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To deal with $\infty$, it's often best to make a change of variables taking $\infty$ to a finite value. In this case if $f(z) = ((z - a_1) \ldots (z - a_n))^{1/m}$, let $w = 1/z$ and $g(w) = 1/f(z) = \frac{w^{n/m}}{(1-a_1 w)\dots(1-a_n w)}^{1/m}$, so if $n/m$ is not an integer there is a branch point at $w=0$, corresponding to $z=\infty$. –  Robert Israel Feb 3 '12 at 3:03
    
@Robert, yes, that's what's done (for $m=2$) at the link I gave. –  Gerry Myerson Feb 3 '12 at 4:33

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