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I was trying to clarify some questions I had about elliptic integrals using

http://websites.math.leidenuniv.nl/algebra/ellcurves.pdf

There they define the map $$\phi\colon w\mapsto \int_0^w\frac{\mathrm{d}z}{\sqrt{1-z^2}}$$ on $\mathbb{C}\setminus[-1,1]$ to get $\phi$ well-defined up to periods of the integral. The choice of the interval $[-1,1]$ is made so that $\sqrt{1-z^2}$ admits a single-valued branch.

Now, I know that the principal branch of the square root $\sqrt{z}$ is discontinuous on the half-line $(-\infty,0)$, so to get a holomorphic map we restrict to $\mathbb{C}\setminus (-\infty,0]$. Substituting $1-z^2$ for $z$ we get that the appropriate branch cuts for the above mapping $\sqrt{1-z^2}$ would be $(-\infty,-1]$ and $[1,\infty)$, which is somewhat the opposite of the suggested interval $[-1,1]$.

From that I conclude that they didn't choose the principal branch, otherwise for e.g. $z=2$ the map would be discontinuous.

My question is: Are both choices possible? Then there must be some way to choose another branch of $\sqrt{1-z^2}$. Is there a good way to see how to choose "elegant" branch cuts and the corresponding holomorphic branches?

A thought of my own: It should be possible to instead integrate on the Riemann sphere, using $\infty$ and not $0$ as a starting point. Then the two intervals would "swap roles". But I don't see how to formalize this.

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    $\begingroup$ You won't need to integrate on the Riemann sphere if you take the branch cut for $\sqrt z$ along $[0,\infty)$ instead. Then $1-z^2\ge 0$ iff $1\ge z^2$ iff $z\in[-1,1].$ $\endgroup$
    – Andrew
    Aug 19, 2012 at 21:19
  • $\begingroup$ @Andrew, Thank you! I didn't think about starting from the other branch. The web of confusion is disentangling now. Please consider posting your comment as an answer, I would like to upvote it. $\endgroup$ Aug 19, 2012 at 23:12
  • $\begingroup$ You're welcome, I'm glad to help! I posted the comment as you suggested. $\endgroup$
    – Andrew
    Aug 19, 2012 at 23:27

2 Answers 2

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When taking a branch of $\sqrt z$, you can choose any ray emanating from the origin. In this case, for $\sqrt{1-z}$ and $\sqrt{1+z}$, we need to choose two rays emanating from $-1$ and $1$, and the author chooses them to be $[-1, \infty)$ and $[1,\infty)$.

This seems to rule out the entire interval $[-1, \infty)$ from being part of the domain. But it can be shown that the "jumps" in the branch cuts of the square root functions cancel on the interval $[1,\infty)$, so we get an analytic function on $\mathbb{C}-[-1,1]$.

It's not too hard, and I invite you to try it as an exercise, that you get an analytic continuation across $[1,\infty)$.

This is how I think about it. Andrew's answer in the comments is fantastic and probably better, though. For a complete theory of making such branch cuts, you need to learn about Riemann surfaces. If I recall correctly, Forster's book on Riemann surfaces has a treatment of such functions, but it requires some background to access.

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  • $\begingroup$ Thank you. Your way of thinking about it helps me understand the underlying mechanism better. In fact, I have read Forster's book some time ago but I didn't have the time to think it all through - it is not very explicit on those concrete issues, iirc. $\endgroup$ Aug 19, 2012 at 23:14
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    $\begingroup$ @GregorBruns You're right, he's not. I'm struggling to think of a better reference, though. There's an interpretation based on "destroying the monodromy" that extends to $\sqrt f(x)$ where $f$ is some polynomial. It's in the second volume of the text on complex variables by Markushevich, but there's probably another more accessible source somewhere. $\endgroup$
    – Potato
    Aug 19, 2012 at 23:51
  • $\begingroup$ @GregorBruns When I asked my professor about this (pretty much the exact question you had), he recommended the little-known Conformal Mapping on Riemann Surfaces by Harvey Cohn. At that point in my life I wasn't ready for that text, and I haven't looked at it since, but I recall there being some very interesting things in it (e.g. applications of Riemann surface theory to electrostatics and hydrodynamics). I also found it unreadable, so be warned. But it is cheap and may be worth looking in to. $\endgroup$
    – Potato
    Aug 19, 2012 at 23:59
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If we choose a branch of $\sqrt{z}$ with branch cut along the positive axis $[0,\infty)$, then we won't need to integrate on the Riemann sphere, since $1-z^2\ge 0$ if and only if $1\ge z^2$, which happens if and only if $z\in [-1,1]$.

This is equivalent to Potato's answer, which is a nice geometrical interpretation of the branch cut.

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    $\begingroup$ But $1-z^2 \ge 0$ also for $z=iy$, y real. So why the line iy, y real, is not part of the cut? $\endgroup$
    – TCL
    Jan 7, 2014 at 12:11
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    $\begingroup$ clearly the statement: $1-z^2\geq 0$ iff $z\in[-1,1]$ is false, so it is not a valid argument. $\endgroup$
    – Chilote
    Mar 21, 2014 at 19:07
  • $\begingroup$ Dear @Chilote, you are correct. The "simple fix" didn't really simplify things after all, since we must now show the branch cut along the $i$-axis can be removed. $\endgroup$
    – Andrew
    Mar 21, 2014 at 21:48

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