3

I have a question about an example in following script: https://math.berkeley.edu/~teleman/math/Riemann.pdf

Here the excerpt:

enter image description here

Why there don't exist a continuous choice of $w$ near $\pm1$ and $\pm k$? Why it does lead to the opposite choice upon return? What do they mean with return?

Above I have drawn a (in my opinion) possible choice for $w$. Why it is wrong?

user267839
  • 7,293

1 Answers1

3

The problem with your diagram is that it shows only real values of $z$ near $-1$. If you let $z$ move in a small circle in the complex plane, centered at $-1$, then the value of $\sqrt{(z^2-1)(z^2-k^2)}$ will continuously change from positive to negative on returning to $z=-1+\epsilon$ ($\epsilon>0$).

For the problem you're having, it's enough to look at the simpler surface $\sqrt{z^2-1}$. Here's a picture I created of part of it around $\pm1$: Riemann surface of $\sqrt{z^1-1} The x-axis runs from the bottom left to the top right. On the left, I've tried to show the "cross-cut" where the surface appears to intersect itself. (It doesn't really intersect itself, that's just an artifact of trying to draw the surface in regular Euclidean space. Like the circle of "self-intersection" in drawings of a Klein bottle.)

On the right hand side, I've sliced the picture along the y-axis, and display only the part with $y>0$. (When I say "x-axis" and "y-axis", I'm assuming $z=x+iy$.) In my drawing, I focussed on the situation near $1$, so let's work with that; the story near $-1$ is basically the same.

Now consider a small circle in the $z$-plane centered at $1$. Let $z$ go around that circle, starting $w$ on the upper surface. As the point passes passes through the cross-cut, it moves to the lower surface. When $z$ returns to its original position, $w$ is now minus its original value.

You can sort of see this algebraically. In $w=\sqrt{z^2-1}=\sqrt{(z-1)(z+1)}$, for $z$ close to 1, $z+1\approx 2$. So $\sqrt{(z-1)(z+1)}\approx\sqrt{2(z-1)}$. So near $1$, the surface should look a lot like the surface of the square root, except with the branch point at $1$ instead of $0$. (There are good pictures of the Riemann surface of $\sqrt{z}$ all over the web.)

There is a reason Telemann is looking at the more complicated surface $\sqrt{(z^2-1)(z^2-k^2)}$. (Spoiler warning!) I've shown a portion of the surface of $\sqrt{z^2-1}$ double-covering a part of the $z$ plane near the origin. The double covering can be extended to a double covering of the entire Riemann sphere. (The branch points need special attention, but I'll leave that to the textbooks.) The surface $\sqrt{z^2-1}$ is topologically less interesting than the surface of $\sqrt{(z^2-1)(z^2-k^2)}$.

Finally, a recommendation: the first chapter of George Springer's Introduction to Riemann Surfaces has lovely pictures and a fine discussion. It's an old book, but well worth digging up.

Michael Weiss
  • 8,813
  • You mean in the sense analogous to the behavior of the normal vector of Möbius strip when you walk around? – user267839 Apr 03 '18 at 22:05
  • Well, sort of. I added a lot to my answer; see if the new version makes things clearer. – Michael Weiss Apr 03 '18 at 22:50
  • Thank you for the detailed answer. One question to your remarks at the end: The extending of the covering to the entire Riemann sphere means that to do it you have compactify the $z$-plane by $\infty$ and consider the RS as branched covering? Or is there another reason to compactify? – user267839 Apr 03 '18 at 23:52
  • You're exactly right. Adding the point at infinity and the branch points gives a compact covering. Generally speaking, compact Riemann surfaces have rather different properties from noncompact ones. – Michael Weiss Apr 04 '18 at 00:13
  • But what kind of problem occure if one don't compactify? So if one just take /find branch points and say that over other points of $z$-plane the covering (=RS) is unramified... or in other words over them the number of preimages is constant – user267839 Apr 04 '18 at 00:25
  • No problems occur. But if a Riemann surface is compact, then you can prove a lot of nice theorems about it, using the compactness. One of big ones is the first theorem of Lecture 14 in Teleman's notes (p.47). Another is the Riemann-Hurwitz formula (pp.18-19). One of the first examples you'll encounter in the notes is Theorem 3.5 (p.11); also look at 4.13 and Theorem 5.1 (p.16). – Michael Weiss Apr 04 '18 at 12:59
  • Here's an analogy. A polynomial of degree $n$ usually has $n$ distinct roots in the complex plane. We could just say that some polynomials have fewer than $n$ roots, but it's more useful to introduce the concept of repeated roots, and count roots with multiplicity. (This is actually more than an analogy, but I don't want to go into detail in a comment.) – Michael Weiss Apr 04 '18 at 13:12