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Anyone a Riemann Surface guru?

by T-O7
Tags: guru, riemann, surface
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Oct28-04, 03:16 PM
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So I'm supposed to describe the riemann surface of the following map:
I can sort of understand the basic idea and derivation behind the riemann surfaces of [tex]w=e^z[/tex] and [tex]w=\sqrt{z}[/tex],
but ask me a question about another mapping, and I really don't know where to begin. How does one attack a problem like this? i.e. how do you know where exactly the map is single-valued?
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Oct25-05, 08:34 PM
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so i guess no one knows about this....?
Oct25-05, 09:01 PM
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MAT354? ew is single valued on those strips, and it maps those strips onto the plane - {0}, so you want to take the plane - {0} and make a cut such that log(z) maps the cut onto the lines separating those strips. wn is single valued on the sectors, and it maps each sector onto a plane, so you want to take a plane and make a cut such that z1/n maps the cut onto the lines separating those sectors. 1/2(w + 1/w) is single valued outside the unit circle, and inside the unit circle, and it maps each of those regions onto the plane, so you want to take the plane and make a cut such that [itex]z - \sqrt{z^2-1}[/itex] maps the cut onto the boundary separating those 1-1 regions, namely the unit circle. This function maps the interval [-1,1] to that circle, so that's where you make the cut. In the log example, there are infinitely many 1-1 regions (fundamental regions) so you have infinitely many planes that you cut and glue together. In the z1/n example, there are n fundamental regions so you have n planes. You want to make z1/n single-valued, so for points where it's many valued, you want to make a Riemann surface that makes this one piont into many points, so that instead of one point corresponding to many points, you have many points corresponding to many, in a 1-1 way. But 0 is a point that is not many-to-1, so when you have your n-separate planes, you still only want 1 point on the entire Riemann surface corresponding to 0.

In this problem you're doing, -1 and 1 are the only single valued points of the function. So although you have two planes planes, you want their -1 points to be attached, and their 1 points to be attached, because you don't need to make two different copies of 1. The rest of the points are two-valued, so you need two sheets which each have a point corresponding to that one point.

Oct25-05, 10:01 PM
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mathwonk's Avatar
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Anyone a Riemann Surface guru?

this is a multipvalued function, and the riemann surface should have one poinr for each value. but look at your function, it is at most 2 valued sionce it is a square root.

so look where it ahs actually only one value, that would be at 1 and -1. so your riemann surface is a 2 to 1 covewring of the z plane, with only one point oevr 1 and one point over -1. so if you include the point at infinity, it is some compact surface mapping 2 to 1 onto the sphere, and branched at 1 and -1 and possibly at infinmiyt.

now can you figure out:

1) is infinitya ctually a branch point?

2) what is the topology of the covering surface?

answers upside down at bottom of page.
Oct25-05, 10:24 PM
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bottom of page: (encrypted): the function u -->u^2 is 2 tp 1 and branched over 0 and infin, and the function z --> (z-1)/(z+1) takes 1 to 0 and -1 to infin, so the composition of these two and the inverse of the latter takes 1 to 1 and -1 to -1 and is branched at both. thus the map taking t to
(t^2 + 1)/2t = z, makes your function single valued. since
then z - sqrt(z^2-1) =
(t^2+1)/2t - (t^2-1)/2t = 1/t.

and since (t^2 + 1)/2t = z, note then t^2 -2zt + 1 = 0, so

t = +2z + sqrt(4z^2 - 4)/2 = z + sqrt(z^2 -1), as desired.

or maybe i have it upside down.
Oct28-05, 12:26 PM
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Incidentally, can you see that the covering space is a torus?

Well at least I think its a torus at least at first glance. You'd want to glue the two sheets together around the branch points -1 and 1. You have to check infinity as that can be a problem as well.

As an aside, if it *is* a torus (and im pretty sure it is) the problem would have been solvable via other means. Jacobi theta functions and Weierstrass Z functs (as well as PDEs) to name a few
Oct29-05, 01:43 AM
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Actually I've changed my mind, (i'd need to sit down and do this with a pen and paper and glue things together properly) but I now think the topology is simply S^2.

The Riemann surface then would be equivalent to something like W = sqrt ((z-1)(z+1))

To get a torus you'd need another root (I think)

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