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Anyone a Riemann Surface guru? 
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#1
Oct2804, 03:16 PM

P: 55

So I'm supposed to describe the riemann surface of the following map:
[tex]w=z\sqrt{z^21}[/tex] 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 singlevalued? 


#2
Oct2505, 08:34 PM

P: 1

so i guess no one knows about this....?



#3
Oct2505, 09:01 PM

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MAT354? e^{w} 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. w^{n} 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 z^{1/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^21}[/itex] maps the cut onto the boundary separating those 11 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 11 regions (fundamental regions) so you have infinitely many planes that you cut and glue together. In the z^{1/n} example, there are n fundamental regions so you have n planes. You want to make z^{1/n} singlevalued, 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 11 way. But 0 is a point that is not manyto1, so when you have your nseparate 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 twovalued, so you need two sheets which each have a point corresponding to that one point. 


#4
Oct2505, 10:01 PM

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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. 


#5
Oct2505, 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 > (z1)/(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^21) = (t^2+1)/2t  (t^21)/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. 


#6
Oct2805, 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 


#7
Oct2905, 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 ((z1)(z+1)) To get a torus you'd need another root (I think) 


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