
Re: This Week's Finds in Mathematical Physics (Week 229)
Posted:
May 16, 2006 10:32 PM


In article <1145101189.273980.185650@i40g2000cwc.googlegroups.com>, Squark <top.squark@gmail.com> almost wrote:
>Hello John and everyone!
Hello! Long time no see! How are you doing? I showed up at the Perimeter Institute yesterday, and I should be getting a talk ready:
http://math.ucr.edu/home/baez/quantum_spacetime
but I'm goofing off.
>> There's a simple topological interpretation of the element of the >> rational projective line associated to a rational tangle. I don't know >> how to use this to prove the theorem, and I don't know a reference for >> it (maybe it is in one of the references you cited). Anyway, regard a >> rational tangle as a twocomponent curve C in the 3ball B^3 whose four >> boundary points are on the 2sphere S^2. Consider the double branched >> cover of B^3 along C.
>What is "_the_ double branched cover"? Is there a way to choose a >canonical one, or is there only one in this case, for some reason?
Good point. I hope there's a specially nice one.
To pick a branched cover of B^3 along C, it's necessary and sufficient to pick a homomorphism from the fundamental group of B^3  C to Z/2. This says whether or not the two sheets switch places as we walk around C following some loop in B^3  C.
>In the case of a sphere with 4 points removed it should be easy to >check.
Yes.
>The fundamental group has 4 generators  a, b, c, d (loops around each >of the points) and one relation abc = d (since we're on a sphere). Hence, >it is freely generated by a, b, c (say).
[I changed your presentation slightly here, for my own convenience.]
Right, the fundamental group of the fourpunctured sphere is the free group on 3 generators, F_3. I believe the "specially nice" homomorphism
f: F_3 > Z/2
is the one that sends each generator to 1, where I'm thinking of multiplicatively:
Z/2 = {1, 1}
One reason this homomorphism is especially nice is that it also sends d = abc to 1.
So, if you walk around ANY of the four punctures, the two sheets switch!
This is just what you want for the Riemann surface of an elliptic integral, as someone else pointed out in another post: there are four branch points each like the branch of point of sqrt(z). It's also the most symmetrical, beautiful thing one can image.
Now let's see if and how this branched cover extends to a branched cover of the ball B^3 with C (two arcs) removed. The fundamental group of B^3  C is the free group on two generators, say X and Y.
The inclusion of the 4punctured sphere in B^3  C gives a homomorphism
g: F_3 > F_2
as follows
a > X b > X^{1} c > Y d > Y^{1}
So, to extend our branched cover, we need to write our homomorphism
f: F_3 > Z/2
as
f = hg
for some homomorphism
h: F_2 > Z/2
The obvious nice thing to try for h is
X > 1 Y > 1
It works, and it's unique!

