Also available at http://math.ucr.edu/home/baez/week233.html
May 20, 2006
This Week's Finds in Mathematical Physics (Week 233)
John Baez
On Tuesday I'm supposed to talk with Lee Smolin about an idea he's
been working on with Fotini Markopoulou and Sundance BilsonThompson.
This idea relates the elementary particles in one generation of the
Standard Model to certain 3strand framed braids:
1) Sundance O. BilsonThompson, A topological model of composite preons,
available as hepph/0503213.
2) Sundance O. BilsonThompson, Fotini Markopoulou, and Lee Smolin,
Quantum gravity and the Standard Model, hepth/0603022.
It's a very speculative idea: they've found some interesting relations,
but nobody knows if these are coincidental or not.
Luckily, one of my hobbies is collecting mysterious relationships between
basic mathematical objects and trying to figure out what's going on.
So, I already happen to know a bunch of weird facts about 3strand braids.
I figure I'll tell Smolin about this stuff. But if you don't mind, I'll
practice on you!
So, today I'll try to tell a story connecting the 3strand braid group,
the trefoil knot, rational tangles, the groups SL(2,Z) and PSL(2,Z),
conformal field theory, and Monstrous Moonshine.
I've talked about some of these things before, but now I'll introduce
some new puzzle pieces, which come from two places:
3) Imre Tuba and Hans Wenzl, Representations of the braid
group B_3 and of SL(2,Z), available as math.RT/9912013.
4) Terry Gannon, The algebraic meaning of genuszero, available as
math.NT/0512248.
You could call it "a tale of two groups".
On the one hand, the 3strand braid group has generators
  
\ / 
A = / 
/ \ 
  
and
  
 \ /
B =  /
 / \
  
and the only relation is
ABA = BAB
otherwise known as the "third Reidemeister move":
     
\ /   \ /
/   /
/ \   / \
 \ / \ / 
 / = / 
 / \ / \ 
\ /   \ /
/   /
/ \   / \
On other hand, the group SL(2,Z) consists of 2x2 integer matrices
with determinant 1. It's important in number theory, complex analysis,
string theory and other branches of pure mathematics. I've described
some of its charms in "week125", "week229" and elsewhere.
These groups look pretty different at first. But, there's a
homomorphism from B_3 onto SL(2,Z)! It goes like this:
1 1
A >
0 1
1 0
B >
1 1
Both these matrices describe "shears" in the plane. You may enjoy
drawing these shears and visualizing the equation ABA = BAB in these
terms. I did.
I would like to understand this better... and here are some clues.
The center of B_3 is generated by the element (AB)^3. This
element corresponds to a "full twist". In other words, it's
the braid you get by hanging 3 strings from the ceiling, grabbing
them all with one hand at the bottom, and giving them a full 360degree
twist:
  
\ / 
/  A
/ \ 
 \ /
 / B
 / \
\ / 
/  A
/ \ 
 \ /
 / B
 / \
\ / 
/  A
/ \ 
 \ /
 / B
 / \
  
This full twist gets sent to 1 in SL(2,Z):
1 0
(AB)^3 >
0 1
So, the double twist gets sent to the identity:
1 0
(AB)^6 >
0 1
In fact, Tuba and Wenzl say the double twist *generates* the kernel
of our homomorphism from B_3 to SL(2,Z). So, SL(2,Z) is isomorphic
to the group of 3strand braids modulo double twists!
This reminds me of spinors... since you have to twist an electron
around *twice* to get its wavefunction back to where it started.
And indeed, SL(2,Z) is a subgroup of SL(2,C), which is the double
cover of the Lorentz group. So, 3strand braids indeed act on the
state space of a spin1/2 particle, with double twists acting
trivially!
(For more on this, check out Trautman's work on "Pythagorean spinors"
in "week196". There's also a version where we use integers mod 7,
described in "week219".)
If instead we take 3strand braids modulo full twists, we get the
socalled "modular group":
PSL(2,Z) = SL(2,Z)/{+1}
Now, SL(2,Z) is famous for being the "mapping class group" of the torus 
that is, the group of orientationpreserving diffeomorphisms, modulo
diffeomorphisms connected to the identity. Similary, PSL(2,Z) is famous
for acting on the rational numbers together with a point at infinity
by means of fractional linear transformations:
az + b
z > 
cz + d
where a,b,c,d are integers and adbc = 1. The group PSL(2,Z) also
acts on certain 2stand tangles called "rational tangles". In
"week229", I told a nice story I heard from Michael Hutchings,
explaining how these three facts fit together in a neat package.
But now let's combine those facts with the stuff I just said!
Since PSL(2,Z) acts on rational tangles, and there's a homomorphism
from B_3 to PSL(2,Z), 3strand braids must act on rational tangles.
How does that go?
There's an obvious guess, or two, or three, or four, but let's just
work it out.
I just said that the 3strand braid A gets mapped to this shear:
1 1
A >
0 1
In "week229" I said what this shear does to a rational tangle.
It gives it a 180 degree twist at the bottom, like this:
   
   
   
 
 T  >  T 
 
  \ /
  /
  / \
Next, Tuba and Wenzl point out that
0 1
ABA = BAB >
1 0
which is a quarter turn. From "week229" you can see how this quarter
turn acts on a rational tangle:
   
  ____  
  / \  
   
 T  >   T  
   
    \____/
   
   
It gives it a quarter turn!
From these facts, we can figure out what the braid B does to a
rational tangle. So, let me do the calculation.
Scribble, scribble, curse and scribble.... Eureka!
Since we know what A does, and what ABA does, we can figure out
what B must do. But, to make the answer look cute, I needed a
sneaky fact about rational tangles, which is that A *also* acts
like this:
  \ /
  /
  / \
 
 T  >  T 
 
   
   
   
This is proved in Goldman and Kauffman's paper cited in "week228".
With the help of this, I can show B acts like this:
   
   ___ 
   / \ 
  / 
 T  > \  T 
 / \ 
   \___/ 
   
   
And this is *great*! It means our action of 3strand braids on
rational tangles is really easy to describe. Just take your tangle
and let the upper left strand dangle down:

____ 
/ \ 
 
  T 
 
  
  
  
To let a 3strand braid act on this, just attach it to the bottom of
the picture!
(That's why there were *four* obvious guesses about this would work:
one can easily imagine four variations on this trick, depending on
which strand is the "odd man out"  here it's the upper right. It's
just a matter of convention which we use, but my conventions give this.)
In fact, even the group of 4strand braids acts on rational tangles in
an obvious way, but the 3strand braid group is enough for now.
Let me summarize. The 3strand braid group B_3 acts on rational tangles
in an obvious way. The subgroup that acts trivially is precisely the
center of B_3, generated by the full twist. Using stuff from "week229",
it follows that the quotient of B_3 by its center acts on the projectivized
rational homology of the torus. We thus get a topological explanation
of why B_3 mod its center is PSL(2,Z).
But there's more.
For starters, the 3strand braid group is also the fundamental group of
S^3 minus the trefoil knot!
And, S^3 minus the trefoil knot is secretly the same as SL(2,R)/SL(2,Z)!
In fact, Terry Gannon writes that the 3strand braid group can be regarded
as "the universal central extension of the modular group, and the universal
symmetry of its modular forms". I'm not completely sure what that means,
but here's *part* of what it means.
Just as PSL(2,C) is the Lorentz group in 4d spacetime, PSL(2,R) is the
Lorentz group in 3d spacetime. This group has a double cover SL(2,R),
which shows up when you study spinors. But, it also has a universal
cover, which shows up when you study anyons. The universal cover has
infinitely many sheets. And up in this universal cover, sitting over
the subgroup SL(2,Z), we get... the 3strand braid group!
In math jargon, we have this commutative diagram where the rows are
short exact sequences:
1 > Z > B_3 > SL(2,Z) > 1
  
  
v v v
1 > Z > SL(2,R)~ > SL(2,R) > 1
Here SL(2,R)~ is the universal cover of SL(2,R). Since pi_1(SL(2,R)) = Z,
this is a Zfold cover. You can describe this cover explicitly using the
Maslov index, which is a formula that actually computes an integer for any
loop in SL(2,R), or indeed in any symplectic group.
But anyway, fiddling around with this diagram and the long exact sequence
of homotopy groups for a fibration, you can show that indeed:
pi_1(SL(2,R)/SL(2,Z)) = B_3
This also follows from the fact that SL(2,R)/SL(2,Z) looks like S^3 minus
a trefoil.
Gannon believes that number theorists should think about all this stuff,
since he thinks it's lurking behind that weird network of ideas called
Monstrous Moonshine (see "week66").
And here's the basic reason why. I'll try to get this right....
Any rational conformal field theory has a "chiral algebra" A which acts
on the leftmoving states. Mathematicians call this sort of thing a
"vertex operator algebra". A representation of this on some vector
space V is a space of states for the circle in some "sector" of our
theory. Let's pick some state v in V. Then we can define a
"onepoint function" where we take a Riemann surface with little
disk cut out and insert this state on the boundary. This is a number,
essentially the amplitude for a string in the give state to evolve like
this Riemann surface says.
In fact, instead of chopping out a little disk it's nice to just
remove a point  a "puncture", they call it. But, we get an ambiguous
answer unless we pick coordinates at this point, or in the lingo of
complex analysis, a choice of "uniforming parameter". Then our
onepoint function becomes a function on the moduli space of Riemann
surfaces equipped with a puncture and a choice of uniformizing parameter.
If we didn't have this uniformizing parameter to worry about, we'd
just have the moduli space of tori equipped with a marked point,
which is nothing but the usual moduli space of elliptic curves,
H/PSL(2,Z)
where H is the complex upper halfplane. Then our onepoint function
would have nice transformation properties under PSL(2,Z).
But, with this uniformizing parameter to worry about, our onepoint
function only has nice transformation properties under B_3. This
is somehow supposed to be related to how B_3 is the "universal
central extension" of PSL(2,Z): in conformal field theory, all sorts
of naive symmetries hold only up to a phase, so you have to replace
various groups by central extensions thereof... and here that's what's
happening to PSL(2,Z)!
That last paragraph was pretty vague. If I'm going to understand this
better, either someone has to help me or I've got to read something
like this:
5) Yongchang Zhu, Modular invariance of characters of vertex operator
algebras, J. Amer. Math. Soc 9 (1996), 237302. Also available at
http://www.ams.org/jams/1996901/S0894034796001828/home.html
But I shouldn't need any conformal field theory to see how the moduli
space of punctured tori with uniformizing parameter is related to the
3strand braid group! I bet this moduli space is X/B_3 for some space
X, or something like that. There's something simple at the bottom of
all this, I'm sure.

Addenda: Another relation between the trefoil and the punctured torus:
the trefoil has genus 1, meaning that it bounds a torus minus a disc
embedded in R^3. You can see this in the lecture "Genus and knot sum"
in this course on knot theory:
6) Brian Sanderson, The knot theory MA3F2 page,
http://www.maths.warwick.ac.uk/~bjs/MA3F2page.html
This course also has material on rational tangles.
The fact that B_3 is a central extension of PSL(2,Z) by Z, and the
quantummechanical interpretation of a central extension in terms
of phases, plays an important role here:
7) R. Voituriez, Random walks on the braid group B^3 and magnetic
translations in hyperbolic geometry, Nucl. Phys. B621 (2002), 675688.
Also available as http://arxiv.org/abs/mathph/0103008.
Among other things, he points out that the homomorphism B_3 > SL(2,Z)
described above is the "Burau representation" of B_3 evaluated at t = 1.
In general, the Burau representation of B3 is given by:
t 1
A >
0 1
1 0
B >
t t
(Conventions differ, and this may not be the best, but it's the one he
uses.) The Burau representation can also be used to define a knot
invariant called the Alexander polynomial. I believe that with some
work, one can use this to explain why Conway could calculate the
rational number associated to a rational tangle by taking the ratio
of Alexander polynomials of two links associated to it, called its
"numerator" and "denominator". In fact he computed this ratio of
polynomials and then evaluate it at a special value of t  presumably
the same special value we're seeing here (modulo differences in convention).
Another issue: I wrote
> For starters, the 3strand braid group is also the fundamental group of
> S^3 minus the trefoil knot!
>
> And, S^3 minus the trefoil knot is secretly the same as SL(2,R)/SL(2,Z)!
The first one is pretty easy to see; you start with the "Wirtinger
presentation" of the fundamental group of S^3 minus a trefoil, and
show by a fun little calculation that this isomorphic to the braid
group on 3 strands. A more conceptual proof would be very nice, though.
(See "week261" for such a proof  and much more on all this stuff.)
What about the second one? Why is S^3 minus the trefoil knot diffeomorphic
to SL(2,R)/SL(2,Z)? Terry Gannon says so in his paper above, but doesn't
say why. Some people asked about this, and eventually some people found
some explanations. First of all, there's a proof on page 84 of this book:
8) John Milnor, Introduction to Algebraic Ktheory, Annals of Math.
Studies 72, Princeton U. Press, Princeton, New Jersey, 1971.
Milnor credits it to Quillen. Joe Christy summarizes it below.
I can't tell if this proof is essentially the same as another sketched
below by Swiatoslaw Gal, which exhibits a diffeomorphism using functions
called the Eisenstein series g_2 and g_3. They are probably quite
similar arguments.
Joe Christy writes:
I wouldn't be surprised if this was known to Seifert in the 30's,
though I can't lay my hands on Seifert & Threfall at the moment to
check. Likewise for Hirzebruch, Brieskorn, Pham & Milnor in the 60's in
relation to singularities of complex hypersurfaces and exotic spheres.
When I was learning topology in the 80's it was considered a warm up
case of Thurston's Geometrization Program  the trefoil knot complement
has PSL_2(R) geometric structure.
In any case, peruse Milnor's Annals of Math Studies for concrete
references. There is a (typically) elegant proof on p.84 of
"Introduction to Algebraic Ktheory" [study 72], which Milnor credits
to Quillen. It contains the missing piece of John's argument:
introducing the Weierstrass Pfunction and remarking that the
differential equation that it satisfies gives the diffeomorphism to
S^3trefoil as the boundary of the pair (discriminant of diffeq,
C^2 = (P,P')space).
This point of view grows out of some observations of Zariski, fleshed
out in "Singular Points of Complex Hypersurfaces" [study 61]. The
geometric viewpoint is made explicit in the paper "On the Brieskorn
Manifolds M(p,q,r)" in "Knots, Groups, and 3manifolds" [study 84].
It is also related to the intermediate case between the classical
Platonic solids and John's favorite Platonic surface  the Klein quartic
http://www.math.ucr.edu/home/baez/klein.html. By way of a hint, look to
relate the trefoil, qua torus knot, the sevenvertex triangulation of
the torus, and the dual hexagonal tiling of a (flat) Clifford torus in S^3.
Joe
Swiatoslaw Gal writes:
In fact the isomorphism is a part of the modular theory:
Looking for
f: GL(2,R)/SL(2,Z) > C^2  {x^2=y^3}
(there is an obvious action of R+ on both sides:
M > tM for M in GL(2,R), x > t^6 x, y > t^4 y,
and the quotient is what we want).
GL(2,R)/SL(2,Z) is a space of lattices in C.
Such a lattice L has classical invariants
g_2(L) = 60 sum_{z in L'} z^{4},
and
g_3(L) = 140 sum_{z in L'} z^{6},
where L'=L{0}
The modular theory asserts that:
1. For every pain (g_2,g_3) there exist a lattice L
such that g_2(L) = g_2 and g_3(L) = g_3 provided that
(g_2)^3 is not equal to 27(g_3)^2.
2. Such a lattice is unique.
Best,
S. R. Gal
The quantity g_2^3  27 g_3^2 is called the "discriminant" of
the lattice L, and vanishes as the lattice squashes down to being
degenerate, i.e. a discrete subgroup of C with one rather than two
generators.

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