I find that Polyakov model I described last time to be a great example of all sort of things: solitons, instantons, anyons, nonlinear sigma models, gauge theories, and topological quantum field theories all in one! But I want to get back to braids plain and simple and introduce the Yang-Baxter equations. I'll tone down the math for a while so that people who know only a tad of group theory can at least get the definition of the braid group.
So: braids with strands form a group, called the braid group . (Let's take as an example.) Multiplication is just defined by gluing one braid onto the bottom of another. For example this braid:
times this one:
equals this:
The identity braid is the most boring one:
I leave it as a mild exercise to show that every braid has an ``inverse'' such that .
The braid group has special elements , where is the braid where the -th strand goes over and to the right of the -st. For example, equals
Now, the interesting thing about these ``elementary braids'' is that
and commute if (check it!) but and get
tangled up in each other. They satisfy a simple equation, however,
the Yang-Baxter equation. (This was known ages before Yang and Baxter
came along, and its importance was perhaps discovered by Artin.)
Namely,
Draw it for yourself and check that these are topologically equivalent braids (i.e., you can get from one to the other by a little stretching and bending).
Okay, now I'll start turning the math level back up. The braid group is
obviously generated by the elementary braids , but the extremely
non-obvious fact is that the braid group is isomorphic to the group with
the as generators and the relations
It's easy to determine the one-dimensional unitary representations of
the braid group. In such a representation each gets mapped to a
complex number of unit magnitude, which we'll just call . Then
the Yang-Baxter equation shows that , so all the 's are
equal. The other braid group relations are automatic. Thus for each
angle there is a unitary rep of the braid group with
Okay, how about reps of the braid group that are not necessarily
one-dimensional? Well, suppose is a vector space and we want to get
a rep of on the -th tensor power of , which I'll call . Here's
a simple way: just map the -th elementary braid to the linear map
taking
Now, Baxter invented this equation when he was looking for problems in 2d statistical mechanics that were exactly solvable (i.e. one could calculate the partition function). A simple example is the ``six-vertex'' or ``ice-type'' model, so-called because it describes flat square ice! (Theoretical physicists are never afraid of simplifying the world down to the point where they can understand it.) But there are zillions of examples -- take a look at his book on exactly solvable statistical mechanics models! Similarly, Yang ran into the equation while trying to cook up 2d quantum field theories for which one could exactly calculate the S-matrix. I won't get into how this equation does the trick right now; I just want to note that a whole industry developed of finding solutions to the Yang-Baxter equations. This eventually led to the discovery of quantum groups...but for now I'll leave you with a nice solution of the Yang-Baxter equations: let be spanned by two vectors and , and define by
If you are interested in learning about quantum groups, the Yang-Baxter equations, and braids, I suggest taking a look at the following books:
L. Kauffman, Knots and Physics, World Scientific, New Jersey, 1991.
© 1992 John Baez
baez@math.removethis.ucr.andthis.edu