[continued from previous message]   
      
   boring classical dynamics, because the action is constant on each   
   connected component of the path space.  (Or, in physics lingo, the   
   Lagrangian is a total divergence.)  But they can give nontrivial   
   dynamics after quantization, because of phase effects.  In fact, the   
   simplest example of this sort of deal is the Bohm-Aharonov effect.   
   The particle can go around an obstacle in either of two ways so the path   
   space consists of two components.  Classically, a term in the action   
   that is constant on each component doesn't do anything.  But   
   quantum-mechanically it leads to interference due to a shift of phase.   
      
   These days the ultrasophisticated mathematical physicists and   
   topologists love talking to each other about "topological quantum field   
   theories" in which the Lagrangian is a diffeomorphism invariant.  The   
   term with action equal to the integral of A^dA is called the "U(1)   
   Chern-Simons theory", because a 1-form may be regarded as a connection on a   
   U(1) bundle.  This is a very simple theory; the more interesting ones   
   use nonabelian gauge groups.  Witten showed (in his rough-and-ready   
   manner) that just as the linking number is related to the U(1)   
   Chern-Simons theory, the Jones polynomial is related to the SU(2)   
   Chern-Simons theory.  (Many people have been trying to make this more   
   rigorous.  Right now my friend Scott Axelrod is working with Singer on   
   the perturbation theory for Chern-Simons theory, which should make the   
   story quite precise.)   
      
      
   The Yang-Baxter Equation   
      
   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 n strands form a group, called the braid group B_n.   
   (Let's take n = 4 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 x has an "inverse"   
   y such that xy = yx = 1.   
      
   The braid group has special elements s_1, s_2, ..., s_{n-1}, where   
   s_i is the braid where the ith strand goes over and to the right of the   
   (i+1)st.  For example, s_2 equals   
      
   |   \   /   |   
   |    \ /    |   
   |     \     |   
   |    / \    |   
   |   /   \   |   
      
   Now, the interesting thing about these "elementary braids" is that   
   s_i and s_j commute if |i - j| > 1 (check it!) but s_i and s_{i+1} 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,   
      
   s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}.   
      
   Let me draw it for the braid group on three strands, where i = 1:   
      
   \   /   |	|    \ /   
    \ /    |	|     \   
     \     |	|    / \   
    / \    |	|    /  \   
   /   \   |	\   /   |   
   |    \ / 	 \ /    |   
   |     \     =	  \     |	   
   |    / \           \    |   
   |    |	\	 /  \   |   
   \   /   |	|    \ /   
    \ /    |	|     \   
     \     |  	|    / \   
    / \    |	|   /   \   
   /   \   |	|  /     \   
      
      
   That's about the ugliest picture of this beautiful identity that I've   
   ever seen!  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 s_i, but the extremely   
   non-obvious fact is that the braid group is isomorphic to the group with   
   the s_i as generators and the relations   
      
   s_i s_j = s_j s_i       |i-j| > 1   
   s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}.   
      
   Note that there's a homomorphism from the braid group onto the symmetric   
   group with s_i mapped to the ith elementary transposition.  In fact the   
   symmetric group may be obtained by throwing in the relations   
      
   s_i s_i = 1   
      
   in with the above ones.  In other words -- and this has a lot of   
   physical/philosophical significance -- the braid group may be thought of   
   as the obvious generalization of the group of permutations to a   
   situation in which "switching" two things' places twice does NOT get you   
   back to the original situation.  In physics, of course, this happens   
   with anyons.   
      
   It's easy to determine the one-dimensional unitary representations of   
   the braid group.  In such a representation each s_i gets mapped to a   
   complex number of unit magnitude, which we'll just call s_i.  Then   
   the Yang-Baxter equation shows that s_i = s_{i+1}, so all the s_i's are   
   equal.  The other braid group relations are automatic.  Thus for each   
   angle theta there is a unitary rep of the braid group with   
      
   s_i  --->   exp(i theta)   
      
   and these are all the one-dimensional unitary reps.  In physics lingo   
   these are all "fractional statistics."  Note that for one of these reps   
   to quotient through the symmetric group we must have exp(i theta) = 1   
   or -1; these correspond to bosonic and fermionic statistics   
   respectively.   
      
   Okay, how about  reps of the braid group that are not necessarily   
   one-dimensional?   Well, suppose V is a vector space and we want to get   
   a rep of B_n on the nth tensor power of V, which I'll call V^n.  Here's   
   a simple way: just map the ith elementary braid s_i to the linear map   
   taking   
      
   v_1 x v_2 x ... x v_n   
      
   (here x means tensor product) to   
      
   v_1 x .... R(v_i x v_{i+1}) x ... x v_n   
      
   where R is an invertible linear transformation of V^2.  In other words,   
   we use R to "switch" the ith and (i+1)st factors, and leave the rest   
   alone.   It's easy to see that this defines a rep of the braid group as   
   long as we can get the Yang-Baxter equation to hold, and this is   
   equivalent to having   
      
   (R x I)(I x R)(R x I) = (I x R)(R x I)(I x R)   
      
   where again  "x"  means tensor product and I is the identity on V.   
   Actually, *this* equation is more usually known as the Yang-Baxter   
   equation.   
      
   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   
      
   [continued in next message]   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)