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# The Quantum Plane

Ever so slowly I'm creeping towards a description of my own research on the relation between braids and quantization. There are so many inviting byways...Anyway, recall where we were last time. (I never knew I'd become so professorial in my old age.)

Given an 1-1 and onto linear transformation , the Yang-Baxter equations say that where is the identity on . The most famous example of a solution of the Yang-Baxter equations is where is spanned by two vectors and , and Now, given any solution of the Yang-Baxter equations, or "Yang-Baxter operator," we can define a "quantum vector space" as follows. Well, wait -- mathematicians like to ladle on the definitions and explain the point later, but I really should give you some clue about "quantum vector spaces" before proceeding.

Quantum vector spaces were basically invented by Manin in his book, "Noncommutative Geometry and Quantum Groups," and I will only be talking about a subclass of his quantum vector spaces. To get at the concept, you have to think like an algebraic geometer -- or a quantum field theorist. To an algebraic geometer, what counts about a space is the algebra of functions on that space (to be a bit more precise, algebraic functions from that space to the complex numbers...and please, algebraic geometers, don't bother telling me how much I'm oversimplifying -- I know). To a field theorist, what counts about spacetime is the fields on spacetime, which are again functions from spacetime to the complex numbers. (And again, I'm vastly oversimplifying things.) There's a common theme here: for the algebraic geometer points may be regarded as secondary in importance to functions, while for the physicist particles may be regarded as secondary in importance to fields. This makes a lot of sense when you pay attention to what you're doing as you calculate: most of the time you are playing around with functions, and only at the end (if ever) do you bother to evaluate them at a given point.

Well, regardless of how true this actually is, it's a fruitful point of view! It gives rise to noncommutative geometry, in which points are not the point; rather, all attention is concentrated on the algebra of "functions on space," which is allowed to have a noncommutative product. Here I should mention Alain Connes, whose paper "Non-commutative differential geometry" really got the subject rolling in the early 1980's, though the notion of replacing functions on an honest space with a noncommutative algebra is as old as Heisenberg's matrix mechanics. In quantum mechanics, phase space disappears, but the analog of the functions on phase space, the algebra of observables (typically operators on a Hilbert space) lives on.

So now let us define our quantum vector space: what we'll really be defining is an abstract algebra, but in the back of our heads we'll always secretly pretend it consists of functions on some (nonexistent) space, a "quantum space."

First, notice that a Yang-Baxter operator on gives rise to a Yang-Baxter operator on the dual space -- the inverse of the adjoint of . Recall that is none other than the linear functions on . Normally we define the algebra of polynomial functions on , or the "symmetric algebra" , to be the quotient of the tensor algebra by the ideal generated by elements of the form In other words, we take all (formal) products of linear functions on and impose on them the "commutation relations" that times should equal times . Note that here we are switching and ; in our quantum vector space we will use the Yang-Baxter operator to do this switching. Thus we define the "r-symmetric quotient of " by the ideal generated by elements of the form That's all there is to it -- the philosophy is that we should never switch two objects blindly in the usual manner, since they might be "anyonic": we should always switch things with the appropriate Yang-Baxter operator.

Now, all these duals and adjoints really just clutter things up a bit; let's just define the r-symmetric algebra to be modulo the ideal generated by So if we use the example of a Yang-Baxter operator given up above, that I said was so famous, we get the following r-symmetric algebra, called the quantum plane: it has coordinates and , which do not commute, but satisfy !!!

The less jaded of you will be dazzled by the mere appearance of this bizarre entity. The more jaded will wonder what it's good for. Patience, patience. Let me just say for now that many papers have been written about this thing, in respectable physics journals, so it must be good for something. (?)

One can do all sorts of geometry on the quantum plane just as one does on the good old classical plane. There is a "quantum group" of "quantum matrices" which acts as linear transformations of the quantum plane. One can differentiate and integrate functions on the quantum plane, and so on. In fact, MIT just installed new blackboards in some of the math classrooms which have a dial with which the instructor can select the value of .

All but the last sentence of the previous paragraph is true. In fact, the simplest quantum group, , consists of 2-by-2 "quantum matrices" with "quantum determinant" equal to 1. It is a prototype for the quantum groups . People are having a ball these days coming up with -analogs of everything they know about. For example, all semisimple Lie groups admit quantizations of this general type, and such quantum groups turn out to arise as symmetries of field theories in 2 or 3 dimensions (where anyonic statistics arise), and to be crucial for understanding the HOMFLY polynomial and other knot invariants. I'm running ahead of myself here and should quit and go to bed; next time I'll either define quantum matrix algebras and , or start on what I'm really interested in, the differential geometry of certain "quantum spaces" (more precisely, algebras equipped with Yang-Baxter operators).

Next: r-Commutative Geometry Up: Braids and Quantization Previous: The Yang-Baxter Equation