While there are plenty of things to say about the quantum plane, and
quantum groups, I think I'll home in on my main topic at last:
r-commutative geometry. This is a particular approach to noncommutative
geometry that generalizes what mathematicians and physicists call
"supergeometry". So first I should say a brief word about supergeometry.
First, recall that the plain old geometry of manifolds can be cast into the
language of commutative algebra by considering not the manifold itself as a
set of points, but the *algebra* of (smooth, complex) functions
*on* the manifold. (In this language, for example, vector fields are
derivations, vector bundles are projective modules, and so on -- every
geometric construct has an algebraic analog.) Now in the 1970's, physicists
really caught on to the fact that considering only commutative algebras was
horribly unfair to fermions, which like to anticommute: i.e. they have

instead of

So while the phase space of bosonic system is a manifold, the phase space of a system containing both bosonic and fermionic degrees of freedom is a "supermanifold" -- a (particular kind of) algebra which has "even" and "odd" elements, such that the even, or bosonic, elements commute, while the odd elements anticommute. (The even elements commute with the odd elements, by the way.) Generalizing lots of concepts from commutative algebras to supercommutative algebras simply amounts to sticking in appropriate minus signs! The rule of thumb is that whenever one switches two elements and , one should stick in a factor of , where if is even (bosonic) and if is odd (fermionic). One may extend the notions of vector field, differential form, metric, curvature, and all your favorite concepts from geometry to "supergeometry" in this manner.

This turned out to be fascinating (it's a bit premature to say "useful")
in particle physics, where it goes by the name of supersymmetry. The idea
is that there should be a symmetry between bosons and fermions. While this
is *not at all* observed in nature, it would be nice if it *were*
true, so people have developed clever ways of rigging up their theories so
that you never see the "superpartners" every particle has: for the
photon, the photino, for the gluon, the gluino, for the leptons,
schleptons, for the quarks, squarks... you get the pattern. (Don't
complain to *me* if you think this is silly, it wasn't my idea!)
Superstrings are the latest of these "super" ideas in particle physics.

Supersymmetry has actually proven itself in a more practical manner in nuclear physics, where it lets one model resonances in nuclei, relating the properties of fermionic and bosonic nuclei.

Where supergeometry really shines, though, is in mathematics. For example, Ed Witten came up with beautiful proofs of the Atiyah-Singer index theorem and the positive mass theorem (a theorem about general relativity) using supergeometry. (As usual, he left it to others to make his arguments rigorous.)

While I personally don't think that bosons and fermions were created
equal in the manner postulated by supersymmetry, I do favor
an approach to physics which doesn't take bosons, or commuting variables,
to be somehow superior to fermions, or anticommuting variables. This
demands supergeometry. For example, a decent treatment of *classical*
fermions requires a supermanifold for the "phase space".

My own personal twist (motivated by the work of many people on anyons, the
braid group, quantum groups, etc.) is to try to take a look at what
geometry would be like if one wanted to be fair to *anyons* as well as
bosons and fermions. This is "r-commutative geometry."

Recall that given a vector space , an invertible linear transformation
, the Yang-Baxter equations say that

where is the identity on . The idea is that "switches" two elements of , mapping to , and if we draw as a "crossing," as follows:

The Yang-Baxter equations say that

Now suppose that our vector space is really an algebra -- let's call it
. (I mean an associative algebra with unit.) The product in the
algebra defines a multiplication map
, given by

We can draw this as the

Associativity simply says that , or in terms of diagrams,

Now these diagrams aren't really braids because of the "fusion of strands" that's taking place, but they fit in well with the braid group philosophy. For example, there is a sense in which the two trees I've drawn above are topologically the same, just as the Yang-Baxter equation expresses a topological identity. There are generalizations of braids, e.g. the "ribbon graphs" of Reshetikhin and Turaev, that make all this precise.

Now: an "r-algebra" is an algebra equipped with a solution of the Yang-Baxter equations such that:

1) and for all in -- i.e., one "switches" the identity 1 with in the usual manner;

2) the following conditions hold -- I'll draw them pictoriallly:

and

If you like equations instead of pictures, these are

and

as maps from to , respectively. The first one tells you how to compute , and the second one tells you how to compute , so both tell you how to switch a product of two elements past a third element. These are called the "quasitriangularity" conditions and are crucial in the theory of quantum groups (usually in a slightly disguised form).

Now commutativity says that -- in other words, you can multiply
two elements of , or you can switch them first and then multiply them,
and you'll get the same result. Generalizing this, we say that an
r-algebra is "r-commutative" if

as maps from to . In diagrams:

In words: "switch, then multiply, equals multiply."

It turns out that one can do a fair amount of geometry for r-commutative
algebras. And there are lots of examples of r-commutative algebras. In
fact, many of the algebras obtained by *quantization* are r-commutative:
for example, the Clifford algebra, the Weyl algebra, noncommutative
tori, the quantum plane (and all other r-symmetric algebras), and
quantum groups. Also, all supercommutative algebras are automatically
r-commutative.

In fact, I believe that there is a deep relation between braids (or r-commutativity) and quantization. This goes back to Kauffman's observation that the skein relation for the Conway polynomial

and the canonical commutation relations

It was made more clear by the discovery of the relation between quantum groups and knot invariants (which I haven't really touched upon yet - here I highly recommend Louis Kauffman's book

and , write down the Yang-Baxter equation for , and collect all terms of order . This equation is identical to the equation that a "Poisson structure" must satisfy. For details, see Drinfeld's famous review paper on quantum groups. The point here is that a Poisson structure, which defines the Poisson bracket of classical observables, is a semiclassical limit of the commutator which appears in quantum mechanics. Thus an infinitesimal deformation of the usual twist map that is required to satisfy the Yang-Baxter equation is

More later...

© 1992 John Baez

baez@math.removethis.ucr.andthis.edu