Counting with q (or if q=1, Fun with F_un)

First think about natural numbers. They list out the elements of sets. I have 1 thing in my set or maybe 137.
Now think of counting something else, namely lines in a space of dimension n over the field of q elements. There are q^n choices for a point in the space. q^n-1 if you want it to be a nonzero vector. But there are q-1 of them along that line, so there are (q^n-1)/(q-1)=1+q+...+q^(n-1) lines in n dimensional q space. If you set q=1, you just get n.
Let's do something with this now. Take the flag of F^n describing points on lines inside planes.... and so on. How many points are there in this flag when F is of size q? (This gives a non-stupid way to work with a field of size 1.) Well we first pick a line L in F^n, then we pick a line in F^n/L and keep going. This will be nq for the first step [n-1]q for the next step and so on. That means it is the q-factorial, where you just replace each number by its q deformed self.
What's cool about this is when we want to do manifolds instead over R or C instead, we can do the same trick. Of course now instead of getting numbers of points we are going to get the number of k-cells in the manifold. For the flag manifold of C^4 we have 4! q version which is

1 + 3q + 5q2 + 6q3 + 5q4 + 3q5 + q6

 That tells us there are 1 0-cells, 3 complex dimension 1-cells, 5 complex dimension 2-cells and so on. This tells us the chain complex but not the differentials. However in this case the differentials have to be 0 because there are no odd real dimensional cells. There are a bunch of holes in the chain complex. Now we know the complex and differentials so we can just read off homology from the coefficients of the polynomial. For example, the sixth homology group would just be from reading the q^3 coefficient 6. Therefore the answer is Z^6 to describe combinations of those 6 6-cells. 6 6-cells try that again. Of course we see the symmetry between coefficients as is required by Poincare. We just calculated the homologies without having any idea what those cells looked like thanks to the commonalities between projective geometry over finite fields and complex numbers.

Affine Lie Algebras

Start off with your favorite lie algebra g. Now take functions from an S^1 to g. Thinking of the S^1 as the unit circle in the complex plane, this becomes the space . This is an infinite Lie algebra, so unlike finite dimensional Lie algebras which we can't do much with, we can mess with this one. We can put in an anomaly c. This makes the commutation relations  instead of the boring one we had before. What is the Cartan for this system? It includes the Cartan for g as g*t^0 and c. But there is a problem with this, we have no dependence on the power of t. This makes all the eigenspaces of the Cartan acting with ad(X) infinitely dimensional because we can multiply by any power of t to get something else in the eigenspace. This is why we extend the algebra once more by an operator that reads of the power of t like t*d/dt. This gives the full Cartan of our untwisted affine Lie algebra as what we had before and this thing which reads off the power of t. So we got rid of the degeneracy and we are set. We specify a weight by saying what its eigenvalue is under each of the ad(X)'s in the Cartan. Of course we said c is central so ad(c) should have eigenvalue 0 for the weights of the adjoint representation called roots. So roots are of the form (l,0,n) where l is a root of g. This means they are part of an r+1 dimensional lattice. r from g and one more from n. We can pick the simple roots to be (b,0,0) with b a simple root of g and one extra (-t,0,1) where t is the highest root of g. Doing the same trick of Dynkin diagrams as with ordinary Lie algebras indicates we just need to add one more node that represents that last affine root.

The additional node is shown in green. As usual the connections indicate values of the Cartan matrix which is now just one dimension bigger. We could even do more if we twisted by not insisting on a genuine map from the circle to the algebra, but one that differed by some automorphism upon going around the circle. A Z_2 automorphism would be like choosing antiperiodic boundary conditions instead of periodic ones, but no one can tell the difference between + and - so we could have chosen either.