Geometric Representation Theory (Lecture 10)
Posted by John Baez
In my last lecture, I explained that when we simultaneously wave the magic wands of $q$deformation and categorification over the humble binomial coefficient
$\binom{n}{k} = \frac{n!}{k! (nk)!}$
it transforms into a marvelous thing: the Grassmannian of $k$dimensional subspaces of $F_q^n$, where $F_q$ is the field with $q$ elements.
This time in the Geometric Representation Theory seminar, I sketch what happens when we work the same magic on the binomial formula
$(x + y)^n = \sum_{k = 0}^n \binom{n}{k} y^k x^{nk}$
We’re soon led into deep waters: categorified quantum groups!
To $q$deform Pascal’s triangle amounts to putting it in a constant magnetic field, so a little electrically charged ball rolling down from the apex to a given point in the triangle picks up a phase depending on the path it takes. The $q$binomial coefficient
$\binom{n}{k}_q = \frac{(n)_q!}{(k)_q! (nk)_q!}$
is then the sum of these phases over all paths from the apex to the $k$th slot in the $n$th row of the triangle. This is a baby version of a path integral.
If we use $x$ to denote the process of rolling one step down and to the left, and $y$ for rolling one step down and to the right, we have
$x y = q y x$
So, these variables satisfy the $q$deformed binomial formula:
$(x + y)^n = \sum_{k = 0}^n \binom{n}{k}_q y^k x^{nk}$
We can think of $x$ and $y$ as coordinates on the ‘quantum plane’ — a mysterious object from the land of noncommutative geometry. The symmetries of the quantum plane are then the ‘quantum group’ $GL_q(2,k)$. So, if we succeed in categorifying the $q$deformed binomial formula, we should be well on our way towards categorifying this quantum group!
But, to really explain all this, I needed to review the basics of noncommutative geometry. And then I needed to pose the question: what do the variables $x$ and $y$ really mean here? And — a closely connected question — how do we categorify them?

Lecture 10 (Oct. 30)  John Baez on the $q$deformed Pascal’s triangle
and the quantum group $GL_q(2,k)$. Putting Pascal’s triangle
in a magnetic field, we obtain the $q$deformed Pascal’s triangle. Now
the operation of moving down and to right (called $x$) and the operation
of moving down and to the left (called $y$) no longer commute, but
instead satisfy:
$x y = q y x$
This relation implies the $q$deformed binomial formula:
$(x + y)^n = \sum_{k = 0}^n \binom{n}{k}_q y^k x^{nk}$
Picking a field $k$, the ‘algebra of functions on the quantum plane’, $k_q[x,y]$, is the associative algebra over $k$ generated by variables $x$ and $y$ satisfying the relation $x y = q y x$. The symmetries of the quantum plane form the quantum group $GL_q(2,k)$ The basic philosophy of algebraic geometry. The functor from geometry to algebra. Noncommutative geometry as a mutant version of algebraic geometry. Hopf algebras, and how they ‘coact’ on algebras.
A sketch of how we’ll simultaneously $q$deform and categorify the following structures:
 binomial coefficients (to obtain Grassmanians)
 the variables $x$ and $y$ showing up in the binomial theorem (to obtain certain Hecke operators)
 the group $GL(2,k)$ (to obtain a categorified version of the quantum group $GL_q(2,k)$)

Streaming
video in QuickTime format; the URL is
http://mainstream.ucr.edu/baez_10_30_stream.mov  Downloadable vi deo
 Lecture notes by Alex Hoffnung
 Lecture notes by Apoorva Khare
In case you’re wondering, I’m writing the binomial formula in this funny way:
$(x + y)^n = \sum_{k = 0}^n \binom{n}{k} y^k x^{nk}$
because of slightly suboptimal conventions I chose concerning the $q$deformed Pascal’s triangle. Actually, all the conventions I could think of seemed slightly suboptimal one way or another. But it’s no big deal.
Re: Geometric Representation Theory (Lecture 10)
You’d think that the $q = 0$ deformation would be quite simple. The contribution of a path in which there is left move followed by a right is $0$. There’s only one path between the apex and any given slot, so all entries of the deformed Pascal’s triangle are $1$.
Then concerning your earlier question about $q$deformed Gaussians, in the $q = 0$ case we’d have the limit of a uniform distribution as the range increases.
One could easily be led to believe that $q = 0$ is a bit boring. But then crystal bases and free probability seem rich enough. What, for instance, does the latter have to do with noncrossing partitions?