## November 6, 2007

### Geometric Representation Theory (Lecture 8)

#### Posted by John Baez

This week in the Geometric Representation Theory seminar we take another of my favorite things — Pascal’s triangle — and wave two magic wands over it: the wand of categorification, and the wand of $q$-deformation! When we do this, the humble binomial coefficient

$\binom{n}{k}$

magically transforms into the Grassmannian

$\binom{n}{k}_F$

namely the set of all $k$-dimensional subspaces of an $n$-dimensional vector space over the field $F$. And, the famous recursive formula for the binomial coefficients:

$\binom{n}{k} \; = \; \binom{n-1}{k} \;+ \; \binom{n-1}{k-1}$

mutates into an interesting fact about Grassmannians… but one we’ll only understand fully when we bring Hecke operators into the game.

I wrote about some of this back in week188 of This Week’s Finds. But, it’s much nicer to see it as part of the grand program we’re engaged in now: systematically categorifying and $q$-deforming huge tracts of mathematics with the help of geometric representation theory.

By the way: has anybody ever plotted the ‘$q$-deformed Gaussian’ you’d get by graphing the $q$-binomial coefficients in the $n$th row of the $q$-deformed Pascal’s triangle and then taking a suitably rescaled limit as $n \to \infty$? I’d like to see it, but I’m too busy right now to fire up Mathematica and plot it myself. And surely someone has already done this.

• Lecture 8 (Oct. 23) - John Baez on the q-deformed Pascal’s triangle. Categorifying and q-deforming the recursion relation for binomial coefficients. If $\binom{n}{k}_F$ stands for the set of $k$-dimensional subspaces of the vector space $F^n$, we have: $\binom{n}{k}_F \; \cong \; \binom{n-1}{k}_F \; + \; F^{n-k} \times \binom{n-1}{k-1}_F$ so in particular, taking $F$ to be the field with $q$ elements, we obtain this relation for $q$-binomial coefficients: $\binom{n}{k}_q \; = \; \binom{n-1}{k}_q \;+ \; q^{n-k} \binom{n-1}{k-1}_q$ Using this to compute the $q$-deformed Pascal’s triangle. Symmetries of the $q$-deformed Pascal’s triangle. Why the binomial coefficient $\binom{n}{k}$ is the number of combed Young diagrams with ≤ $k$ columns and ≤ $n-k$ rows. Why the $q$-binomial coefficient $\binom{n}{k}_q$ is the sum over such Young diagrams $D$ of $q^{# \; of \; boxes \; of \; D}$ Why each term in this sum corresponds to a specific Bruhat class in the Grassmannian of $k$-dimensional subspaces of $F^n$. The relation between Young diagrams and matrices in reduced row-echelon form.

Posted at November 6, 2007 7:01 PM UTC

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### Re: Geometric Representation Theory (Lecture 8)

…has anybody ever plotted the ‘q-deformed Gaussian’?

I always wanted to understand what those free probability people were on about when they say something about the Wigner semicircular distribution being a $q = 0$ deformation of Gaussian distributions. E.g., section 2.6 of this.

Posted by: David Corfield on November 8, 2007 10:18 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

David, I am not an expert on free probability theory and I don’t know the answer to your question, however, you might be interested to know that there was a more recent conference on the subject than the conference you cite. This conference was two months ago at the Fields Institute.

If you want to you can read the abstracts for the conference. Perhaps one of these experts might know something more about your question.

Posted by: Charlie Stromeyer Jr on November 11, 2007 3:41 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

David raises a good point, which I would like to amplify:

WHAT THE HECK IS ‘FREE PROBABILITY THEORY’ REALLY ABOUT?

Or perhaps I should say: ‘what the Hecke?’

According to the paper by Marek Bożejko and Roland Speicher which David’s link refers to, free probability theory is related to the $q \to 0$ limit of a version of Fock space where the annihilation and creation operators satisfy $q$-deformed commutations relations like this:

$a_i a^*_j - q a^*_j a_i = \delta_{ij}$

Precisely these $q$-deformed commutation relations show up in the Joyal–Street–Hecke theory of categorified $q$-deformed Fock space, described in week185. We’ll talk about that quite soon in this seminar.

But, what happens in the $q \to 0$ limit? — and how does all this work on ‘free probability theory’, completely mysterious to me, actually connect to things I understand?

Posted by: John Baez on November 9, 2007 6:40 PM | Permalink | Reply to this

### Free probability

I’m not an expert on free probability (and I know nothing about the specific questions being raised here), but I know enough to tell you that different experts will give very different answers if you ask what free probability is “really” about. This is true to the extent that some people will make it sound like a subfield of combinatorics, or of probability (but not combinatorial probability!), or of operator algebra theory.

The approach to the subject which I suspect would be most to the liking of people here is unfortunately not usually put up-front because it’s buried behind a lot of motivating material on operator algebras and/or random matrices. The lecture notes here may be a good place to start.

Posted by: Mark Meckes on November 11, 2007 3:17 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

WHAT THE HECK IS ‘FREE PROBABILITY THEORY’ REALLY ABOUT?

It appears to have something to do with Random matrices, planar algebras and subfactors.

Posted by: David Corfield on December 19, 2007 3:54 PM | Permalink | Reply to this

### Free probability and entropy; Re: Geometric Representation Theory (Lecture 8)

I know even less. But free probability is indeed a mathematical theory which studies non-commutative random variables, where “freeness” is the analogue of the classical notion of independence, and it is connected with free products.

This theory was initiated by Dan Voiculescu around 1986 in order to attack the free group factors isomorphism problem, an important unsolved problem in the theory of operator algebras.

Given a free group on some number of generators, we consider the von Neumann algebra generated by the group algebra, which is a type II_1 factor. Are these isomorphic for different numbers of generators? Nobody even knows if any two free group factors are isomorphic.

Theorists, driven by hope and aesthetics, want to construct new invariants of von Neumann algebras. In that context, free dimension is anointed as a reasonable candidate for such an invariant. The main tool used for the construction of free dimension is free entropy.

There’s a PDF online about why and when Voiculescu received the NAS award in mathematics, which includes a relatively readable description of free probability.

Posted by: Jonathan Vos Post on December 19, 2007 6:44 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

John, as I am actually an expert on free probability theory, let me try to straight out some of the questions raised in your post.

Just to get started, let me explain in which sense the semicircle is the $q=0$ version of the classical Gaussian distribution.

Consider the relation $a a^* - q a^* a=1$ together with a vacuum vector $\Omega$, defined by $a\Omega=0$. This data determines all moments of $a+a^*$ with respect to the vacuum expectation state, i.e., all numbers $\langle \Omega,(a+a^*)^n\Omega\rangle.$ Whereas for q=1 these numbers are the moments of a classical Gaussian distribution, it turns out that for $q=0$ they are the moments for the semicircle (which, by the way, are the Catalan numbers). In this sense the semicircle is the $q=0$ version of the classical ($q=1$) Gaussian.

Actually, for all $-1\leq q\leq 1$, the moments of $a+a^*$ are the moments of a probability distribution - that’s what I would address as a q-Gaussian distribution. By the way, for $q=-1$, one gets just the symmetric Bernoulli distribution with equal mass at $-1$ and $+1$.

Up to this point there has no free probability shown up in the picture. This comes in if one considers not just one, but several operators. Take the relations $a_i a_j^* - q a_j^* a_i = \delta_{ij},$ again with a vacuum vector $\Omega$ with $a_i\Omega=0$ for all $i$. Then, for $q=1$, the operators $a_1+a_1^*$ and $a_2+a_2^*$ are independent with respect to the vacuum expectation state, whereas for $q=0$ they are free in the sense of free probability theory.

A. Nica, R. Speicher: Lectures on the combinatorics of free probability. London Mathematical Society Lecture Note Series, Vol. 335, Cambridge University Press, 2006

Posted by: Roland Speicher on December 21, 2007 3:20 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

Thanks, Roland, for a very clear explanation of what’s going on. The way the semicircle shows up is very beautiful.

Since I used to work on quantum field theory, I’m familiar with the $q = 1$ and $q = -1$ cases of what you’re talking about… and since I like $q$-deformation, I’m eager to learn about the other cases too. I’ll try to get ahold of your book, but here’s a little question for now:

Then, for $q=1$, the operators $a_1+a_1^*$ and $a_2+a_2^*$ are independent with respect to the vacuum expectation state, whereas for $q=0$ they are free in the sense of free probability theory.

What is that sense? Is it the usual algebraic sense? Namely that $a_1 + a_1^*$ and $a_2 + a_2^*$ satisfy no relations, so they generate a free associative algebra on 2 generators?

Actually, I could answer this myself if I knew what commutation relations you impose for the annihilation operators $a_i$ and $a_j$. In the $q = \pm 1$ cases the natural choice would be

$a_i a_j - q a_j a_i = 0$

But this doesn’t look consistent for other values of $q$, unless we require that $i \lt j$. I suppose you could also not impose any relations on the annihilation operators… but in the $q = \pm 1$ cases, people usually do.

Posted by: John Baez on December 21, 2007 6:19 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

I will write next time in which sense $a_1+a_1^*$ and $a_2+a_2^*$ are free; let me for now just comment on whether one can impose commutation relations between the annihilation operators.

As I said before, the q-commutation relations together with the fact that the vacuum is annihilated by the annihilation operators determines all moments of the $a_i$ and $a_j^*$ with respect to the vacuum expectation state, and for general $q$ these are not compatible with any kind of relation between $a_i$ and $a_j$. For $q=+1$ it turns out that the vacuum expectation vanishes on the ideal generated by $a_i a_j - a_j a_i$, thus we can add the relation $a_i a_j -a_j a_i =0$ without any harm; for $q=-1$ it turns out that the vacuum expectation vanishes on the ideal generated by $a_i a_j + a_j a_i$, thus we can add the relation $a_i a_j+ a_j a_i=0$. For all the other $q$, however, there is no such ideal which vanishes under the vacuum expectation, so in all those cases (including the “free” case $q=0$) there are no relations between the annihilation operators.

[The same can be rephrased in the following way: whereas the relations for $q=1$ can be realized on the symmetric Fock space and for $q=-1$ on the anti-symmetric Fock space, for all the other $q$ we have realizations on the full Fock space without dividing out anything (in particular, without using any form of q-deformed symmetrization).]

So in particular, for $q=0$ this means that our free operators satisfy no algebraic relation and are free in the algebraic sense. However, freeness in the sense of free probability is a much stronger condition than just that; in a similar way as independence of two random variables does not just mean that they commute, but is a very precise condition on moments of the variables. I will say something more precise on freeness next time.

Posted by: Roland Speicher on December 22, 2007 3:57 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

Roland wrote:

whereas the relations for $q=1$ can be realized on the symmetric Fock space and for $q=−1$ on the anti-symmetric Fock space, for all the other $q$ we have realizations on the full Fock space without dividing out anything (in particular, without using any form of $q$-deformed symmetrization).

Okay, great — that’s very clear, especially since I know that the ‘full Fock space’ is what mathematicians call ‘the tensor algebra’ $T V$ on a vector space $V$ (or the Hilbert space completion of $T V$, if $V$ is a Hilbert space).

But, I’m also fond of certain $q$-deformed Fock spaces that we get when $V$ is equipped with a Hecke $R$-matrix, that is, an operator

$R: V \otimes V \to V \otimes V$

satisfying the Yang–Baxter equations

$(R \otimes 1)(1 \otimes R)(R \otimes 1) = (1 \otimes R)(R \otimes 1)(1 \otimes R)$

and the Hecke relations — some quadratic equation ensuring that $R^2$ has just two eigenvalues, say $q$ and $-q^{-1}$, though nobody can agree on the precise normalization. These relations let us define $R$-symmetrization’ and then an $R$-symmetrized Fock space’. A lot of people think about things like this.

Can your $q$-deformed annihilation and creation operators be represented on some $q$-deformed Fock space, or only on the full Fock space?

(I’m just trying to see if free probability theory is connected to this $q$-deformed Fock space game, or not.)

Posted by: John Baez on December 24, 2007 2:18 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

The q-deformed Fock space for the q-deformed commutation relations is the full Fock space, equipped with some q-deformed inner product. There is a Yang-Baxter operator behind this construction (which is used to define the deformed inner product), but it does not satisfy the Hecke relations; actually it is just given by flipping the two factors in the tensorproduct and multiplying by q. This can be generalized to more general Yang-Baxter operators; for this see

Bozejko M., Speicher R.: Completely positive maps on Coxeter groups, deformed commutation relations, and operator spaces. Math. Ann. 300 (1994), 97-120

and

Positive Representations of general commutation relations allowing Wick ordering (P.E.T. Jørgensen, L.M. Schmitt, R.F.Werner; J. Funct. Anal. 134 (1995) 33-99)

Posted by: Roland Speicher on January 7, 2008 4:06 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 8)

I wonder, as I think I did somewhere else, whether $q \to 0$ behaviour might depend on how you go to $0$.

Closer to the lectures is the kind of work gently conveyed here by Jonathan Brundan on crystals and Young diagrams.

Posted by: David Corfield on November 10, 2007 6:31 PM | Permalink | Reply to this
Read the post Geometric Representation Theory (Lecture 10)
Weblog: The n-Category Café
Excerpt: Simultaneously categorifying and q-deforming Pascal's triangle will lead us to a categorified quantum group. Here we take the first steps in that direction.
Tracked: November 12, 2007 9:43 PM

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