## October 26, 2007

### Geometric Representation Theory (Lecture 6)

#### Posted by John Baez

Where would a wizard be without his magic wands?

In mathematics, a ‘magic wand’ is any systematic process that you can apply to big chunks of interesting mathematics and get new, more interesting mathematics. Or — more magical still — it’s a mysterious bunch of tricks that feel like they’d be part of a systematic process if only we understood them better.

What are some magic wands? One of the most famous was stolen from physicists: it’s called quantization. Muttering one of several cryptic spells, you can wave this wand over any mathematical concept related to classical mechanics, and hope that — POOF! — it suddenly transforms into an analogous concept related to quantum mechanics. We’ve had huge success with this over the last century, but it’s still poorly understood.

Another magic wand is categorification: replacing any number by a set with that number of elements, replacing any set by a category whose set of isomorphism classes it is, and so on. You could almost say this blog is a shrine to categorification. It too, is still poorly understood. Perhaps when a magic wand’s powers become fully understood, it ceases to count as ‘magic’!

Yet another magic wand is $q$-deformation — closely related to quantization but not the same. It’s a way of modifying mathematical entities that depends on a parameter $q$. Sometimes this parameter has the physical meaning of $exp(i\hbar)$… but sometimes it’s better to think of it as a power of a prime number! In fact, $q$-deformation was discovered by Gauss long before the quantum was a twinkle in Planck’s eye.

When you have two magic wands at your disposal, you can ask if they commute. First wave one, then the other. First wave the other, then the one. Does the same magic occur? Or at least isomorphic magics?

In lecture 6 of the Geometric Representation Theory seminar, I wave two magic wands — categorification and $q$-deformation — at a humble mathematical entity: the binomial coefficient. It seems they commute. But, puzzles abound!

• Lecture 6 (Oct. 16) - John Baez on categorifying and $q$-deforming the theory of multinomial coefficients. A surprising fact: $q$-multinomial coefficients are actually polynomials in $q$ with natural number coefficients. It suffices to prove this for $q$-binomial coefficients, since any $q$-multinomial coefficient is a product of $q$-binomial coefficients. Since a $q$-binomial coefficient is the number of points in a Grassmanian over $F_q$ (the field with $q$ elements), it’s enough to decompose this Grassmannian into ‘Bruhat classes’, and show that each of these is isomorphic (as a set) to $F_q^k$ for some k. For this, we show that each Bruhat class corresponds to a set of matrices in reduced row echelon form. Example: the Young diagram
for which $D$-flags on $F_q^4$ correspond to 2d subspaces of $F_q^4$, or equivalently, lines in projective 3-space.

Another surprising fact: any $q$-multinomial is actually a ‘palindromic’ polynomial in $q$. The closure of a Bruhat class is called a Schubert cell, and this palindromic property follows from Poincaré duality, since Schubert cells are a basis for the cohomology of the Grassmannian over $\mathbb{C}$.

Posted at October 26, 2007 9:39 PM UTC

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### Re: Geometric Representation Theory (Lecture 6)

Have you ever thought of using magic wands from electrical engineering [EE]?

Paul AM Dirac had no degree in physics, but a BS in EE and PhD in mathematics. This allowed him to do things that were considered not to be rigorous.

http://nobelprize.org/nobel_prizes/physics/laureates/1933/dirac-bio.html

Richard Bellman developed dynamic programming or optimization.

http://en.wikipedia.org/wiki/Dynamic_programming

When a colleague remarked that this was not rigorous, Bellman reportedly responded, “Of course not. It’s not even precise. A good principle should guide the intuition.”

Posted by: Doug on October 26, 2007 11:09 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

Notes not available on this server??

Posted by: jim stasheff on October 27, 2007 1:28 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

John, your lecture looks interesting and I will try to find some time to watch it, but before I do can you please clarify something:

Is there a good mathematical reason why you consider the classical cohomology of a complex Grassmannian as opposed to its quantum cohomology? See, for example, the paper Quantum Schubert Calculus by Aaron Bertram.

Posted by: Charlie Stromeyer Jr on October 27, 2007 2:44 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

The reason I’m talking about the ordinary cohomology of Grassmannians instead of the quantum cohomology is that it’s simpler, and it provides a little mystery that we’re perfectly positioned to solve at this early stage in the course.

Take a Grassmannian over the complex numbers:

$\binom{n}{k}_{\mathbb{C}} = \{ k-dimensional subspaces of \mathbb{C}^n \}$

Compute its rational cohomology. Let $b_i$ be the dimension of its $2i$th cohomology group, also known as its $2i$th Betti number:

$b_i = dim H^{2i}(\binom{n}{k}_{\mathbb{C}}, \mathbb{Q})$

Form this generating function, also known as a Poincaré polynomial:

$\sum_i b_i q^i$

Lo and behold! This turns out to be the $q$-binomial coefficent:

$\sum_i b_i q^i = \binom{n}{k}_q$

which is also the number of points of the Grassmannian over the field with $q$ elements:

$\binom{n}{k}_q = |\binom{n}{k}_{F_q}|$

Now the fun part: explain why!

It’s a tiny baby version of the Weil conjectures, but for now it’s mainly an excuse to learn about Schubert cells. Try week188 for a quick sketch of the story.

Later, someday, maybe, we can get into the Schubert calculus, which describes the ring structure of the cohomology of a Grassmannian. Even later, someday, maybe, we could talk about its $q$-deformed analogue — the modified ring structure which shows up in the quantum cohomology of a Grassmannian. I would need to learn a lot more to do a good job of this. Luckily, this course may last two years.

But, there’s too much to do! This is just one branch of a many-branched tree — and our main goal is to climb a branch most people don’t know about yet, called The Tale of Groupoidification.

Posted by: John Baez on October 27, 2007 6:08 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

It was sort of fun watching you guys battle your way through the last bit, using the row echelon business to prove that $|Gr_{n, k}(F_q)|$ belongs to $\mathbb{N}[q]$. But your row echelon forms looked a little “left-handed” to me, and I’m wondering whether that made it any more confusing to the people out in TV Land, who are used to reduced row echelon forms “done right”.

I’d like to replay what you said in slightly different words (obviously with the same mathematical content). My preference would be to start with what “almost always” happens when you carry out the reduction process (i.e., start with the “generic” case), and then get more and more special.

Almost always, you can finagle your rank two $(2 \times n)$-matrix so that the $(11)$-entry is nonzero, which we then rescale to make it 1, and then kill off the (21)-entry. Then, almost always the (22)-entry will be nonzero; we rescale to make that 1, and then kill off the (12)-entry. There’s nothing more to be done: we’re in reduced form, and the class of matrices here looks like

$1 0 a a$ $0 1 a a$

(where $a = *$ stands for ‘anything’). Or maybe in the second sentence of the last paragraph, we had 0 for the (22)-entry, but something nonzero for the (23)-entry. Rescale to make that 1, then kill off the (13)-entry, and then you’re done. The class of matrices here looks like

$1 a 0 a$ $0 0 1 a$

Or maybe both the (22)- and (23)-entries are 0; then the last (24)-entry has to be nonzero, and eventually we reduce to

$1 a a 0$ $0 0 0 1$

Or, maybe when you start, you have 0’s in the first column (but from there on it’s generic):

$0 1 0 a$ $0 0 1 a$

Or, it might be

$0 1 a 0$ $0 0 0 1$

Or, in the most unlikely case of all, you have nothing but 0’s in the first two columns, and in reduced form you wind up with

$0 0 1 0$ $0 0 0 1$

This is what reduced row echelon forms usually look like (at least in textbooks I am familiar with). Now, in hindsight, we can figure out the flag we use to get the corresponding Bruhat cells: I think it’s

$\langle e_4 \rangle \subset \langle e_4, e_3 \rangle \subset \langle e_4, e_3, e_2 \rangle \subset \langle e_4, e_3, e_2, e_1 \rangle$

where $e_i$ is the vector with 1 in the $i$-th slot, and zeroes elsewhere. The cases where the second row is $e_4$ should be where the given projective line passes through the point of the flag. So, in order of the reduced row echelon forms as given above, I think we get

1. generic
2. Line $L$ intersects line $L_0$ of flag
3. Line $L$ intersects point $P_0$ of flag
4. Line $L$ in plane $\Pi_0$ of flag
5. Line $L$ in plane $\Pi_0$ and through point $P_0$ of flag
6. $L = L_0$

So it look like my order is opposite to yours in every respect!

Posted by: Todd Trimble on October 27, 2007 6:37 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

Thanks for writing that explanation! Written material makes a nice complement to the videos… especially when the video involves a lot of class participation, so a neat written “summary” can clarify what the upshot actually was. I can easily imagine people watching our class and then saying “Umm… so what just happened there?”

Personally I have trouble remembering the difference between left and right, since nobody ever told me how they’re defined. (Okay, I could rip open my chest and find my heart…) Actually the problem is that I’m left-handed, but almost ambidextrous, so which side is called “left” and which is called “right” is easy to forget. Visits to England only made things worse. It was incredibly devious of them to switch everything around like that after the US ceased to be a colony.

I’ve also never done much row reduction of matrices, so I don’t have an ingrained tendency to do it one way or another. Somehow row reduction is one of the things I managed to skip, mostly, in school.

Since the first standard basis vector is usually called (1,0,0,…), I set up my row reduction in this class so the ‘best possible’ — or in your words, ‘least likely’ — Bruhat class looks like

$1000$ $0100$

But, you’re right — for some reason most people do it upside down and backwards compared to this. Is there some reason, other than tradition?

Posted by: John Baez on October 27, 2007 6:52 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

But, you’re right – for some reason most people do it upside down and backwards compared to this. Is there some reason, other than tradition?

With the usual row reduction, you arrange to get columns $e_1$, $e_2$, … to appear as soon as you can, as you move in traditional left to right direction. So in the generic case, the $e_1$, $e_2$, … $e_k$ appear, in order, at the far left in a row-reduced rank $k$ matrix. In the video, the way you set things up forces you, in the generic case, to defer the columns $e_1$, $e_2$, … to the end, i.e., to the far right. It’s a little curious, I agree, but that’s the way the ball bounces!

Posted by: Todd Trimble on October 28, 2007 8:12 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

There should be a nice variant of ‘reduced row echelon form’ that describes the Schubert cells not just for Grassmannians, but also more general flag varieties. The difference should be something like this: you group the rows of your matrix into ‘bands’, and you’re only allowed to add a multiple of one row to another if it lies in the same or higher band.

Does this idea have a name?

As we’ll see next time, the Schubert cells for Grassmannians correspond to Young diagrams in a nice way — and you can instantly read off the Young diagram from the reduced row echelon form!

So: what’s the correspondingly cute notation for the Schubert cells of a general (i.e., partial) flag variety?

And while I’m at it: what’s the most fun and readable treatment of the Schubert calculus? What’s the most thorough one? What’s the most notation-ridden and tedious one?

Posted by: John Baez on October 27, 2007 11:50 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

I don’t know the answer to your question about Schubert cells but you might want to look at the generalized Plucker coordinates which are used to distinguish different Schubert cells that are quasi-projective subvarieties of a generalized flag variety.

I can better answer your question about Schubert calculus which was developed initially as the calculus of enumerative geometry. In this sense, the most “fun” intro is Scubert Calculus by S.L. Kleiman and D. Laskov in Amer Math Monthly v79(10) (1972) 1061-82.

For the shortest possible description of Schubert calculus see this paper , and for a more thorough treatment try P. Griffiths and J. Harris, Principles of Algebraic Geometry, Wiley-Interscience (1978).

In the last 20 years, the term “Schubert calculus” has come to mean the study of geometric, algebraic and combinatoric aspects of the Schubert basis in various cohomological settings and its relation to the rest of maths.

In this more recent sense of the term, I don’t know what the best intro would be but if you have more specific questions then I could try to answer them by consulting the almighty Google. I love Google but sometimes I secretly wish that it would not take over the entire known universe !-)

Posted by: Charlie Stromeyer Jr on October 28, 2007 4:28 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

On “why cohomology and not quantum cohomology?” the answer would be “they have the same Betti numbers, just different product structures. And the q-stuff here is only about the Betti numbers.”

On the right notation for Schubert classes in partial flag manifolds, the standard thing is to pull them back to the full flag manifold and then use permutations. Concretely, a permutation pi_1 pi_2 … pi_n comes from a partial flag manifold of d_1,d_2,…,d_k planes if the “descents” pi_i > pi_{i+1} only occur in places {d_j}. A “Grassmannian permutation” is one with only one descent.

There is a very nice way to index using the inverse permutations, but if I tell you it you’re likely to confuse yourself when you read most people’s papers.

Here’s where to read about modern Schubert calculus.

Posted by: Allen Knutson on October 29, 2007 6:55 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

I don’t know if you’d call this a “cute” notation, but here is how I like to think about the Schubert cells of G/P (for G=GL_n). Let’s suppose that G/P parameterizes partial flags of dimension (0, d_1, d_1+d_2, …, d_1+d_2+…+d_r), where d_1+d_2+…+d_r=n. So the Grassmannian is r=2, with (d_1,d_2)=(d,n-d) and the full flag variety is r=n with (d_1,d_2,…,d_n)=(1,1,…,1). Then Schubert cells of G/P are indexed by ways of coloring {1,2,..,n} with r colors, with the kth color used d_k times.

Here is how to write down the corresponding Schubert cell. We’ll record a Schubert cell by giving an nxn matrix whose first d_1 rows are a basis for the first subspace, whose next d_1+d_2 rows are a basis for the next subspace, and so forth. To help keep track of this, I find it helpful to color the first d_1 rows one color, the next d_2 rows another and so forth. (Or, if colored pencils aren’t avalable, to draw heavy lines seperating the blocks.) Our input data is an assignment of a color to each column of our matrix.

For the kth color, there are k rows of that color, and k columns. These intersect in a square submatrix; write an identity matrix in this square. We now need to figure out what to write in the other cells of our nxn matrix. For each cell, look at the row that it is in and the column. Each of them has a color, and those colors are different. If the row color is lower, write a *. If the column color is lower, write a 0. Then the Schubert cell is described by all ways to replace the *’s by elements of whathever field we are working over.

If anyone knows a prettier version, let me know!

Posted by: David Speyer on October 29, 2007 7:02 PM | Permalink | Reply to this

### An elementary proof that q-multinomials are palindromes in N[q]

I was amused to watch your rather painful check that the $q$-multinomials are in fact palindromes in $N[q]$. Here’s an argument that a middle-school student could follow.

I begin with an easy observation: if $m$ divides $n$, then $[m]_q$ divides $[n]_q$. Indeed, let $n/m = l$. Then

(1)$[n]_q = \frac{q^n - 1}{q-1} = \frac{(q^m)^l - 1}{q^m-1} \times \frac{q^m-1}{q-1} = [l]_{q^m} [m]_q$

So, now, let’s consider some binomial coefficient $n$-choose-$k$. From combinatorics, we know that the integer binomial is in fact an integer: this observation gives us a collection of explicit cancellations between numbers in the denominator of $\frac{n!}{k!(n-k)!}$.

So, say you want to evaluate a $q$-multinomial. You do the integer calculation, which consists first of a series of divisions. Each division can be promoted into a division of polynomials in $q$ as above. So at the end of the day, a $q$-multinomial is a product of palindromes in $N[q]$. Explicitly, it’s a product of “integers” of the form $[l]_{q^m}$ for various $l$ and $m$. The product of these is definitely also in $N[q]$.

Moreover, there’s an easy algebraic description of palindromic polynomials. Let $P(q)$ be a polynomial of degree $d$. Then it is a palindrome iff

(2)$P(q^{-1}) = q^{-d} P(q)$

I.e. $P(q)$ is a palindrome iff $q^{-d/2} P(q)$ is symmetric under $q \mapsto -q$. But the product of symmetric functions is itself symmetric, and we can keep track of degrees: the product of palindromic polynomials is palindromic.

Not that the Bruhat classes aren’t themselves interesting, but $q$-deformed arithmetic works for more elementary reasons.

(Incidentally, is there a good way to get the Mathbb fonts here? I’d love to write “{\mathbb N}” rather than $N$.)

Posted by: Theo on October 28, 2007 2:52 AM | Permalink | Reply to this

### Re: An elementary proof that q-multinomials are palindromes in N[q]

I was amused to watch your rather painful check that the $q$-multinomials are in fact palindromes in $N[q]$.

You read Miss Manners, do you?

(Incidentally, is there a good way to get the Mathbb fonts here? I’d love to write “{\mathbb N}” rather than N.)

Posted by: Todd Trimble on October 28, 2007 7:08 AM | Permalink | Reply to this

### itex

Incidentally, is there a good way to get the Mathbb fonts here? I’d love to write “{\mathbb N}” rather than N.)

Blackboard letters are obtained the LaTeX way: \mathbb{N} produces $\mathbb{N}$. And binomial coefficients are also produced the LaTeX way: \binom{n}{k} produces $\binom{n}{k}$. More generally, see the itex command summary for what’s supported.

Posted by: Jacques Distler on October 28, 2007 7:13 AM | Permalink | PGP Sig | Reply to this

### Re: An elementary proof that q-multinomials are palindromes in N[q]

So, say you want to evaluate a $q$-multinomial. You do the integer calculation, which consists first of a series of divisions. Each division can be promoted into a division of polynomials in $q$ as above. So at the end of the day, a $q$-multinomial is a product of palindromes in $N[q]$. Explicitly, it’s a product of “integers” of the form $[l]_{q^m}$ for various $l$ and $m$. The product of these is definitely also in $N[q]$.

I’m not quite sure you’ve nailed it. For example, say you try to implement this procedure in the case $\binom{8}{4}_q$. Your algorithm asks us to carry out a series of divisions in

$\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}$

in ordinary arithmetic, and then promote these to cancellations in $q$-arithmetic. Alright, suppose we first perform the cancellation

$\frac{8}{4 \cdot 2} = 1$

in ordinary arithmetic. But this doesn’t promote too well to $q$-arithmetic, because we are left with

$\frac{q^4 + 1}{q + 1}.$

You might counter and say that I chose an unfortunate way of cancelling in ordinary arithmetic; I should have used $\frac{6}{2 \cdot 3} = 1$ instead. But then the burden is back on you, to show there’s always a good way to cancel.

But I think the larger point is not that we’re trying to find the fastest symbolic proofs of results in $q$-arithmetic, but rather that these symbolic manipulations are really the shadows of some deeper, actually more concrete manipulations that we can perform directly on the structures whose cardinalities we’re counting. This procedure is well-known to combinatorialists, who often speak of giving “bijective proofs” of combinatorial identities. And that’s what we’re doing here: promoting equations to isomorphisms between structures. It’s called “categorification”, and it can lead to a lot of insight.

Posted by: Todd Trimble on October 28, 2007 4:02 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

The comment by Theo above that there are direct ways of establishing that $|Gr_{n, k}(F_q)|$ lies in $\mathbb{N}[q]$ does remind me of another tack to take, which has a nice combinatorial (or categorified) interpretation.

Namely, there’s a $q$-Pascal’s triangle for $\binom{n}{k}_q$:

$\binom{n}{k}_q = \binom{n-1}{k}_q + q^{n-k}\binom{n-1}{k-1}_q.$

To prove this combinatorially, say you have a $k$-plane in $F_{q}^n$. Let $x_n = 0$ be a coordinate hyperplane of dimension $n-1$. Then either your $k$-plane is contained in $x_n = 0$ (and there are $\binom{n-1}{k}_q$ ways that can happen), or it intersects $x_n = 1$ in an (affine) $(k-1)$-dimensional space. These affine spaces can be described in “slope-intercept” form: there are $\binom{n-1}{k-1}_q$ possible slopes (where “slope” means “parallel translate of the $(k-1)$-space through a chosen origin of $x_n = 1$”), and $(n-1)-(k-1) = n-k$ degrees of freedom in choosing the “intercept”. This gives the Pascal formula above.

We then immediately infer by induction that $\binom{n}{k}_q$ belongs to $\mathbb{N}[q]$.

Posted by: Todd Trimble on October 28, 2007 5:19 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

You just gave away part of lecture 8, which is all about the $q$-deformed Pascal’s triangle.

But that’s fine… we can call it ‘foreshadowing’.

As for Theo’s observation that we can simplify $q$-fractions using the identity

$n\ell = m \; \Rightarrow \; [n]_q = \frac{q^n - 1}{q-1} = \frac{(q^m)^\ell - 1}{q^m-1} \times \frac{q^m-1}{q-1} = [\ell]_{q^m} [m]_q$

the interesting thing to me is not this sort of ‘elementary’ argument per se, but whether it categorifies.

Since $[n]_q$ is the number of points in the projective space $P(F_q^n)$, this identity seems to want to relate projective spaces over two different fields, $F_q$ and $F_{q^m}$:

$n\ell = m \; \Rightarrow \; P(F_q^n) \cong P(F_{q^m}^\ell) \times P(F_q^m)$

Of course a product decomposition is more than we need: a ‘twisted product’, or bundle, would suffice.

So: is there some bundle with $P(F_q^n)$ as total space, with one of the spaces $P(F_{q^m}^\ell)$ and $P(F_q^m)$ as base space, and the other as fiber? That would give a nice ‘bijective proof’ that

$n\ell = m \; \Rightarrow \; |P(F_q^n)| = |P(F_{q^m}^\ell)| \times |P(F_q^m)|$

The fun part is how more than one field at a time seems to be getting into the act here. Maybe the Frobenius would come in handy.

Posted by: John Baez on October 28, 2007 8:01 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

You just gave away part of lecture 8, which is all about the $q$-deformed Pascal’s triangle.

Sorry about that. Please feel free to delete the comment in that case.

Posted by: Todd Trimble on October 28, 2007 8:38 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

No reason to delete it! We’ll probably keep coming around back to these Grassmannians and over and over throughout the course, using them as examples of all sorts of ideas.

Posted by: John Baez on October 28, 2007 8:57 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 6)

Of course a product decomposition is more than we need: a ‘twisted product’, or bundle, would suffice.

Interesting! There seems to be something a bit Hopf-fibration-y about the sequence

$E^{\star}/k^{\star} \to \mathbb{P}^{l m-1}(k) \to \mathbb{P}^{l-1}(E)$

where $E$ is a degree $m$ extension of a ground field $k$, that may be one way to think of this.

Posted by: Todd Trimble on October 28, 2007 9:47 PM | Permalink | Reply to this

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