The n-Category Café

Notes not available on this server??

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:

$${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_{ℂ}=\{k-\mathrm{dimensional}\mathrm{subspaces}\mathrm{of}{ℂ}^n\}$$

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

$$b_i=\dim H^{2i}({\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_{ℂ},\mathbb{Q})$$

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:

$$\sum _i{b}_i{q}^i={\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q$$

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

$${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q=\mid {\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_{F_q}\mid$$

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.

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?

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.

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!

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.)

Try “\mathbb{N}” instead.

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 $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. More generally, see the itex command summary for what’s supported.

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 ${\left(\genfrac{}{}{0ex}{}{8}{4}\right)}_q$. 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.

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

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

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.