## November 7, 2016

### Linear Algebraic Groups (Part 6)

#### Posted by John Baez

When you’re doing math, if you ever want to keep things from getting too wispy and ethereal, it’s always good to count something. In fact, even if counting were good for nothing else — a strange counterfactual, that — mathematicians might have invented it for this purpose. It’s a great way to meditate on whatever structures one happens to be studying. It’s not the specific numbers that matter so much, it’s the patterns you find.

Last time we introduced Grassmannians as a key example of Klein’s approach to geometry, where each type of geometrical figure corresponds to a homogeneous space. Now let’s count the number of points in a Grassmannian over a finite field. This leads to a $q$-deformed version of Pascal’s triangle. Then, if we categorify the recurrence relation defining the $q$-binomial coefficients, we’ll understand the Bruhat cells for Grassmannians over arbitrary fields!

• Lecture 6 (Oct. 11) - Proof that the cardinality of $\mathrm{Gr}(n,j)$ over $\mathbb{F}_q$ is $\binom{n}{j}_q$. The $q$-deformed version of Pascal’s triangle. Bruhat cells for the Grassmanian. How to count the total number of Bruhat cells, which is just the ordinary binomial coefficient $\binom{n}{j}$, and how to count the number of cells of any given dimension.
Posted at November 7, 2016 1:00 AM UTC

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### Re: Linear Algebraic Groups (Part 6)

In $q$-Pascal’s lemma, I think equation (1) needs fixing.

Posted by: David Roberts on November 10, 2016 10:08 AM | Permalink | Reply to this

### Re: Linear Algebraic Groups (Part 6)

Toward the bottom of page 3, you want to say

$\frac{|k^{n-1}|}{|L \cap k^{n-1}|} = \frac{q^{n-1}}{q^{j-1}} = q^{n-j}$

or something similar.

Posted by: Todd Trimble on November 10, 2016 12:36 PM | Permalink | Reply to this

### Re: Linear Algebraic Groups (Part 6)

Thanks, David and Todd! John Simanyi has done a stellar job of producing good LaTeXed notes at a fast rate… but clearly there are typos to be found and fixed.

Fixed.

Posted by: John Baez on November 11, 2016 3:58 PM | Permalink | Reply to this

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