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 -deformed version of Pascal’s triangle. Then, if we categorify the recurrence relation defining the -binomial coefficients, we’ll understand the Bruhat cells for Grassmannians over arbitrary fields!
- Lecture 6 (Oct. 11) - Proof that the cardinality of over is . The -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 , and how to count the number of cells of any given dimension.
Re: Linear Algebraic Groups (Part 6)
In -Pascal’s lemma, I think equation (1) needs fixing.