[Math] Schubert calculus, as lowbrow as possible

ag.algebraic-geometrymathematics-educationschubert-calculus

Starting in a week I'm going to be an instructor at a summer program for exceptionally mathematically talented high school students, and I'm going to be teaching a class on Schubert calculus. The students will definitely know linear algebra, and many (but not all) of them will know what a ring is. They will definitely all have enough exposure to rigorous mathematics to know how to recognize a rigorous proof when they see it. They will also have seen quite a bit about the theory of symmetric functions, including Schur functions and the Pieri rule.

I want to present the basics of Schubert calculus in as believable a way as possible. I'll have maybe four or five days, one hour per day, to get to, say, the Pieri rule for Schubert cycles, or to any other way to connect Schubert calculus to symmetric functions.

I definitely plan to talk about the Grassmannian and the Plücker embedding and to convince them that what we're looking for is "just" the number of points on the Grassmannian (and therefore in projective space) that satisfy some polynomial equations. I'd like to talk about the product on the cohomology ring of the Grassmannian, but probably without calling it that, and with as little hand-waving as possible.

So my question is:

What is the best way to convince this audience that multiplying Schubert cycles is a sensible thing to do, that it counts solutions to Schubert problems, and that it formally looks like multiplying Schur functions?

I know it probably won't be possible to be rigorous about everything, but I'd like the presentation to be as clean and convincing as possible because otherwise I think it will be very unsatisfying.


EDIT: Some of the responses have inspired me to be more specific. I'll definitely be able to show that if you intersect two Schubert varieties of complementary dimension with respect to opposite flags, you get one point if they're dual and no points otherwise. Brushing under the rug the question of the multiplicity of that point, this means I could show the Pieri rule by counting the number of points in the intersection of three Schubert varieties w.r.t. flags in general position, one of which is a "special" Schubert corresponding to a partition with one row, provided I could convince them that:

  • There is some notion of "generic" such that, given finitely many subvarieties of the Grassmannian, I can take generic $GL_n$ translates of each of them and intersect them, and there's some meaningful equivalence relation such that the thing I get is "equivalent" provided the translates were "generic" enough.
  • Schuberts span the cohomology ring. Happy to just state this without proof if I can find a way to.

Clearly I have to black-box something; I'm not going to develop the entire modern theory of intersection cohomology for high-schoolers in two hours. The question is how to make it not seem like I'm pushing 100% of the motivation under the rug.

Best Answer

You could try Kleiman and Laksov's paper Schubert Calculus, The American Mathematical Monthly , Vol. 79, No. 10 (Dec., 1972), pp. 1061-1082. This is an excellent introduction to the subject, with several examples from enumerative geometry, and it also manages to avoid too much technical details (it's Monthly article).

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