Schubert calculus, as lowbrow as possible

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.

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In my summer school for undergrads last summer, I took a completely different approach, notes here: math.cornell.edu/~allenk/schubnotes.pdf I started with symmetric polynomials and that they are a polynomial ring, and raised the question of how far we could go with partly-symmetric polynomials. This motivated divided difference operators, and from there, Schubert polynomials. That led back to Schur polynomials. No geometry whatsoever. But it gave them lots and lots to calculate on their own, as you can see in those notes. – Allen Knutson Jun 22 '13 at 1:34
I am intrigued: at what summer program are you meeting high school students who know about Schur functions? – KConrad Jun 22 '13 at 2:13
Dear Nicolas, I don't have any good suggestions related to your overall plan, but I would make sure that you get as far as giving a believable proof that there are two lines meeting 4 generic skew lines in space. I think this is one of the first results in cohomology-type computations on Grassmanians which isn't immediately obvious (at least, it wasn't to me when I first saw it), and at the same time is immediately comprehensible. So if you can make sure that your students appreciate this question and its answer, and it may help them with the necessary suspension of disbelief that will ... – Emerton Jun 22 '13 at 23:42
People always prove that $\int [X_\box]^4 = 2$ statement by invoking the statement that degree 2 hypersurfaces are ruled. Don't. Instead, start from the special case that the four lines are not generic: $|L_1 \cap L_2| = |L_3 \cap L_4| = 1$. In that case one can name the two lines: the one connecting $L_1 \cap L_2$ to $L_3 \cap L_4$, and the intersection of the planes $\langle L_1,L_2 \rangle$ and $\langle L_3,L_4 \rangle$. In general, of course, one can't name the lines individually because a Galois group exchanges them. – Allen Knutson Jun 23 '13 at 2:01
Two comments here. 1) The primary advantages of cohomology over algebraic intersection theory are that it's (somewhat) easier to develop rigorously and that in many cases, students will already have seen cohomology, so the theory is already given to you. Since neither of these seems applicable to your case, you might as well go with algebraic cycles rather than cohomology for your setting; the definitions are actually a bit more intuitive than those of cohomology, as long as you black-box the moving lemma. 2) Since the Grassmannian is homogeneous, you can use Kleiman's Theorem for genericity. – Charles Staats Jun 23 '13 at 19:42