# Schubert calculus

Let $$X = Gr(2,4)$$ the complex Grassmannian of $$2$$-planes in $$V = \Bbb C^4$$ and $$S$$ the tautological bundle, $$Q$$ the quotient bundle. The cohomology ring is generated by $$c_1(S), c_2(S)$$ with relations $$c(S)c(Q) = 1$$, coming from the short exact sequence of vector bundles $$0 \to S \to V \to Q \to 0$$.

One should get $$c_1(S)^4 = 1$$ purely from these relations but I'm not able to do so. Can someone explain how to do it ?

• If only to remember the two lines meeting four given lines in 3-space: In schubert in maple: grass(2,4,c) integral(c1^4) In Schubert2 in Macaulay2 see help integral G = flagBundle {2,2} dim G A = intersectionRing G f = (chern_1 OO_G(1))^4 integral f Sep 30, 2020 at 13:22

First of all, $$c_1(S)^4 = 2$$, not 1. The computation itself is quite easy. Let me denote $$a_i := c_i(S)$$, $$b_i = c_i(Q)$$. Then the relations are $$a_1 + b_1 = a_2 + a_1b_1 + b_2 = a_1b_2 + a_2b_1 = a_2b_2 = 0.$$ The first gives $$b_1 = -a_1$$, the second gives $$b_2 = a_1^2 - a_2$$, and the last two give $$a_1^3 = 2a_1a_2, \qquad a_1^2a_2 = a_2^2.$$ A combination of the last two equalities gives $$a_1a_2^2 = a_1(a_1^2a_2) = (a_1^3)a_2 = 2a_1a_2^2,$$ hence $$a_1a_2^2 = 0$$. This means that the cohomology ring is spanned over $$\mathbb{Z}$$ by $$1$$, $$a_1$$, $$a_2$$, $$a_1a_2$$, $$a_2^2$$, and that $$a_1^4 = 2a_2^2$$ which translates into the required equality $$c_1(S)^4 = 2$$.
• sorry, I mean 2 and not 1. How did you get $a_2^2 = 1$ ? It's easy if you describe geometrically $a_2$, but I feel confused because I saw written somewhere that $H^*(X) = \Bbb Z[c_1,c_2]/(c(S)c(Q) = 1)$ but I can't get $a_2^2 = 1$ with the relation. Sorry, maybe my question is not very interesting and you need to add this relation. Sep 30, 2020 at 15:40
• It doesn't make sense to say $a_2^2 = 1$ --- the ring is graded and the elements $1$ and $a_2^2$ live in different degrees. What makes sense is to say that the element $a_2^2$ generates the top degree component of the ring, and this is clear from the computation above. Sep 30, 2020 at 15:53