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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 ?

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  • $\begingroup$ 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 $\endgroup$ Sep 30, 2020 at 13:22

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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 $c_1(S)^4 = 2$.

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  • $\begingroup$ 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. $\endgroup$
    – curious
    Sep 30, 2020 at 15:40
  • $\begingroup$ 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. $\endgroup$
    – Sasha
    Sep 30, 2020 at 15:53
  • $\begingroup$ you are right and I was silly, thank you for the explanations. $\endgroup$
    – curious
    Sep 30, 2020 at 18:49

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