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I've the following problem:

I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be $$1-\sigma_1+\sigma_{1,1}$$ where $\sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $\Sigma_{i_1,i_2}=\{\Lambda\in G\mid\dim(V_{2-i_j+j}\cap\Lambda)\geq j\;\forall j\}$ (here $\{V_j\}\subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?

Thank you!

Matt Samuel
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User3773
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    The natural map $CH(G)\rightarrow H^*(G,\mathbb{Z})$ is a ring isomorphism. –  Jul 15 '14 at 14:28
  • I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-\sigma_1$? And that of $\sigma_{1,1}$? I never did such a calculation.. – User3773 Jul 17 '14 at 09:25

1 Answers1

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The cohomology ring of $G(k,n)$ is isomorphic to $$\frac{\mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$ where $T$ and $Q$ are respectively the tautological and the quotient bundle. The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:

  • $c_i(T) = (−1)^i\sigma_{1,...,1}$,
  • $c_i(Q) = \sigma_i$.
Puzzled
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    In the displayed formula, you should have a $\mathbb{Z}$ where you have a "coefficient field" $k$. – Jason Starr Jul 15 '14 at 17:15
  • You are right. I corrected it. – Puzzled Jul 15 '14 at 19:16
  • @F_L I have seen the formula for cohomology ring of Grassmannian in Bott & Tu‘s book, but I think the proof is not correct or I missing something. Can you give me some other reference for the formula? Thanks. – MiGang Apr 24 '22 at 23:50