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Let $G_2(\mathbb{C}^{n+1})$ be the complex grassmannian.

Then the cohomology ring $H^*(G_2(\mathbb{C}^{n+1});\mathbb{C})=\mathbb{C}[c_1,c_2]/(f_n,f_{n+1})$, where $f_k=\sum_{i=0}^{[k/2]}(-1)^{k-i}{{k-i}\choose i}c_1^{k-2i}c_2^i$.

I want to compute the cup-length of the first Chern class $c_1$, i.e., the smallest positive integer $l$ such that $c_1^l=0$. (For the real case, I found a paper Cup products in Grassmannians, R.E. Stong, Topology and its Applications, Volume 13, Issue 1, 1982, Pages 103–113).

By considering the dimension, I have $l\leq 2n-1$. Tested by computer, I obtain $l=2n-1$ for $n\leq 30$.

Are there any references or method to prove $l=2n-1$? This question is equivalent to the following question: a problem about ideals of polynomial rings. Thanks!

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    $\begingroup$ $G_2(\mathbb C^{n+1})$ is a real symplectic manifold of complex dimension $d=2(n+1-2)$. Its $2$-form $\omega$ is a representative of $c_1$, isn't it? If I'm not mistaken in that, you get as an immediate consequence that $c_1^{2n-2} = [\omega^{\wedge d}]\ne0$, hence $l\geqslant 2n-1$. $\endgroup$
    – Francois Ziegler
    Commented May 6, 2015 at 3:22
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    $\begingroup$ Any Grassmannian is a Fano variety, that is, $c_1$ is an ample class. This implies that your "cup-length" is the dimension plus one, in your case $2n-1$. $\endgroup$
    – abx
    Commented May 6, 2015 at 4:25
  • $\begingroup$ Dear Prof. Francois Ziegler, could you give any references? Thanks! $\endgroup$
    – Shiquan Ren
    Commented May 6, 2015 at 5:49
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    $\begingroup$ @RSQ. For the theorem abx states, you can look up "Hard Lefschetz Theorem" in Griffiths and Harris, among others. There are, of course, simpler proofs for the Grassmannian. I am sure that the description of the complex Grassmannian in Griffiths and Harris includes your result. In particular, $c_1^{2n-2}$ is computed as a Catalan number. $\endgroup$
    – Jason Starr
    Commented May 6, 2015 at 8:09
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    $\begingroup$ @RSQ: For the (indeed, true) relation $c_1 = [\omega]$, see e.g. S. S. Chern, Complex manifolds without potential theory, p. 82. $\endgroup$
    – Francois Ziegler
    Commented May 11, 2015 at 6:20

1 Answer 1

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The mere fact that the Grassmannian $G=G_k(\mathbb C^{n+k})$ can be embedded as a variety in some projective space $\mathbb P^N$(for example by the Pluecker embedding) implies that there is a class $c\in H^2(G)$ such that the top power $c^{kn}$ is nonzero, namely the restriction of a generator of $H^2(\mathbb P^N)$. In the case of the Grassmannian, $H^2$ is one-dimensional so every nonzero class $c\in H^2(G)$ will have this property as soon as one such class does.

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