I'm trying to understand the Hodge-Index Theorem at the moment. What does it say explicitly for the case of $\mathbb{CP}^2$?
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That the inner product on $H^2(\mathbb{CP}^2)$ has signature $(1,0)$.
answered Nov 16, 2009 at 18:36
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$\begingroup$ .... and for the complex Grassamnnian Gr(n,k)? $\endgroup$ Commented Nov 16, 2009 at 18:40
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$\begingroup$ Don't have time to do G(k,n) in general. Here's a small example: H^4(G(2,4)) is spanned by two Schubert classes: X:=[20] and Y:=[11]. The pairing on them has matrix ((0,1),(1,0)). Taking omega to be the chern class of the standard Plucker embedding, omega^2 H^0 is spanned by X+Y. Also, omega H^1 is spanned by X+Y. The statement should be that the pairing is positive definite on omega^2 H^0 and negative definite on the orthogonal complement of omega H^1 (namely, X-Y). I think I might have gotten some of the signs wrong in my previous write up of this; I'll check later. $\endgroup$ Commented Nov 16, 2009 at 19:09
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$\begingroup$ @John: it might be a good idea to post another question -- more people will be able to answer that way. $\endgroup$ Commented Nov 16, 2009 at 20:28