3
$\begingroup$

I'm trying to understand the Hodge-Index Theorem at the moment. What does it say explicitly for the case of $\mathbb{CP}^2$?

$\endgroup$
4
$\begingroup$

That the inner product on $H^2(\mathbb{CP}^2)$ has signature $(1,0)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ .... and for the complex Grassamnnian Gr(n,k)? $\endgroup$ – John McCarthy Nov 16 '09 at 18:40
  • $\begingroup$ 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. $\endgroup$ – David E Speyer Nov 16 '09 at 19:09
  • $\begingroup$ @John: it might be a good idea to post another question -- more people will be able to answer that way. $\endgroup$ – Ilya Nikokoshev Nov 16 '09 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.