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I am trying at the moment to understand Schubert calculus, and have taken the simple example of the complex projective line ${\mathbb CP}^1$ as a guide. Now in the simplest formulation I know, we have that $$ \Omega^0({\mathbb CP}^1) \simeq \Omega^2({\mathbb CP}^1) \simeq {\cal O}({\mathbb CP}^1), $$ and $$ \Omega^{(0,1)}({\mathbb CP}^1) \simeq L_{-2}, ~~~ \Omega^{(1,0)}({\mathbb CP}^1) \simeq L_{2}, $$ where $L_{-2}$ and $L_{2}$, are the vector bundles corresponding to $-2$ and $2$ in the standard classification of the line bundles over ${\mathbb CP}^{1}$ in terms of ${\mathbb Z}$.

Can anyone give me a concrete presentation of the cohomology groups $H^{p,q}$ in terms of this description? Moreover, what are the Schubert cell generators of the groups?

A concrete presentation of how all this works for ${\mathbb CP}^2$ and ${\mathbb CP}^3$ would also be very welcome.

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For $\mathcal{P}^n$, $H^{qp}=\mathcal{C}$ if $0\le p=q\le n$. Otherwise, you get $0$. Your bundle isomorphisms would have to be in the $C^\infty$ category to makes sense. Probably that's what you meant. – Donu Arapura Sep 21 '10 at 17:21
I can't edit above. Sub $\mathcal{P}$ -> $\mathbb{P}$, $\mathcal{C}$-> $\mathbb{C}$. – Donu Arapura Sep 21 '10 at 17:23
@Donu: Thanks for your help. Why do my isomorphisms need to be in the smooth category. – Abtan Massini Sep 21 '10 at 17:34
Abtan: Ah so that wasn't what you meant. Here a couple of reasons why has to be like that. (1) $\Omega^{(0,1)}$ has no holomorphic structure. (2) when I saw your first isomorphism, it looked wrong to me since there are no holomorphic $2$-forms. But then I realized that it does make sense if $\Omega^2$ is the bundle of $C^\infty$ $2$-forms. The first map would be $f\mapsto fdvol$. – Donu Arapura Sep 21 '10 at 18:03
@Donu: I've just realised what you meant: smooth as opposed to holomorphic (I understood smooth as opposed to polynomial). Yes, of course that's what I mean ($\Omega(0,1)$ isn't called the anti-holomorphic tangnet bundle for nothing!). Really, I'm thinking of all this in terms algebraic geometry (as opposed to differential), that's why I tagged it as such. – Abtan Massini 0 secs ago – Abtan Massini Sep 21 '10 at 20:28

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