Schubert Cells for the Projective Line - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:30:49Z http://mathoverflow.net/feeds/question/39526 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39526/schubert-cells-for-the-projective-line Schubert Cells for the Projective Line Abtan Massini 2010-09-21T16:46:05Z 2010-09-21T16:46:05Z <p>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 ${\mathbb Z}$. </p> <p>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? </p> <p>A concrete presentation of how all this works for ${\mathbb CP}^2$ and ${\mathbb CP}^3$ would also be very welcome.</p>