Schubert problems to cycle class in Grassmanian - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:42:46Z http://mathoverflow.net/feeds/question/84399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84399/schubert-problems-to-cycle-class-in-grassmanian Schubert problems to cycle class in Grassmanian Ruke 2011-12-27T18:00:15Z 2011-12-29T17:23:47Z <p>Say I have a family of linear spaces, and that I can solve all Schuber problems of that family (that is, how many members of the family pass through a set $S$ of linear spaces, where we consider all possible $S$). How can I go from these solutions to the cycle class of the family in the corresponding Grassmanian. I would appreciate any reference in literature. A computer program is even better.</p> http://mathoverflow.net/questions/84399/schubert-problems-to-cycle-class-in-grassmanian/84530#84530 Answer by Alexander Woo for Schubert problems to cycle class in Grassmanian Alexander Woo 2011-12-29T17:23:47Z 2011-12-29T17:23:47Z <p>For simplicity, I am supposing your family is a pure-dimensional (though not necessarily irreducible) subvariety $V$ in the Grassmannian.</p> <p>As you probably know, given a fixed $i$, the classes $[X_\lambda]$ of Schubert subvarieties $X_\lambda$, where $\lambda$ is a partition which has $i$ boxes and fits inside a $k \times n-k$ rectangle, form a basis of $H^{2i}(G_{k,n})$. (The Chow ring and cohomology ring are the same for Grassmannians over $\mathbb{C}$.) (I am assuming a particular indexing convention for Schubert varieties; with a different indexing convention, $\lambda$ should have $k(n-k)-i$ boxes.)</p> <p>Under the intersection pairing $\langle \cdot, \cdot\rangle$ between $H^{2i}$ and $H^{2[k(n-k)-i]}$, the Schubert bases are dual to each other. To be precise, $\langle[X_\lambda],[X_\mu]\rangle = 1$ if $\lambda^*=\mu$ and $\langle[X_\lambda],[X_\mu]\rangle=0$ if <code>$\lambda^*\neq\mu$</code>. Here <code>$\lambda^*$</code> is the box-complement to $\lambda$. Take all the squares in the $k\times (n-k)$ rectangle which are not part of $\lambda$, rotate 180 degrees, and you have the partition <code>$\lambda^*$</code>. In notation, <code>$\lambda^*_i = n-k+1-\lambda_{k+1-i}$</code>.</p> <p>This means that the class $[V]$ is given by $$[V]=\sum_{\lambda} \langle [V], [S_{\lambda^*}]\rangle [S_\lambda],$$ where the sum is over all partitions $\lambda$ fitting inside a $k\times n-k$ rectangle with $\mathrm{codim} V$ boxes.</p> <p>There cannot be any easier method because it takes $d$ pieces of information to determine an element in a vector space of dimension $d$.</p>