Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Let me try to rephrase. You have a subvariety $V$ of the Grassmannian $G_{k,n}$, and you know $[V][W]\in H^*(G_{k,n})$ whenever $W$ is such that $[V][W]$ is a multiple of the class of a point (i.e. the top dimensional class). You want to know $[V]\in H^*(G_{k,n})$ (presumably either in terms of Schubert classes or in terms of the (anti)tautological line bundles). Correct? –  Alexander Woo Dec 27 '11 at 21:12
    
Yes, and also $W$ has to be enumerative. –  Ruke Dec 27 '11 at 23:24
    
I'm certain this can be done in theory, but I want to know if there is an easy/fast way to do this? Once I found the class of $V$, I would intersect with other classes to get numbers. It does not matter what basis one uses, as long as it is fast/easy. –  Ruke Dec 27 '11 at 23:28

1 Answer 1

up vote 2 down vote accepted

For simplicity, I am supposing your family is a pure-dimensional (though not necessarily irreducible) subvariety $V$ in the Grassmannian.

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.)

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 $\lambda^*\neq\mu$. Here $\lambda^*$ 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 $\lambda^*$. In notation, $\lambda^*_i = n-k+1-\lambda_{k+1-i}$.

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.

There cannot be any easier method because it takes $d$ pieces of information to determine an element in a vector space of dimension $d$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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