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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.

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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
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 $d$.

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