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.



Schubert problems to cycle class in Grassmanian
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 $d$.


