For Grassmannians, the Schubert cells can be indexed by certain Young Tableaux, whose partition determines the dimensions of intersections of the chosen subspace with the standard complete flag. For example, up to change in convention, in $Gr(2,4)$, the subspace $V = \langle e\_2,e\_4\rangle$ may belong to $X\_{(1,0)}$, since, if $F\_\bullet$ is the standard complete flag, then $dim(V \cap F\_j) = (0,1,1,2)$, while for a generic 2-dimensional subspace, $V\_{gen}$, $dim(V\_{gen} \cap F\_j) = (0,0, 1,2)$. One scheme for associating the cells to partitions is to record the difference (between $V$ and $V\_{gen}$) between the position of the first appearance of 1 in the "intersection dimension vector" as the first element of the partition, and the difference between the position of the first appearance of 2 as the second entry, hence $(1,0)$. (This was explained to me only one time in person, so please let me know if I've got some aspect of this process wrong.) So if I am given the algebraic data (partition), I can determine the geometric data (intersection dimension vector) and vice versa.


For complete flags, the Schubert cells are naturally indexed by $S\_n$, which admits a bijection to pairs of same-shape standard Young tableaux via Robinson-Schensted correspondence. Is there a geometric understanding of these SYT's in a way analogous to the above? i.e. For $Fl(m)$, we have a $2\times m-1$ ($m-1$ if we ignore the final column, which will always just be $\begin{pmatrix} m \\ m \end{pmatrix}$) matrix of intersection dimensions, $dim(E\_i\cap F\_j)$ for $i,j \in \{1,2,\dots,m-1\}$. If I know the intersection dimension matrix of a certain flag, can I determine the pair of SYT of its Schubert cell or vice versa?



geometric meaning to pairs of SYT indexing for the basis of cohomology ring of full flag variety
Since, surprisingly, there are still no answers or even comments, let me note that the answer to the last question is well known to be "yes": the Schubert cell containing a flag $(E\_1,\dots,E\_m)$ is indeed determined by the values $\dim(E\_i\cap F\_j)$.


Specifically, if $\sigma$ is the corresponding permutation, then $\dim(E\_i\cap F\_j)$ is the number of values no greater than $j$ among $\sigma(1),\dots,\sigma(i)$. It is evident that this data determines the permutation uniquely, so the boring method here is to determine the permutation and then apply the insertion algorithm. The question whether there is a substantially more direct relation between the dimensions and the pair of tableaux is curious but I wouldn't count on it.


