From the standard results on Springer fibers of type A, we know that given a Springer fiber, say $\mathcal{B}_\lambda,$ its irreducible components are all equidimensional and parametrized by standard Young tableaux of the Young diagram associated to partition $\lambda$.
Now the question: Given two standard Young Tableaux $T_1^{\lambda}$ and $T_2^{\lambda},$ of Young diagram $\lambda,$ is there a combinatorial way to find out whether the corresponding components, say $\mathcal{B}^{T_1}_\lambda$ and $\mathcal{B}^{T_2}_\lambda,$ have nonempty intersection?
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2$\begingroup$ I think that one way to approach your question is using the torus fixed points in $ \mathcal B_\lambda $.. The intersection of any two components is torusinvariant and thus if nonempty must contain a fixed point. We have a bijection between these torus fixed points and the set of rowstrict (but not necc. column strict) tableaux of shape $ \lambda $. If you can figure how for which rowstrict tableaux $ U$, we have $ x_U \in \mathcal B_\lambda^{T} $, then you can answer the question. $\endgroup$– Joel KamnitzerCommented Jul 2, 2019 at 13:22

$\begingroup$ Thank you @Joel. Now that I have finally understood your answer, I would ask the thing that you have mentioned: Is there a combinatorial algorithm that, given a standard tableau $T$, tells you for which rowstrict tableau $U$ (except for ones whose standardization is T) $x_U \in \mathcal{B}_{\lambda}^T$ ? $\endgroup$– FilipCommented Apr 9, 2020 at 17:02
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