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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 non-empty intersection?

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    $\begingroup$ 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 torus-invariant and thus if non-empty must contain a fixed point. We have a bijection between these torus fixed points and the set of row-strict (but not necc. column strict) tableaux of shape $ \lambda $. If you can figure how for which row-strict tableaux $ U$, we have $ x_U \in \mathcal B_\lambda^{T} $, then you can answer the question. $\endgroup$ – Joel Kamnitzer Jul 2 '19 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 row-strict tableau $U$ (except for ones whose standardization is T) $x_U \in \mathcal{B}_{\lambda}^T$ ? $\endgroup$ – Filip92 Apr 9 at 17:02

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