Suppose we have a Young diagram $\lambda$ inside an $r \times n$ rectangular grid, i.e. $\lambda \subset [r] \times [n]$. If I were to add just one more box to $\lambda$, obtaining a new partition (Young diagram) $\lambda'$, would $\lambda'$ still lie inside the $r \times n$ rectangular grid? I'm asking this question because I'm trying to show that the complement of Schubert cells $\Omega_\lambda ^o$ inside a Schubert variety $\Omega_\lambda$ is the union (not necessarily disjoint) of all Schubert varieties $\Omega_{\lambda'}$, where $\lambda'$ is obtained from $\lambda$ by adding one box. I think symbolically, what I'm trying to show is that $\Omega_{\lambda}- \Omega_{\lambda}^o= \bigcup_{\lambda' \supset \lambda} \Omega_{\lambda'}$. Anyhow, I will try to prove it myself, but would like some clarification as to whether my intuition is correct.
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1$\begingroup$ This obviously depends on $\lambda$ and, sometimes, on the extra cell. $\endgroup$– Alex DegtyarevMar 1, 2016 at 23:28
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1$\begingroup$ What if $\lambda$ is the entire $r \times n$ grid? Then you cannot add an additional box... $\endgroup$– Per AlexanderssonMar 1, 2016 at 23:28
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$\begingroup$ I'm relatively new to Young tableaux, especially from a combinatorial standpoint. But I think I can prove the statement I'm trying to show by usage of simple definitions of Schubert cells and Schubert varieties, and analysis of jump sequences related to a reference flag. $\endgroup$– LibertronMar 1, 2016 at 23:47
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1$\begingroup$ Your statement is correct once you make it "by adding one box, while staying inside the $r\times n$ rectangle". $\endgroup$– Allen KnutsonMar 2, 2016 at 2:29
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