Let $\lambda$ and $\mu$ be two Young diagrams, such that $\lambda$ can be obtained from $\mu$ by extending one single column with additional $b$ boxes. Let $\Sigma^\lambda U$ and $\Sigma^\mu U$ denote the corresponding Schur functors (that we consider as representations of $GL(U)$) and let $a
\geq 0$ be a integer number.


By Pieri's rule $\Sigma^\lambda U$ can be considered as an irreducible subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U$. Thus, one may think of $\Sigma^\lambda U\otimes\Lambda^a U$ as a subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$.


At the very same time, $\Lambda^{b+a}U$ is naturally an irreducible subrepresentation of $\Lambda^b U\otimes\Lambda^a U$. Thus, one may consider $\Sigma^\mu U\otimes\Lambda^{b+a} U$ as a subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$.


I'm interested in computing the intersection of these two subrepresentations.




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Let us decompose $\Sigma^\mu U\otimes\Lambda^{b+a} U = \oplus\_{\nu\in P} \Sigma^\nu U$ into irreducibles. Here $\nu$ is running over the set of Young diagrams such that $\nu/\mu$ consists of $a+b$ boxes and is of width $1$ (every row of $\mu$ can be extended with at most one box).


Define the submodule $W=\oplus\_{\nu\in Q}\Sigma^\nu U$ consisting of those $\nu\in P$, such that $\nu\supset\lambda$ (recall that $\lambda$ can be obtained from $\mu$ by extending $b$ consecutive rows with a single box). It's easy to see that every such $\nu\in Q$ appears in the decomposition of $\Sigma^\lambda U\otimes\Lambda^a U$ into irreducibles and these are the only ones that can. Thus, the intersection should be contained in $W$.


**Conjecture:** the intersection coincides with $W$.


The main problem is the following one: despite every irreducible factor $\Sigma^\nu U$ in $W$ being distinct and appearing in both $\Sigma^\lambda U\otimes\Lambda^a U$ and $\Sigma^\mu U\otimes\Lambda^{b+a} U$ with multiplicity one, most of the time its multiplicity in the ambient representation $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$ is quite big. Thus, one should somehow use the actual form of Pieri's embedding as counting multiplicities is not enough.



Iterated Pieri's rule, Schur functors and intersection of subrepresentations
It turned out to be a rather non-trivial statement and follows from an underknown work of Olver. An accessible reference would be *Pieri resolutions for classical groups* by Sam and Weyman (<http://arxiv.org/abs/0907.4505>).


