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.

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.


1 Answer 1


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).


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