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The Saturation Conjecture (proved by Knutson-Tao) asserts that $c_{n\mu,n\nu}^{n\lambda}\neq 0\Rightarrow c_{\mu,\nu}^{\lambda} \neq 0$, where $c$ denotes a Littlewood-Richardson coefficient and $n$ is a positive integer. Let $\mathfrak{o}$ be a (commutative) discrete valuation ring with a finite residue field $k=\mathfrak{o}/\mathfrak{p}$. Let $M$ be a finite $\mathfrak{o}$-module. (For instance, we can take $\mathfrak{o}$ to be the $p$-adic integers $\mathbb{Z}_p$, in which case $M$ is a finite abelian $p$-group.) Thus there is a unique partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_r)$ such that $$ M \cong \bigoplus_{i=1}^r (\mathfrak{o}/\mathfrak{p}^{\lambda_i}). $$ We call $\lambda$ the type of $M$. A fundamental result of Philip Hall is that if $M$ has type $\lambda$, then there exists a submodule $N$ of type $\mu$ and cotype $\nu$ (i.e., $M/N$ has type $\nu$) if and only if $c^{\lambda}_{\mu\nu}\neq 0$. See Macdonald, Symmetric Functions and Hall Polynomials, Chapter II, (4.3).

Now suppose that $M$ has type $n\lambda$, and $N$ has type $n\mu$ and cotype $n\nu$. Is there is a submodule $L$ of $M$ of type $\lambda$, such that $L\cap N$ has type $\mu$ and cotype (with respect to $L$) $\nu$?

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  • $\begingroup$ Is there a combinatorial interpretation for some special case (similar to Yamanouchi tableaux for the LR-coeffs)? The saturation conjecture implies the analogous statement for Kosktka and skew Kosktka coefficients, so is there a "Kostka"-analogue of this question, which should be easier to prove? $\endgroup$ Jul 26, 2015 at 19:25
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    $\begingroup$ A note on terminology (and then I will think about this interesting question). I usually understood the Hall algebra to refer to the ring whose elements were formal sums of isomorphism classes of $\mathfrak{o}$-modules, and where multipication was given by counting submodules of given type and co-type. See, for example, arxiv.org/abs/math/0611617 . Also, as I imagine you know, Derksen and Weyman gave a far reaching generalization of the saturation conjecture in terms of this sort of Hall algebra -- see ams.org/journals/jams/2000-13-03/S0894-0347-00-00331-3 $\endgroup$ Jul 26, 2015 at 20:06
  • $\begingroup$ @DavidSpeyer You are right about the definition of Hall algebra. I have fixed this. $\endgroup$ Jul 26, 2015 at 22:52

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