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One variant of Fulton's K-saturation conjecture is as follows:

$K_{\lambda/\mu,w} > 0 \Leftrightarrow K_{n\lambda/n\mu,n w} > 0$ for any integer $n>0.$

Here $K_{\lambda/\mu,w}$ denotes the Kostka numbers (number of skew SSYT of shape $\lambda/\mu$ and weight $w.$

This has been proved in various ways, (Knutson, Tao), so it is no longe a conjecture, but to me the proofs are quite involved. The proof shows the similar statement for Littlewood-Richardson coefficients, using K-hives and the above follows as a corollary.

Question: Is there an elementary proof of the above statement? Could one expect a short proof of this?

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  • $\begingroup$ To my mind, the elementary reason that saturation might be true for ordinary (not skew) Kostka numbers is that the ring $\oplus_n (V_{n\lambda})[w$ weight space$]$ might be generated in degree $1$. Moreover, that would explain Fulton's conjecture (that if the degree $1$ piece is $1$-d, then so is every piece), which we also proved. But in general this ring is not generated in degree $1$! I forget Fulton's counterexample in which the degree $2$ part is larger than $sym^2$ of the degree $1$ part, or something like that, but there is one. $\endgroup$
    – Allen Knutson
    Commented May 6, 2013 at 3:52
  • $\begingroup$ Oops, that's supposed to be the $[nw$ weight space$]$. $\endgroup$
    – Allen Knutson
    Commented May 6, 2013 at 3:53
  • $\begingroup$ Correct me if I'm wrong, but for the non-skew version, $K_{\lambda,w}>0$ iff $\lambda \geq_d w$ in dominance order. And I suppose that $\lambda \geq_d w \Leftrightarrow n\lambda \geq_d nw$ is easy to show... $\endgroup$
    – Per Alexandersson
    Commented May 6, 2013 at 8:12
  • $\begingroup$ You're right, I was confusing myself. Anyway perhaps the simplest, or at least best-motivated, proof is that saturation is equivalent to Horn's conjecture (this is a simple, but great, observation of Belkale) and Horn's conjecture has a very natural proof due to Purbhoo and Sottile. $\endgroup$
    – Allen Knutson
    Commented May 6, 2013 at 13:43

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I answer this question here.

The proof is of combinatorial nature and quite short, using quite elementary arguments.

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