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Jul 16, 2013 at 7:45 vote accept Per Alexandersson
Jul 16, 2013 at 7:45 answer added Per Alexandersson timeline score: 1
May 6, 2013 at 13:43 comment added Allen Knutson 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.
May 6, 2013 at 8:12 comment added Per Alexandersson 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...
May 6, 2013 at 3:53 comment added Allen Knutson Oops, that's supposed to be the $[nw$ weight space$]$.
May 6, 2013 at 3:52 comment added Allen Knutson 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.
May 5, 2013 at 16:10 history asked Per Alexandersson CC BY-SA 3.0