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replaced the link to the arXiv front end; see https://meta.mathoverflow.net/questions/5124/is-it-time-to-replace-links-to-the-ucdavis-arxiv-frontend
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Martin Sleziak
Martin Sleziak
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If $\lambda,\mu,\nu$ are considered as vectors in $\mathbb Z^n$, then the set of triples forms a convex cone, given by a finite (for each $n$) list of inequalities. (The corresponding statement does not hold for general Lie groups.) The unique minimal list is knownknown. All known descriptions of it involve combinatorics, and I daresay that will continue.

If you want just necessary or sufficient, you can do better, e.g. $\nu = \lambda + \mu$ is a sufficient condition, and any single one of the inequalities (e.g. $\lambda_1 + \mu_1 \geq \nu_1$) is necessary.

If $\lambda,\mu,\nu$ are considered as vectors in $\mathbb Z^n$, then the set of triples forms a convex cone, given by a finite (for each $n$) list of inequalities. (The corresponding statement does not hold for general Lie groups.) The unique minimal list is known. All known descriptions of it involve combinatorics, and I daresay that will continue.

If you want just necessary or sufficient, you can do better, e.g. $\nu = \lambda + \mu$ is a sufficient condition, and any single one of the inequalities (e.g. $\lambda_1 + \mu_1 \geq \nu_1$) is necessary.

If $\lambda,\mu,\nu$ are considered as vectors in $\mathbb Z^n$, then the set of triples forms a convex cone, given by a finite (for each $n$) list of inequalities. (The corresponding statement does not hold for general Lie groups.) The unique minimal list is known. All known descriptions of it involve combinatorics, and I daresay that will continue.

If you want just necessary or sufficient, you can do better, e.g. $\nu = \lambda + \mu$ is a sufficient condition, and any single one of the inequalities (e.g. $\lambda_1 + \mu_1 \geq \nu_1$) is necessary.

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Allen Knutson
Allen Knutson
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If $\lambda,\mu,\nu$ are considered as vectors in $\mathbb Z^n$, then the set of triples forms a convex cone, given by a finite (for each $n$) list of inequalities. (The corresponding statement does not hold for general Lie groups.) The unique minimal list is known. All known descriptions of it involve combinatorics, and I daresay that will continue.

If you want just necessary or sufficient, you can do better, e.g. $\nu = \lambda + \mu$ is a sufficient condition, and any single one of the inequalities (e.g. $\lambda_1 + \mu_1 \geq \nu_1$) is necessary.