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Nov 25, 2015 at 18:34 comment added Terry Tao Actually, the inequalities in the compact operator case are simply the union of the inequalities in the finite dimensional cases: see for instance the paper of Bercovici, Li, and Timotin in ams.org/mathscinet-getitem?mr=2567501 , together with the references therein.
Nov 25, 2015 at 18:25 comment added Terry Tao I'd like to mention that the initial solution to Horn's conjecture also relies heavily on Klyachko's paper "Stable bundles, representation theory and Hermitian operators", Selecta Math. (N.S.) 4 (1998), no. 3, 419–445, in addition to my paper with Knutson. (There have since been subsequent proofs of Horn's conjecture that avoid the difficult GIT machinery of Klyachko, though.)
Nov 25, 2015 at 17:53 answer added user75274 timeline score: 3
Nov 25, 2015 at 15:50 comment added Denis Serre @Zoalroshd. For a reference, see my edits. The polytope his an exponentialcomplexity as $n\rightarrow+\infty$. Hence I doubt that it adapts easily to the case of a Hilbert space.
Nov 25, 2015 at 15:48 history edited Denis Serre CC BY-SA 3.0
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Nov 25, 2015 at 15:32 comment added BigM I'm not expert in this field by no means and have a question rather than an insightful comment. Can you provide a link to Knutson & Tao's paper. I wonder if their answer changes if one replaces matrices with two self-adjoint compact operators on a separable Hilbert space.
Nov 25, 2015 at 14:58 history asked Denis Serre CC BY-SA 3.0