Timeline for Is there "Schur-Weyl duality" for infinite dimensional unitary group?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2012 at 4:06 | comment | added | Theo Johnson-Freyd | Of course, I should also mention that everything I know about GL(t) I learned from @Noah Snyder. | |
| Mar 16, 2012 at 4:06 | comment | added | Theo Johnson-Freyd | @Noah: Yes, of course. GL = U, not SU. | |
| Mar 7, 2012 at 21:19 | vote | accept | Michał Oszmaniec | ||
| Mar 7, 2012 at 21:19 | history | bounty ended | Michał Oszmaniec | ||
| Mar 7, 2012 at 21:18 | answer | added | Michał Oszmaniec | timeline score: 2 | |
| Mar 7, 2012 at 3:40 | history | edited | John Pardon | CC BY-SA 3.0 |
fixed grammar in first sentence
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| Mar 4, 2012 at 18:23 | comment | added | Noah Snyder | Theo, certainly it'd be U(t) not SU(t), because you haven't killed a determinant. | |
| Mar 4, 2012 at 13:20 | answer | added | Todor Tsankov | timeline score: 7 | |
| Mar 4, 2012 at 10:33 | answer | added | Anatoly Kochubei | timeline score: 3 | |
| Mar 4, 2012 at 8:09 | comment | added | Alexander Chervov | +1. I doubt duality exists for U(H), usually people work with limit of U(n), n->inf and consider this as an "correct" analog of U(n) for finite n. By the way what are irreps of U(H) ? As a Hilbert space H^k = H , so I am afraid some trouble like that will happen with representations constructed from H^k. | |
| Mar 4, 2012 at 7:29 | history | bounty started | Michał Oszmaniec | ||
| Mar 3, 2012 at 22:09 | history | edited | Michał Oszmaniec |
edited tags
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| Mar 2, 2012 at 12:13 | comment | added | Michał Oszmaniec | Thanks Theo for your comment. Unfortunately I have a very little background in category theory so I cannot understand the details. Yet, it is good to know that there are connections of this kind. As for myself, I got interested in this problem because of physics. For my purposes it would be sufficient if $\mathcal{H}^{\otimes k}$ would decompose "nicely" onto irreducible components just like in the finite dimensional case. | |
| Mar 2, 2012 at 3:25 | comment | added | Theo Johnson-Freyd | ... $t$ is an integer, this category has morphisms all of whose partial traces (after composing with anything) are zero, but that are not themselves zero, and on modding out by such morphisms you recover the category of finite-dimensional representations of GL(|t|). Anyway, the category does still have a Schur-Weyl theorem, in that the $n$th power of the generator decomposes exactly as is given by the Schur functors for $S_n$. So in some sense your question is whether this category can be modeled in terms of part of the representation theory of $U(H)$ $\otimes$-generated by $H$. | |
| Mar 2, 2012 at 3:21 | comment | added | Theo Johnson-Freyd | I think this question is a great one, and I hope you get useful answers. +1. As a category and representation theorist, I would comment on one way to think about the question. Namely, there is a category that I've seen called "GL(t)", although perhaps "SU(t)" is a better name, which is essentially the free category with some part of Schur-Weyl duality. Namely, you take the free symmetric monoidal category on a dualizable generator, and you impose the condition that its dimension is some scalar $t$, and then Karoubi-complete. When $t$ is transcendental, this category is semisimple. When ... | |
| Mar 2, 2012 at 1:17 | history | asked | Michał Oszmaniec | CC BY-SA 3.0 |