Is there "Schur-Weyl duality" for infinite dimensional unitary group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:56:38Z http://mathoverflow.net/feeds/question/90006 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90006/is-there-schur-weyl-duality-for-infinite-dimensional-unitary-group Is there "Schur-Weyl duality" for infinite dimensional unitary group? Michal Oszmaniec 2012-03-02T01:17:40Z 2012-03-07T21:18:47Z <p>To what extent does the relation between the diagonal representation of $SU(n)$ in $(\mathbb{C}^n)^{\otimes k}$ and representations of the symmetric group $S_k$ remain valid when instead of the group $SU(n)$ we take the general unitary group $U(\mathcal{H})$ ($\mathcal{H}$ is some separable Hilbert space) and its diagonal representation in $\mathcal{H}^{\otimes k}$? In particular is it true that irreducible representations of of $U(\mathcal{H})$ in $\mathcal{H}^{\otimes k}$ correspond to young diagrams of $S_k$ just like in the finite dimensional case? Is it possible to get any irreducible representation of $U(\mathcal{H})$ by considering decompositions of $\mathcal{H}^{\otimes k}$ for sufficiently big $k$?</p> http://mathoverflow.net/questions/90006/is-there-schur-weyl-duality-for-infinite-dimensional-unitary-group/90192#90192 Answer by Anatoly Kochubei for Is there "Schur-Weyl duality" for infinite dimensional unitary group? Anatoly Kochubei 2012-03-04T10:33:01Z 2012-03-04T14:18:12Z <p>As far as I know, there is no representation theory for the group $U(H)$, it is in a way "too big". However there is a rich representation theory for its subgroups admitting some kind of approximation by finite dimensional groups, in particular for the inductive limit group $U(\infty )$. For the latter, I quote the paper <a href="http://annals.math.princeton.edu/wp-content/uploads/annals-v161-n3-p05.pdf" rel="nofollow">http://annals.math.princeton.edu/wp-content/uploads/annals-v161-n3-p05.pdf</a> by Borodin and Olshanski:</p> <p>"It is worth noting that the similarity of theories for the two groups $S(\infty )$ and $U(\infty )$ seems to be a striking phenomenon. In addition, as mentioned above, this can be traced in the geometric construction of the â€˜naturalâ€™ representations and in probabilistic properties of the corresponding point processes. At present we cannot completely explain the nature of this parallelism (it looks quite different from the well-known classical connection between the representations of the groups S(n) and U(N))".</p> <p>Thus, at the present state of this theory, the answer to your question seems negative.</p> <p>EDIT. For the subgroup of unitary operators differing from $I$ by compact ones, the answer is positive. See the papers referred to in Todor Tsankov's answer.</p> http://mathoverflow.net/questions/90006/is-there-schur-weyl-duality-for-infinite-dimensional-unitary-group/90198#90198 Answer by Todor Tsankov for Is there "Schur-Weyl duality" for infinite dimensional unitary group? Todor Tsankov 2012-03-04T13:20:45Z 2012-03-04T16:35:17Z <p>The answer to all of your questions is yes. This is a theorem announced by Kirillov in</p> <p>Kirillov, A. A. Representations of the infinite-dimensional unitary group. <em>Dokl. Akad. Nauk. SSSR</em>, 1973, 212, 288-290</p> <p>and proved by Olshanski in</p> <p>Olshanski, G. I. Unitary representations of the infinite-dimensional classical groups $U(p,\infty)$, $SO_0(p,\infty )$, $Sp(p,\infty)$, and of the corresponding motion groups. <em>Funktsional. Anal. i Prilozhen.</em>, 1978, 12, 32-44, 96</p> <p><strong>Edit</strong>: This answer applies to continuous unitary representations of $U(\mathcal{H})$, where the latter group is equipped with the strong operator topology (which is the usual topology on this group). However, if you are only interested in representations on a separable Hilbert space, the continuity assumption can be dropped, as is shown in arXiv:1109.1200.</p> http://mathoverflow.net/questions/90006/is-there-schur-weyl-duality-for-infinite-dimensional-unitary-group/90505#90505 Answer by Michal Oszmaniec for Is there "Schur-Weyl duality" for infinite dimensional unitary group? Michal Oszmaniec 2012-03-07T21:18:47Z 2012-03-07T21:18:47Z <p>Thank you Anatoly and Tador for your answers - they pointed to the very useful references. Despite to what Anatoly Claims in his answer there is a very rich representation theory of $U(\mathcal{H})$ (not only for separable Hilbert spaces). I am not an expert in this field but as far as I can tell the paper by DOUG PICKRELL: <a href="http://www.ams.org/journals/proc/1988-102-02/S0002-9939-1988-0921009-X/S0002-9939-1988-0921009-X.pdf" rel="nofollow">http://www.ams.org/journals/proc/1988-102-02/S0002-9939-1988-0921009-X/S0002-9939-1988-0921009-X.pdf</a> answers nearly all my questions. It stands that all separable representations (ie. representations in separable Hilbert space) of $U(\mathcal{H})$ ($\mathcal{H}$ - separable) decompose onto irreducible components that correspond to irreducibles that are taken "naivly" from some $\mathcal{H}^{\otimes k}$. It seams that this theory was known for $U(\infty)$ but in this paper author handled the disturbing "Galkin Algebra". I only wonder what happens to the determinant representation of $U(n)$ if we go to $\infty$ limit?</p>