Is there "Schur-Weyl duality" for infinite dimensional unitary group? - MathOverflow

Is there "Schur-Weyl duality" for infinite dimensional unitary group?

2012-03-02T01:17:40Z

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 $k$?

Answer by Anatoly Kochubei for Is there "Schur-Weyl duality" for infinite dimensional unitary group?

2012-03-04T10:33:01Z

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 http://annals.math.princeton.edu/wp-content/uploads/annals-v161-n3-p05.pdf by Borodin and Olshanski:

"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 ‘natural’ 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))".

Thus, at the present state of this theory, the answer to your question seems negative.

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.

Answer by Todor Tsankov for Is there "Schur-Weyl duality" for infinite dimensional unitary group?

2012-03-04T13:20:45Z

The answer to all of your questions is yes. This is a theorem announced by Kirillov in

Kirillov, A. A. Representations of the infinite-dimensional unitary group. Dokl. Akad. Nauk. SSSR, 1973, 212, 288-290

and proved by Olshanski in

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. Funktsional. Anal. i Prilozhen., 1978, 12, 32-44, 96

Edit: 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.

Answer by Michal Oszmaniec for Is there "Schur-Weyl duality" for infinite dimensional unitary group?

2012-03-07T21:18:47Z

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: http://www.ams.org/journals/proc/1988-102-02/S0002-9939-1988-0921009-X/S0002-9939-1988-0921009-X.pdf 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?