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 $k$?
$\begingroup$
$\endgroup$
7
-
3$\begingroup$ 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 ... $\endgroup$– Theo Johnson-FreydCommented Mar 2, 2012 at 3:21
-
3$\begingroup$ ... $t$ 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 $H$. $\endgroup$– Theo Johnson-FreydCommented Mar 2, 2012 at 3:25
-
$\begingroup$ 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. $\endgroup$– Michał OszmaniecCommented Mar 2, 2012 at 12:13
-
$\begingroup$ +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. $\endgroup$– Alexander ChervovCommented Mar 4, 2012 at 8:09
-
1$\begingroup$ Of course, I should also mention that everything I know about GL(t) I learned from @Noah Snyder. $\endgroup$– Theo Johnson-FreydCommented Mar 16, 2012 at 4:06
|
Show 2 more comments
3 Answers
$\begingroup$
$\endgroup$
7
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.
-
2$\begingroup$ As I wrote in my answer, the paper you cite are not about the whole unitary group. As for the results, see the quotation from Olhanski himself in my answer. $\endgroup$ Commented Mar 4, 2012 at 13:34
-
1$\begingroup$ Meanwhile, indeed, for the subgroup considered by Kirillov, there is a relation of the requested kind. $\endgroup$ Commented Mar 4, 2012 at 13:47
-
$\begingroup$ I am not sure I understand your comment. Every representation of the infinite-dimensional unitary group restricts to a representation of the direct limit of finite-dimenssional unitary groups and conversely, each of the representations they consider extends to one of the full unitary group. $\endgroup$ Commented Mar 4, 2012 at 16:19
-
1$\begingroup$ @Todor But restriction and extension does not mean equvivalence of irreps. I mean when you restrict irrep you can get reducible, as well as, when you extend non-equiv. irreps. you can get equiv reps. As far as I remember it is always emphasized that the two groups: 1) U(H) and 2) limit U(n) has very different rep. theory. – Alexander Chervov $\endgroup$ Commented Mar 4, 2012 at 18:34
-
3$\begingroup$ I should have also mentioned that the direct limit of the finite-dimensional groups is dense in $U(\mathcal{H})$. $\endgroup$ Commented Mar 5, 2012 at 9:09
$\begingroup$
$\endgroup$
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.
answered Mar 4, 2012 at 10:33
$\begingroup$
$\endgroup$
2
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?
-
1$\begingroup$ @Michal Thank you for the reference, which is important also for my understanding of the subject. $\endgroup$ Commented Mar 8, 2012 at 6:44
-
$\begingroup$ It seems like this should be the accepted answer? Since there was some back-and-forth debate between the other two, and neither got as far as answering the question as this Pickrell paper does. Or have I misunderstood something? $\endgroup$ Commented Aug 25, 2020 at 19:40