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Consider the semisimple compact group $K=SU(N_1)\times SU(N_2) \times \ldots \times SU(N_S)$ acting naturally on $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2 \otimes \ldots \otimes \mathcal{H}_S$, where $\mathcal{H}_i \approx \mathbb{C}^{N_i}$. Take the diagonal action of this group on $\mathcal{H}^{\otimes m}$. $\mathcal{H}^{\otimes m}$ would decompose onto irreducible components corresponding to collection of young diagrams: $(\lambda_1,\lambda_2,\ldots,\lambda_S)$, where $\lambda_i$ is a Young diagram having $m$ entries and no more than $N_i$ rows. I have two questions concerning the action of $K$ on $Sym^m(\mathcal{H})\subset\mathcal{H}^{\otimes m}$:

  1. Is this representation multiplicity free?

  2. Which irreducible representations of $K$ appear in $Sym^m(\mathcal{H})$?

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    $\begingroup$ As far as I understand all multiplicity free representations of reductive groups are classified - so you can check the list - see e.g. section 5 in F. Knop "Some remarks on multiplicity free actions" 1998 algeo.math.uni-erlangen.de/fileadmin/algeo/users/knop/papers/… (he reprocudes early results by Kac, Brion, BensonRatclif) $\endgroup$
    – Alexander Chervov
    Commented Apr 8, 2012 at 10:17
  • $\begingroup$ I essentially gave the same answer as Victor Prosak that we have an equivalence of representation categories for $K_1, K_2$ compact groups $$Rep(K_1 \times K_2) \cong \Rep(K_1) \otimes \Rep(K_2),$$ but I decided to delete it after your objection that I did not understand the question. I also commented that the result would follow from $Sym^k(H) \subset H^{\otimes k}$. So I guess, you real problem focuses reduces how to detect $Sym^k(H) \subset H^{\otimes k}$, is this correct? $\endgroup$
    – Marc Palm
    Commented Apr 13, 2012 at 7:52
  • $\begingroup$ Exactly - the first part (description of $\mathcal{H}^{\otimes k}$) I already knew. I also know how to handle the case $m=2$ when the group that generates the commutant of the action of $K$ in $\mathcal{H}^{\otimes k}$ is abelian (it is simply the m−fold direct product of $S_2$). In this case $sym^m(\mathcal{H})$ is indeed multiplicity free. $\endgroup$
    – Michał Oszmaniec
    Commented Apr 13, 2012 at 8:25

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First, a terminological nitpick: the Schur-Weyl duality deals with the unitary group $U(N)$ (in the compact formulation) or the general linear group $GL(N)$ (in the algebraic group version) acting on $\mathcal{H}^{\otimes m}$, where $\mathcal{H}=\mathbb{C}^N.$ The duality states that the isotypic components under the $U(N)$ action are irreducible $S_m$-modules. Explicitly,

$$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda}\rho_{U(N)}^\lambda\otimes\rho_{S_m}^{\lambda},$$

where $\lambda$ runs over Young diagrams with $m$ boxes and at most $\operatorname{min}(N,m)$ rows. Observe that for $m\geq 3$ and $N\geq 2$ this decomposition is not multiplicity-free as a $U(N)$-module. However, for $N\geq 2$ the restriction to $SU(N)$ does not introduce new multiplicities: for distinct $\lambda,\mu$ as above, the representations $\rho_{SU(N)}^\lambda$ and $\rho_{SU(N)}^\mu$ are non-isomorphic .

Now for the present question. Consider the $K$-equivariant isomorphism $$\mathcal{H}^{\otimes m}\simeq \mathcal{H}_1^{\otimes m} \otimes\ldots\otimes\mathcal{H}_s^{\otimes m}.$$ Then each factor $SU(N_i)$ of $K$ acts on its own $m$th tensor power space $\mathcal{H}_i^{\otimes m}.$ It is a standard fact in representation theory that the irreducible representations of $K=SU(N_1)\times\ldots\times SU(N_s)$ have the form $V_1\otimes\ldots\otimes V_s,$ where $V_i$ is an irreducible representation of $SU(N_i)$ determined uniquely up to isomorphism. Hence the question is reduced to the case $s=1.$ Explicitly, $$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda_1,\ldots,\lambda_s}\rho_{SU(N_1)}^{\lambda_1}\otimes\ldots\otimes \rho_{SU(N_s)}^{\lambda_s}\otimes\rho_{S_m}^{\lambda_1}\otimes\ldots\otimes\rho_{S_m}^{\lambda_s}.$$

Different $\lambda_i$'s with $m$ boxes are independently chosen, each subject to its own restriction on the number of rows. This is not multiplicity-free unless $m\leq 2$ or all $N_i$ are equal to 1.

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  • $\begingroup$ Dear Victor. Thank you for your answer. My statement of the question was not clear enough. Basically I knew already all that you wrote: "Take the diagonal action of this group on $\mathcal{H}^{\otimes m}$. $\mathcal{H}^{\otimes m}$ would decompose onto irreducible components corresponding to collection of young diagrams: $(\lambda_1,\lambda_2,\ldots,\lambda_S)$, where $\lambda_i$ is a Young diagram having $m$ entries and no more than $N_i$ rows" $\endgroup$
    – Michał Oszmaniec
    Commented Apr 12, 2012 at 23:53
  • $\begingroup$ My TRUE interest is the action of $K$ on $Sym^m(\mathcal{H})$ - I changed the statement of the question to clarify that. $\endgroup$
    – Michał Oszmaniec
    Commented Apr 12, 2012 at 23:55
  • $\begingroup$ Michal, you are welcome; feel free to vote it up. I don't have time to edit my answers in response to changing questions, suffice it to say that the decomposition of the symmetric power can be easily obtained from the decomposition of the tensor power displayed in my answer. $\endgroup$
    – Victor Protsak
    Commented Apr 13, 2012 at 0:19
  • $\begingroup$ Victor - I did not changed the question - I simply clarified it. I think that it was stated clear enough that I meant $Sym^n(\mathcal{H})$. Here is how it looked exactly (I copied it from the history of revisions): 1. Is the representation of $K$ restricted to the m fold symmetrization of $\mathcal{H}^{\otimes m}$ multiplicity free? 2. Which irreducible representations of $K$ appear in this representation? I agree that the point 2 might be misleading. Yet I think my remark in the body of the question suggests that I meant irreps of $K$ in $Sym^m(\mathcal{H})$ $\endgroup$
    – Michał Oszmaniec
    Commented Apr 13, 2012 at 0:37

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