Is there a generalization of Schur - Weyl duality and plethysm for direct product of special unitary groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:47:33Z http://mathoverflow.net/feeds/question/93454 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93454/is-there-a-generalization-of-schur-weyl-duality-and-plethysm-for-direct-product Is there a generalization of Schur - Weyl duality and plethysm for direct product of special unitary groups? Michal Oszmaniec 2012-04-07T20:10:12Z 2012-04-12T23:50:09Z <p>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 $Sym^m(\mathcal{H})\subset\mathcal{H}^{\otimes m}$:</p> <ol> <li><p>Is this representation multiplicity free? </p></li> <li><p>Which irreducible representations of $K$ appear in $Sym^m(\mathcal{H})$? </p></li> </ol> http://mathoverflow.net/questions/93454/is-there-a-generalization-of-schur-weyl-duality-and-plethysm-for-direct-product/93918#93918 Answer by Victor Protsak for Is there a generalization of Schur - Weyl duality and plethysm for direct product of special unitary groups? Victor Protsak 2012-04-12T23:28:15Z 2012-04-12T23:28:15Z <p>First, a terminological nitpick: the Schur-Weyl duality deals with the <em>unitary</em> 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,</p> <p>$$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda}\rho_{U(N)}^\lambda\otimes\rho_{S_m}^{\lambda},$$</p> <p>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 <em>not</em> 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 .</p> <p>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, $$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda_1,\ldots,\lambda_s}\rho_{SU(N_1)}^{\lambda_1}\otimes\ldots\otimes \rho_{SU(N_s)}^{\lambda_s}\otimes\rho_{S_m}^{\lambda_1}\otimes\ldots\otimes\rho_{S_m}^{\lambda_s}.$$</p> <p>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. </p>