Generalization of Schur's lemma (Update) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:43:08Z http://mathoverflow.net/feeds/question/33958 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33958/generalization-of-schurs-lemma-update Generalization of Schur's lemma (Update) Alireza 2010-07-30T23:21:11Z 2010-08-21T00:30:45Z <p>I am not a mathematician nor physicist. I just know the basics of the representation theory. In my research, I realized that there is an orthogonality relation between the unitary group matrix elements as follows:</p> <p>$$I_1 = \int \mathrm{D}\mathbf{U} \; U_{i j}^{(\mathbf{r})} U_{k l }^{*(\mathbf{r}^{\prime})} = \frac{1}{ d_{ \mathbf{r} } } \delta_{\mathbf{r} \mathbf{r}^{\prime} } \delta_{i k} \delta_{j l}$$</p> <p>where $\mathbf{U} \in \mathcal{U}(N)$, $\mathrm{D}\mathbf{U}$ is the standard Haar measure, $U_{ij}^{(\mathbf{r})}$ denotes the $(i,j)$-th element of the representation matrix of $\mathbf{U}$, and $d_{ \mathbf{r} }$ is the dimension of the irreducible representation $\mathbf{r}$. </p> <p>Now, I need to know the answer for this integral:</p> <p>$$I_2 = \int \mathrm{D} \mathbf{U} \; U_{i_1 j_1}^{(\mathbf{r})} U_{ k_1 l_1 }^{ * ( \mathbf{r} ) } U_{ i_2 j_2 }^{(\mathbf{r}^{\prime})} U_{ k_2 l_2 }^{* ( \mathbf{r}^{ \prime \prime } ) }$$</p> <p>I appreciate any help.</p> <p>p.s. Here is my conjecture for the answer:</p> <p><code>I_2 = \delta_{ \mathbf{r}^{\prime} \mathbf{r}^{\prime \prime} } \times \left\{ \eqalign{ \frac{1}{ d_{ \mathbf{r} } d_{ \mathbf{r}^{ \prime } } -1 } \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } ( 1- \delta_{ \mathbf{r} \mathbf{r}^{\prime} } ) \\ + \delta_{ \mathbf{r} \mathbf{r}^{\prime} } \left[ \eqalign{ \frac{ 1 }{ d_{ \mathbf{r} }^2 -1 } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_2 } \delta_{ i_2 k_1 } \delta_{ j_2 l_1 } ) \\ - \frac{ 1 }{ d_{ \mathbf{r} } ( d_{ \mathbf{r} }^2 -1 ) } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_2 } \delta_{ i_2 k_2 } \delta_{ j_2 l_1 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_1 } \delta_{ i_2 k_1 } \delta_{ j_2 l_2 } ) } \right] } \right\}</code></p> <p><strong>UPDATE:</strong></p> <p>I have been advised that it might be helpful if I can find the tensor product of two irreducible representations, $\mathbf{s} = \mathbf{r} \otimes \mathbf{r}^{\prime}$, which most likely leads to a reducible representation, and then I need to decompose $\mathbf{s}$ into its irreducible components (by using the Clebschâ€“Gordan coefficients, according to wikipedia), to be able to use the Schur's lemma to get the answer!!!</p> <p>However, it is hard for me to do this, and needs awful background.</p> http://mathoverflow.net/questions/33958/generalization-of-schurs-lemma-update/35330#35330 Answer by Paul Zinn-Justin for Generalization of Schur's lemma (Update) Paul Zinn-Justin 2010-08-12T09:22:40Z 2010-08-12T09:22:40Z <p>I'm afraid I do not know the answer to your problem, but here's a counterexample to your conjecture that the result is zero if $r'\ne r''$. take $r=(2,1)$ (i.e., the Young diagram with two boxes one the first row and one on the second). then in the decomposition into irreducible representations of $r=(2,1)$ with its dual $r^\star=(...,-1,-2)$ one finds $(2,...,-1,-1)$ and $(1,1,...,-2)$. now these occur naturally as the tensor product of $r'=(2)$ and the dual of $r''=(1,1)$, or vice versa. ergo, $r\otimes r^\star\otimes r'\otimes r''^\star$ contains the trivial representation so that the integral of some of its matrix elements will be non-zero.</p>