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:

$$I_1 = \int \mathrm{D}\mathbf{U} \; U_{i j}^{(\mathbf{r})} U_{k l }^{*(\mathbf{r}^{\prime})} = \frac{1}{ d_{ \mathbf{r} } } \delta_{\mathbf{r} \mathbf{r}^{\prime} } \delta_{i k} \delta_{j l} $$

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 $\mathbf{r}$.

Now, I need to know the answer for this integral:

$$I_2 = \int \mathrm{D} \mathbf{U} \; U_{i_1 j_1}^{(\mathbf{r})} U_{ k_1 l_1 }^{ * ( \mathbf{r} ) } U_{ i_2 j_2 }^{(\mathbf{r}^{\prime})} U_{ k_2 l_2 }^{* ( \mathbf{r}^{ \prime \prime } ) } $$

I appreciate any help.

p.s. Here is my conjecture for the answer:

$$ I_2 = \delta_{ \mathbf{r}^{\prime} \mathbf{r}^{\prime \prime} } \times \left\{ \eqalign{ \frac{1}{ d_{ \mathbf{r} } d_{ \mathbf{r}^{ \prime } } -1 } \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } ( 1- \delta_{ \mathbf{r} \mathbf{r}^{\prime} } ) \\ + \delta_{ \mathbf{r} \mathbf{r}^{\prime} } \left[ \eqalign{ \frac{ 1 }{ d_{ \mathbf{r} }^2 -1 } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_2 } \delta_{ i_2 k_1 } \delta_{ j_2 l_1 } ) \\ - \frac{ 1 }{ d_{ \mathbf{r} } ( d_{ \mathbf{r} }^2 -1 ) } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_2 } \delta_{ i_2 k_2 } \delta_{ j_2 l_1 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_1 } \delta_{ i_2 k_1 } \delta_{ j_2 l_2 } ) } \right] } \right\} $$


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 Clebsch–Gordan coefficients, according to wikipedia), to be able to use the Schur's lemma to get the answer!!!

However, it is hard for me to do this, and needs awful background.

  • $\begingroup$ Why do call this a generalisation of the Schur's Lemma? It seems to me like a generalisation of the Peter-Weyl theorem instead. $\endgroup$ Jul 30, 2010 at 23:30
  • $\begingroup$ I don't know the answer to your question, but this seems to me like the sort of calculation that lattice gauge theorists might know how to do. $\endgroup$ Jul 30, 2010 at 23:32
  • $\begingroup$ Thanks for the hint. p.s. The mathematicians I have talked to so far call I_1 the Schur's lemma. $\endgroup$
    – Alireza
    Jul 31, 2010 at 21:01
  • $\begingroup$ Do you really want two ${\bf r}$'s, one ${\bf r'}$ and one ${\bf r''}$ in your expression? Or is there a typo? $\endgroup$
    – Peter Shor
    Aug 1, 2010 at 15:46
  • 1
    $\begingroup$ I don't think your conjecture can be right. Look at $SU(2)$. Suppose ${\bf r}=3$, ${\bf r}'=5$ and ${\bf r''=7}$. Then the tensor products decompose into irreducible representations as: $R_3\otimes R_5=R_3\oplus R_5 \oplus R_7$ and $R_3\otimes R_7=R_5\oplus R_7\oplus R_9$, where $R_n$ is the irreducible representation of dimension $n$. When you sum over your indices $i_{1,2},\, j_{1,2},\, k_{1,2},\, l_{1,2}$, you should get $12=5+7$, not $0$, since these overlap in $R_5$ and $R_7$. How this sum of $12$ is distributed over $i_{1,2},\,j_{1,2},\,k_{1,2},\,l_{1,2}$ is a mystery to me. $\endgroup$
    – Peter Shor
    Aug 2, 2010 at 20:29

1 Answer 1


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.


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