Great orthogonality theorem

The great orthogonality theorem establishes the orthogonal relation between entries of matrices of irreducible representations of a group. Mathematically, it is expressed as

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)^*_{ij}\Gamma_p(\alpha)_{kl}=\frac{\vert G\vert}{d}\delta_{ik}\delta_{jl}\delta_{pq}\;\;\;\;\;\;\;\;14$

where

1. $\Gamma_p(\alpha)_{kl}$ refers to the matrix entry in $k$ -th row and $l$ -th column of the $\alpha$ -th matrix of the $p$ -th irreducible representation.
2. $\vert G\vert$ is the order of the group $G$ .
3. $d$ is the dimension of the irreducible representation.

The proof of eq14 involves analysing two cases and then combining the results. Consider the matrix

$M=\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\alpha)X\Gamma_q^{-1}(\alpha)\;\;\;\;\;\;\;\;15$

where $\Gamma_p(\alpha)$ is an $n'\times n'$ matrix associated with representation $p$ , $\Gamma_q^{-1}(\alpha)$ is and $n\times n$ matrix associated with representation $q$ and $X$ is an arbitrary matrix with $d_{n'}$ rows and $d_{n}$ columns.

Multiplying eq15 on the left by some matrix $\Gamma_p(\beta)$ associated with representation $p$ ,

$\Gamma_p(\beta)M=\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\beta)\Gamma_p(\alpha)X\Gamma_q^{-1}(\alpha)=\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\beta)\Gamma_p(\alpha)X\Gamma_q^{-1}(\alpha)\Gamma_q^{-1}(\beta)\Gamma_q(\beta)$

Using the matrix identity $(AB)^{-1}=B^{-1}A^{-1}$ ,

$\Gamma_p(\beta)M=\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\beta)\Gamma_p(\alpha)X\biggr(\Gamma_q(\beta)\Gamma_q(\alpha)\biggr)^{-1}\Gamma_q(\beta)$

Question

Proof the matrix identity $(AB)^{-1}=B^{-1}A^{-1}$ .

Answer

$AB=C\implies B=A^{-1}C\implies I=B^{-1}A^{-1}C\implies C^{-1}=B^{-1}A^{-1}$ and so $C^{-1}=(AB)^{-1}=B^{-1}A^{-1}$ .

According to the closure property of a group, $\Gamma_p(\beta)\Gamma_p(\alpha)=\Gamma_p(\gamma)$ and $\Gamma_q(\beta)\Gamma_q(\alpha)=\Gamma_q(\gamma)$ and thus

$\Gamma_p(\beta)M=\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\gamma)X\Gamma_q^{-1}(\gamma)\Gamma_q(\beta)=M\Gamma_q(\beta)\;\;\;\;\;\;\;\;16$

Case 1: $p=q$ .

Eq15 and eq16 becomes $M=\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)$ and $\Gamma_q(\alpha)M=M\Gamma_q(\alpha)$ respectively (we have changed the dummy index from $\beta$ to $\alpha$ in eq16). As every element of a representation can undergo a similarity transformation to a unitary matrix, we shall assume $\Gamma_q(\alpha)$ and its inverse are unitary matrices. According to Schur’s first lemma, eq16 implies that $M=cI$ and eq15 becomes

$cI=\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)\;\;\;\;\;\;\;\;17$

$tr(cI)=tr\biggr\[\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)\biggr\]=\sum_{\alpha=1}^{\vert G\vert}tr[\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)]\\=tr[\Gamma_q(1)X\Gamma_q^{-1}(1)]+tr[\Gamma_q(2)X\Gamma_q^{-1}(2)]+\cdots+tr[\Gamma_q(\vert G\vert)X\Gamma_q^{-1}(\vert G\vert)]$

$\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)$ is a similarity transformation, where $X$ is similar to some other arbitrary matrix. Since the traces of similar matrices are the same,

$tr(cI)=tr(X)+\cdots+tr(X)=\vert G\vert tr(X)$

$cd_q=\vert G\vert tr(X)\;\;\;\;\;\;\;\;18$

where $d_q$ is the identity matrix’s dimension, which is equal to the dimension of the matrix $\Gamma_q(\alpha)$ .

Substitute eq18 in eq17, we have $\frac{\vert G\vert}{d_q}tr(X)I=\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)$ , or in terms of matrix entries,

$\frac{\vert G\vert}{d_q}\biggr\[\sum_{l=1}^n(X)_{ll}\biggr\]\delta_{ik}=\sum_{\alpha=1}^{\vert G\vert}\sum_{l=1}^{n}\sum_{j=1}^{n}\Gamma_q(\alpha)_{kl}(X)_{lj}\Gamma_q^{-1}(\alpha)_{ji}$

The RHS of the above equation is a finite summation of the product of three scalars and their order can be changed. So,

$\sum_{l=1}^{n}\sum_{j=1}^{n}(X)_{lj}\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)_{kl}\Gamma_q^{-1}(\alpha)_{ji}-\frac{\vert G\vert}{d_q}\biggr\[\sum_{l=1}^n(X)_{ll}\biggr\]\delta_{ik}=0$

With $\sum_{l=1}^{n}(X)_{ll}=\sum_{l=1}^n\sum_{j=1}^{n}(X)_{lj}\delta_{jl}$ ,

$\sum_{l=1}^{n}\sum_{j=1}^{n}(X)_{lj}\biggr\[\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)_{kl}\Gamma_q^{-1}(\alpha)_{ji}-\frac{\vert G\vert}{d_q}\delta_{ik}\delta_{jl}\biggr\]$

Since $X$ is an arbitrary matrix, the above equation must satisfy any $(X)_{lj}$ . This is only possible if

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)_{kl}\Gamma_q^{-1}(\alpha)_{ji}=\frac{\vert G\vert}{d_q}\delta_{ik}\delta_{jl}$

Since $\Gamma_q^{-1}(\alpha)$ is unitary,

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_q(\alpha)_{kl}\Gamma_q^{*}(\alpha)_{ij}=\frac{\vert G\vert}{d_q}\delta_{ik}\delta_{jl}\;\;\;\;\;\;\;\;19$

Case 2: $p\neq q$ and $\Gamma_p$ is not equivalent to $\Gamma_q^{-1}$ .

According to Schur’s second lemma, eq15 becomes $\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\alpha)X\Gamma_q^{-1}(\alpha)=0$ , or in terms of matrix entries,

$\sum_{\alpha=1}^{\vert G\vert}\sum_{l=1}^{n}\sum_{j=1}^{n}\Gamma_p(\alpha)_{kl}(X)_{lj}\Gamma_q^{-1}(\alpha)_{ji}=0$

$\sum_{l=1}^{n}\sum_{j=1}^{n}(X)_{lj}\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\alpha)_{kl}\Gamma_q^{-1}(\alpha)_{ji}=0$

Since $X$ is an arbitrary matrix, the above equation must satisfy any $(X)_{lj}$ . This is only possible if

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\alpha)_{kl}\Gamma_q^{-1}(\alpha)_{ji}=0$

Since $\Gamma_q^{-1}(\alpha)_{ji}$ is unitary,

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_p(\alpha)_{kl}\Gamma_q^{*}(\alpha)_{ij}=0\;\;\;\;\;\;\;\;20$

Combining eq19 and eq20, we have the expression for the great orthogonality theorem:

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_q^{*}(\alpha)_{ij}\Gamma_p(\alpha)_{kl}=\frac{\vert G\vert}{d_q}\delta_{ik}\delta_{jl}\delta_{pq}\;\;\;\;\;\;\;\;20a$

which can also be expressed as

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_q^{*}(\alpha)_{ij}\Gamma_p(\alpha)_{kl}=\frac{\vert G\vert}{d}\delta_{ik}\delta_{jl}\delta_{pq}\;\;\;\;\;\;\;\;20b$

because the RHS vanishes if $p\neq q$ , which renders the subscript $q$ unnecessary for $d$ .

Question

What about the case where $p\neq q$ and $\Gamma_p$ is equivalent to $\Gamma_q^{-1}$ ?

Answer

In this case, $\Gamma_p(\alpha)$ can undergo a similarity transformation to become $\Gamma_q^{-1}(\alpha)$ , which in turn can undergo a similarity transformation to become elements of a unitary representation. This implies that both $\Gamma_p(\alpha)$ and $\Gamma_q^{-1}(\alpha)$ can be expressed as the same unitary representation because if $X$ is similar to $Y$ and $Y$ is similar to $Z$ , then $X$ is similar to $Z$ . In other words, we have $\Gamma_r(\alpha)M=M\Gamma_r(\alpha)$ (where is $\Gamma_r(\alpha)$ unitary), which is case 1.

Let’s rewrite eq20b as

$\sum_{\alpha=1}^{\vert G\vert}\Gamma_k^{*}(\alpha)_{ij}\Gamma_{k'}(\alpha)_{i'j'}=\frac{\vert G\vert}{d}\delta_{ii'}\delta_{jj'}\delta_{kk'}\;\;\;\;\;\;\;\;21$

Eq21 has the form of the inner product of two vectors, where $\Gamma_k^*(\alpha)_{ij}$ and $\Gamma_{k'}(\alpha)_{i'j'}$ are $\vert G\vert$ components of vectors $\boldsymbol{\mathit{u}}$ and $\boldsymbol{\mathit{v}}$ respectively in a $\vert G\vert$ -dimensional vector space. We can regard the components of the vectors as a function of three indices $i$ , $j$ and $k$ such that the two vectors are orthogonal to each other when $(i,j,k)\neq (i',j',k')$ . This orthogonal relation of matrix entries is why eq21 is called the great orthogonality theorem.

Question

Verify eq21 for

i) $k=k'=\Gamma_1$ ,
ii) $k=k'=\Gamma_2$ with $i=1,j=2$ , and
iii) $k=\Gamma_1,k'=\Gamma_2$ , with $i'=2,j'=1$ .

Answer

i) $1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1=\frac{6}{1}$
ii) ${0\cdot 0+(-\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})+0\cdot 0+(-\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2})+(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})=\frac{6}{2}}$
iii) ${1\cdot 0+1\cdot (\frac{\sqrt{3}}{2})+1\cdot (-\frac{\sqrt{3}}{2})+1\cdot 0+1\cdot (-\frac{\sqrt{3}}{2})+1\cdot (\frac{\sqrt{3}}{2})=0$

Great orthogonality theorem

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