Advanced Chemistry - Mono Mole

December 7, 2023

Schur’s second lemma

Schur’s second lemma describes the restrictions on a matrix that commutes with elements of two distinct irreducible representations, which may have different dimensions.

Consider an arbitrary matrix $M$ and two irreducible representations of a group, $A_{\alpha}$ of dimension $n$ and $A_{\alpha}'$ of dimension $n'$ , such that

$MA_{\alpha}=A_{\alpha}'M\;\;\;\;\;\;\;\;10$

where $\alpha=1,2,\cdots,\vert G\vert$ .

Taking the conjugate transpose of eq10 and using the matrix identity $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$ , we have $A_{\alpha}^{\dagger}M^{\dagger}=M^{\dagger}A_{\alpha}^{'\dagger}$ . As every element of a representation of a group can be expressed as a unitary matrix $U$ via a similarity transformation without any loss of generality, and as $U^{\dagger}=U^{-1}$ ,

$A_{\alpha}^{-1}M^{\dagger}=M^{\dagger}A_{\alpha}^{'-1}\;\;\;\;\;\;\;\;11$

Since the inverse property of a group states that $A_{i}^{-1}=A_{j}$ and $A_{i}^{'-1}=A_{j}'$ , we can express eq10 as

$MA_{\alpha}^{-1}=A_{\alpha}^{'-1}M\;\;\;\;\;\;\;\;12$

Multiplying eq11 on the left by $M$ and using eq12, we have $MA_{\alpha}^{-1}M^{\dagger}=MM^{\dagger}A_{\alpha}^{'-1}$ and therefore $A_{\alpha}^{'-1}(MM^{\dagger})=(MM^{\dagger})A_{\alpha}^{'-1}$ , which implies that $MM^{\dagger}$ commutes with all elements of an irreducible representation of $G$ . With reference to Schur’s first lemma,

$MM^{\dagger}=cI\;\;\;\;\;\;\;\;13$

where $c$ is a constant and $I$ is the identity matrix.

If we multiply eq11 on the right by $M$ and repeat the steps above, we have

$M^{\dagger}M=cI\;\;\;\;\;\;\;\;14$

Let’s consider the following cases for eq13:

Case 1: $n=n'$

Let the $(i,k)$ -th entry of $M$ be $m_{ik}$ . If $n=n'$ , we can rewrite eq13 in terms of matrix entries: $\sum_{k=1}^nm_{ik}m^*_{ik}=c\delta_{mn}\implies \sum_{k=1}^n\vert m_{ik}\vert^2=c\delta_{mn}$ . If $c=0$ , we have $\sum_{k=1}^n\vert m_{ik}\vert^2=0$ , which implies that $M$ is the zero matrix because $m_{ik}=0$ for all $i,k$ .

Combining eq13 and eq14, we have $MM^{\dagger}=cI=M^{\dagger}M$ or $M(\frac{1}{c}M^{\dagger})=I=(\frac{1}{c}M^{\dagger})M$ . This implies that $M^{-1}=\frac{1}{c}M^{\dagger}$ exists if $c\neq 0$ . We can therefore rewrite eq10 as $A_{\alpha}=M^{-1}A_{\alpha}'M$ , which is a similarity transformation if $c\neq 0$ .

Case 2: $n\neq n'$

If $n\neq n'$ , the arbitrary matrix (denoted by $M'$ ) is an $n'\times n$ matrix with reference to eq10. Suppose $n<n'$ ; we have:

$M'=\begin{pmatrix}m_{11}&m_{12}&\cdots&m_{1n}\\m_{21}&m_{22}&\cdots&m_{2n}\\\vdots&\vdots&\ddots&\vdots\\m_{n'1}&m_{n'2}&\cdots&m_{n'n}\end{pmatrix}$

If we enlarge $M'$ to form an $n'\times n'$ matrix $N$ with the additional elements equal to zero, we have

$N=\begin{pmatrix}m_{11}&m_{12}&\cdots&m_{1n}&0&\cdots&0\\m_{21}&m_{22}&\cdots&m_{2n}&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\m_{n'1}&m_{n'2}&\cdots&m_{n'n}&0&\cdots&0\end{pmatrix}\;\;\;N^{\dagger}=\begin{pmatrix}m_{11}^*&m_{21}^*&\cdots&m_{n'1}^*\\m_{12}^*&m_{22}^*&\cdots&m_{n'2}^*\\\vdots&\vdots&\ddots&\vdots\\m_{1n}^*&m_{2n}^*&\cdots&m_{n'n}^*\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{pmatrix}$

Due to the zeroes, $NN^{\dagger}=MM^{\dagger}=cI$ . Taking the determinants,

$\vert NN^{\dagger}\vert=\vert cI\vert$

Using the determinant identities $\vert AB\vert=\vert A\vert\vert B\vert$ , $\vert cA\vert=c^n\vert A\vert$ and $\vert I\vert=1$ , we have

$\vert N\vert\vert N^{\dagger}\vert=c^n$

Since one of the columns of $N$ is zero, $\vert N\vert=0$ . So, $c=0$ , which implies that $M$ must be a zero matrix according to the results of case 1.

Finally, we can summarise Schur’s second lemma as follows:

Given an arbitrary matrix $M$ and two irreducible representations, $A_{\alpha}$ of dimension $n$ and $A_{\alpha}'$ of dimension $n'$ , where $MA_{\alpha}=A_{\alpha}'M$ , then

1. if $n=n'$ , either $M=0$ or the representations are related by a similarity transformation, i.e. equivalent representations.
2. If $n\neq n'$ , $M=0$ .

Next article: Great orthogonality theorem

Previous article: Schur’s first lemma

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December 7, 2023May 14, 2025

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

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

$\begin{align}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)]\end{align}$

$\Gamma_q(\alpha)X\Gamma_q^{-1}(\alpha)$ 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) ${1\cdot 0+1\cdot (\frac{\sqrt{3}}{2})+1\cdot (-\frac{\sqrt{3}}{2})+1\cdot 0+1\cdot (-\frac{\sqrt{3}}{2})+1\cdot (\frac{\sqrt{3}}{2})=0$

Next article: Little orthogonality theorem

Previous article: Schur’s second lemma