# Lemma 1 ^49716f A [[Matrix]] which [[Commutator|commutes]] with all matrices of an [[Irreducible Representation]] is a constant matrix. That is, if a non-constant matrix exists, the [[Group Representation|representation]] is [[Irreducible Representation|reducible]], if none exists, the representation is irreducible. ## Proof Let $M$ be a matrix which commutes with all the matrices of the [[Irreducible Representation]] $D_1, D_2, \ldots, D_h$ $ M D_\alpha = D_\alpha M. $ Then, taking the adjoint of both sides, $ M^\dagger D_\alpha^\dagger = D_\alpha^\dagger M^\dagger $ Without loss of generality, we can take $D_\alpha$ to be [[Unitary Operator|unitary]] by the theorem on [[Theorem on the Unitarity of Representations|unitarity of representations]]. Thus, we get: $ M^\dagger D_\alpha = D_\alpha M^\dagger, $ i.e, if $M$ commutes with all the $D_\alpha$, then so does $M^\dagger$, and so do the Hermitian matrices $H_1$ and $H_2$ $ \begin{align} H_1 &= M + M^\dagger,\\ H_2 &= i(M - M^\dagger), \\ H_j D_\alpha &= D_\alpha H_j, \quad j = 1, 2 \\ \end{align} $ Now, we perform a [[Similarity Transformation|unitary transformation]] on the matrices of the irrep to get $\hat{D}_\alpha = U^{-1} D_\alpha U$, where $U$ is a matrix that [[Diagonalizable Matrix|diagonalizes]] $H_j$, e.g., $H_1$, to give the diagonal matrix $d$ $ d = U^{-1}H_j U $ Multiplying the equation $H_j D_\alpha = D_\alpha H_j$ by $U^{-1}$ on the left and $U$ on the right yields $ d \hat{D}_\alpha = \hat{D}_\alpha d $ So now, we have a diagonal matrix which commutes with all matrices of the representation. Lets take the $ij$ element of both sides $ d_{ii} (\hat{D_\alpha})_{ij} = (\hat{D}_\alpha)_{ij} d_{jj} $ or $ (\hat{D_\alpha})_{ij} (d_{ii} - d_{jj}) = 0 $ If $d_{ii} \neq d_{jj}$, then the matrix $d$ is not constant. Thus $(\hat{D}_\alpha)_{ij}$ must be $0$ for all $\alpha$, and the unitary transformation has brought all the matrices into a similar block diagonal form, i.e., the representation is reducible. However, this is a contradiction. We have assumed that the representation is an irrep from the get go, and thus $d_{ii} = d_{jj}$ and the matrix is diagonal. In conclusion, if the representation is irreducible and there exists a matrix which commutes with all representation matrices, then the matrix is a constant matrix $\square$. # Lemma 2 Consider two matrix representations of the [[group]] [[Group Element|elements]] $a_1, a_2, \ldots, a_h$: representation $D^{(1)}(a_\alpha)$ of [[dimension|dimensionality]] $\ell_1$ and representation $D^{(2)}(a_\alpha)$ of [[dimension|dimensionality]] $\ell_2$. If there is $\ell_2 \times \ell_1$ matrix ($\ell_1$ columns and $\ell_2$ rows) $M$ such that: $ M D^{(1)}(a_\alpha) = D^{(2)}(a_\alpha)M $ for $\alpha = 1, \ldots, h$, then: 1. If $\ell_1 \neq \ell_2$, $M$ must be the null matrix $\mathbb{0}$. ^fb3244 2. If $\ell_1 = \ell_2$, either 1. $M = \mathbb{0}$ 2. The representations $D^{(1)}(a_\alpha)$ and $D^{(2)}(a_\alpha)$ differ from each other by a [[similarity transformation]], i.e., they are equivalent representations. ^fa8c62 ## Proof ### Preliminaries Using the theorem on [[Theorem on the Unitarity of Representations]], we will assume that the matrices of both representations $D^{(1)}(a_\alpha)$ and $D^{(2)}(a_\alpha)$ have already been made [[Unitary Operator|unitary]]. We assume $\ell_1 \leq \ell_2$ and take the adjoint of the equation above: $ [D^{(1)}(a_\alpha)]^\dagger M^\dagger = M^\dagger [D^{(2)}(a_\alpha)]^\dagger $ By the unitarity of the [[Group Representation|representation]], we have: $ [D^{(j)}(a_\alpha)]^\dagger = [D^{(j)}(a_\alpha)]^{-1} = [D^{(j)}(a_\alpha^{-1})], \quad j = 1,2, \quad \alpha = 1,\ldots, h $ Thus, $ \begin{align} D^{(1)}(a_\alpha^{-1})M^\dagger &= M^\dagger D^{(2)}(a_\alpha^{-1}),\\ M D^{(1)}(a_\alpha^{-1})M^\dagger &= M M^\dagger D^{(2)}(a_\alpha^{-1}) \end{align} $ since $a_\alpha^{-1}$ is also an element of the [[group]], we have: $ \begin{align} M D^{(1)}(a_\alpha^{-1}) = D^{(2)}(a_\alpha^{-1})M \end{align} $ and thus, we conclude: $ D^{(2)}(a_\alpha^{-1}) M M^\dagger = M M^\dagger D^{(2)}(a_\alpha^{-1}) $ To summarize what we have done so far, if $M D^{(1)}(a_\alpha) = D^{(2)}(a_\alpha)M$, then $M M^\dagger$ commutes with all matrices of the representation (2). Then, by [[#Lemma 1|Lemma 1]], it is a constant matrix of [[dimension]] $\ell_2 \times \ell_2$: $ M M^\dagger = c I $ ### Case 1: $\ell_1 = \ell_2$ In this case, $M$ is a [[square matrix]] with an inverse: $ M^{-1} = M^{\dagger}/c, \quad c \neq 0 $ Then, if $M^{-1} \neq \mathbb{0}$, we have: $ D^{(1)}(a_\alpha) = M^{-1} D^{(2)}(a_\alpha)M $ and the representations differ only by a [[similarity transformation]] (case [[Schur's Lemma (Group Theory)#^fa8c62|(2b)]], $\square$). However, if $c = 0$, we consider $M M^\dagger = \mathbb{0}$ directly. $ \begin{align} (M M^\dagger)_{ij} &= \sum_k M_{ik} M^{\dagger}_{kj} = \sum_k M_{ik}M^{*}_{jk} = 0, \\ (M M^\dagger)_{ii} &= \sum_k M_{ik}M^{*}_{ik} = \sum_k |M_{ik}|^2 = 0, \\ \end{align} $ Thus, $M_{ik} = 0$ for all $i, k$ and the matrix $M$ is a null matrix (case [[Schur's Lemma (Group Theory)#^fa8c62|(2a)]], $\square$). ### Case 2: $\ell_1 \neq \ell_2$ Without loss of generality, we can take $\ell_1 < \ell_2$. Then $M$ has $\ell_1$ columns and $\ell_2$ rows. We can make it a square $\ell_2\times\ell_2$ matrix by padding it using $(\ell_2 - \ell_1)$ columns of zeros. $ \begin{align} \begin{pmatrix} & & & 0 & 0 & \ldots & 0 \\ & & & 0 & 0 & \ldots & 0 \\ &M& & 0 & 0 & \ldots & 0 \\ & & & \vdots & \vdots & \ldots & \vdots \\ & & & 0 & 0 & \ldots & 0 \\ \end{pmatrix} \end{align} = N = \text{square } \ell_2 \times \ell_2 \text{ matrix} $ And we have the adjoint as well: $ \begin{align} \begin{pmatrix} & & & & \\ & &M^\dagger & & \\ & & & & \\ 0 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & 0 & \ldots & 0 \\ \end{pmatrix} \end{align} = N^\dagger $ Note that now we have $ N N^\dagger = c I \quad (\text{dimension } \ell_2 \times \ell_2) $ $ \begin{align} (N N^\dagger)_{ii} &= \sum_k N_{ik} N_{ki}^\dagger = \sum_k |N_{ik}|^2 = c \delta_{ii}, \\ \end{align} $ Now in the second of these equations, we see *all* the diagonal entries of $N N^\dagger$ are equal to $c$. Thus, if a single one of those entries is zero, then we can prove that $c = 0$. Multiplying $N N^\dagger$ directly, we get: $ \begin{align} \begin{pmatrix} & & & 0 & 0 & \ldots & 0 \\ & & & 0 & 0 & \ldots & 0 \\ &M& & 0 & 0 & \ldots & 0 \\ & & & \vdots & \vdots & \ddots & \vdots \\ & & & 0 & 0 & \ldots & 0 \\ \end{pmatrix} \begin{pmatrix} & & & & \\ & &M^\dagger & & \\ & & & & \\ 0 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ 0 & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & 0 \\ \end{pmatrix} = \begin{pmatrix} & & &0 &0&0&\ldots &0 \\ &MM^\dagger& &0 &0&0&\ldots &0 \\ & & &0 &0&0&\ldots&0 \\ 0 & 0 &0 &0 &0&0&\ldots&0 \\ \vdots & \vdots &\vdots &\vdots &\vdots&\vdots&\ddots&\vdots \\ 0 & 0 &0 &0 &0&0&\ldots&0 \\ \end{pmatrix} \end{align} $ As we can see, there is still some zeros on the diagonal and we conclude that $c = 0$. Thus, we get $N_{ik} = 0$ for all $i,k$ and the matrix $N$ (and consequently $M$) is a null matrix ([[#^fb3244|case 1]], $\square$).