Schur’s first lemma

Schur’s first lemma states that a non-zero matrix that commutes with all matrices of an irreducible representation of a group is a multiple of the identity matrix.

The proof of Schur’s first lemma involves the following steps:

1. Consider a representation of a group $G$ , i.e. $\Gamma=\{A_1,A_2,\cdots,A_{\vert G\vert}\}$ , where each element of $\Gamma$ is an $n\times n$ matrix, which can be regarded as a unitary matrix $U$ according to a previous article.
2. Proof that a Hermitian matrix $H$ that commutes with the irreducible representation element $A_{\alpha}$ , where $\alpha=1,2,\cdots,\vert G\vert$ , is a constant multiple of the identity matrix.
3. Infer from step 2 that any arbitrary non-zero matrix $M$ that commutes with the irreducible representation element $A_{\alpha}$ is a multiple of the identity matrix.

Step 1 is self-explanatory. For step 2, we begin with a Hermitian matrix $H$ that commutes with $A_{\alpha}$ :

$HA_{\alpha}=A_{\alpha}H$

Multiplying the above equation on the left and right by $U^{-1}$ and $U$ respectively,

$U^{-1}HA_{\alpha}U=U^{-1}A_{\alpha}HU$

Since $UU^{-1}=I$

$U^{-1}HUU^{-1}A_{\alpha}U=U^{-1}A_{\alpha}UU^{-1}HU$

$U^{-1}HU\tilde A_{\alpha}=\tilde A_{\alpha}U^{-1}HU$

where $\tilde A_{\alpha}=U^{-1}A_{\alpha}U$ .

Question

Show that $\tilde A_{\alpha}$ is also a representation of $G$ .

Answer

If $\tilde A_{\alpha}$ is also a representation of $G$ , its elements must multiply according to the multiplication table of $G=\{A_1,A_2,\cdots,A_{\vert G\vert}\}$ . Since $\tilde A_{\alpha}=U^{-1}A_{\alpha}U$ , we have

$\tilde A_{i}\tilde A_j=(U^{-1}A_iU)(U^{-1}A_jU)=U^{-1}A_iA_jU=U^{-1}A_kU=\tilde A_k$

The third equality ensures that the closure property of $G$ is satisfied for $A_{\alpha}$ and hence $\tilde A_{\alpha}$ . In other words, the elements of $G=\{\tilde A_1,\tilde A_2,\cdots,\tilde A_{\vert G\vert}\}$ multiply according to the multiplication table of $G=\{A_1,A_2,\cdots,A_{\vert G\vert}\}$ .

As a Hermitian matrix can undergo a similarity transformation by a unitary matrix to give another Hermitian matrix $D$ which is diagonal, i.e. $U^{-1}HU=D$ , we have

$D\tilde A_{\alpha}=\tilde A_{\alpha}D$

Rewriting $D\tilde A_{\alpha}=\tilde A_{\alpha}D$ in terms of its matrix elements, we have $(D\tilde A_{\alpha})_{mn}=(\tilde A_{\alpha}D)_{mn}$ or $D_{mm}(\tilde A_{\alpha})_{mn}=(\tilde A_{\alpha})_{mn}D_{nn}$ , which can be rearranged to

$(\tilde A_{\alpha})_{mn}(D_{mm}-D_{nn})=0$

Consider the following cases for the above equation:

Case 1: All diagonal elements of $D$ are distinct, i.e. $D_{mm}\neq D_{nn}$ if $m\neq n$ .

We have $(\tilde A_{\alpha})_{mn}=0$ for $m\neq n$ , which means that all off-diagonal elements of $\tilde A_{\alpha}$ are zero. In other words, $\tilde A_{\alpha}$ is an element of a reducible representation that is a direct sum of $n$ elements of one-dimensional matrix representations. Furthermore, the definition of a reducible representation implies that $A_{\alpha}$ is also an element of a reducible representation of $G$ because $\tilde A_{\alpha}=U^{-1}A_{\alpha}U$ .

Case 2: All diagonal elements of $D$ are equal, i.e. $D_{mm}=D_{nn}\;\forall\; m,n$ .

$(\tilde A_{\alpha})_{mn}$ can be any finite number, and consequently $\tilde A_{\alpha}$ may be either an element of a reducible or an irreducible representation. However, the diagonal matrix $D$ must be a multiple of the identity matrix if $D_{mm}=D_{nn}\;\forall\; m,n$ .

Case 3: Some but not all diagonal elements of $D$ are equal.

Instead of considering all possible permutations of equal and unequal diagonal entries in $D$ , we rearrange the columns of $U$ such that equal diagonal entries of $D$ are in adjacent columns of $D$ . This is always possible as the order of the columns of $U$ corresponds to the order of the diagonal entries in $D$ (see this article). Let’s suppose the first $i$ diagonal entries are the same, while the rest are distinct, i.e. $D_{11}=D_{22}=\cdots=D_{ii}\neq D_{i+1,i+1}\neq\cdots\neq D_{nn}$ . With reference to Case 1 and Case 2, $\tilde A_{\alpha}$ must be an element of a reducible representation with the block diagonal form:

$\tilde A_{\alpha}\begin{pmatrix}X&0\\0&Y\end{pmatrix}$

For example, if $D_{11}=D_{22}\neq D_{33}\neq D_{44}$ in the following $4\times 4$ matrix,

then $(\tilde A_1)_{12},(\tilde A_1)_{21}$ can be any finite number, while all other off-diagonal elements are zero.

Combining all three cases, if $\tilde A_{\alpha}$ is an irreducible representation, the diagonal matrix $D$ must be a multiple of the identity matrix. Since $D\tilde A_{\alpha}=\tilde A_{\alpha}D$ , where $D$ is Hermitian, we have proven step 2.

For the last step, let’s consider an arbitrary non-zero matrix $M$ that commutes with $A_{\alpha}$ :

$MA_{\alpha}=A_{\alpha}M\;\;or\;\;A_{\alpha}^{\dagger}M^{\dagger}=M^{\dagger}A_{\alpha}^{\dagger}$

Since $A_{\alpha}$ is unitary, $A_{\alpha}^{\dagger}=A_{\alpha}^{-1}$ and so $A_{\alpha}^{-1}M^{\dagger}=M^{\dagger}A_{\alpha}^{-1}$ , which when multiplied from the left and right by $A_{\alpha}$ gives $M^{\dagger}A_{\alpha}=A_{\alpha}M^{\dagger}$ . This implies that if $M$ commutes with $A_{\alpha}$ , then $M^{\dagger}$ also commutes with $A_{\alpha}$ .

Question

i) Show that if $M$ and $M^{\dagger}$ commutes with $A_{\alpha}$ , then any linear combination of $M$ and $M^{\dagger}$ also commutes with $A_{\alpha}$ .
ii) Show that the linear combinations $H_1=M+M^{\dagger}$ and $H_2=iM-M^{\dagger}$ are Hermitian.
iii) Show that $M=\frac{H_1-iH_2}{2}$ .

Answer

$(aM+bM^{\dagger})A_{\alpha}=aMA_{\alpha}+bM^{\dagger}A_{\alpha}=aA_{\alpha}M+bA_{\alpha}M^{\dagger}=A_{\alpha}(aM+bM^{\dagger})$

ii)

$H_1^{\dagger}=(M+M^{\dagger})^{\dagger}=M^{\dagger}+M=H_1$
$H_2^{\dagger}=(iM-iM^{\dagger})^{\dagger}=-iM^{\dagger}+iM=H_2$
iii) Substitute $H_1=M+M^{\dagger}$ and $H_2=iM-iM^{\dagger}$ in $\frac{H_1-iH_2}{2}$ , we get $M$ .

With reference to step 2, $H_1$ must be a constant multiple of the identity matrix and so must $H_2$ . Therefore, $\frac{H_1-iH_2}{2}$ is also a constant multiple of the identity matrix. This concludes that proof of Schur’s first lemma, which together with Schur’s second lemma, is used to proof the great orthogonality theorem.

Schur’s first lemma

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