- Schur's Lemma in Representation Theory

Schur’s lemma is a fundamental result in representation theory that characterizes the intertwining operators between irreducible representations. The lemma consists of two important parts.

Part I: Intertwining Maps Between Different Irreps #

If $D_a(g)$ and $D_b(g)$ are inequivalent irreducible representations of a group $G$, and $A$ is a linear operator satisfying:

$$D_a(g)A = AD_b(g)$$

for all $g \in G$, then $A = 0$.

Proof: #

1. Suppose there exists a non-zero vector $|\mu\rangle$ such that $A|\mu\rangle = 0$. Let $P$ be the projection operator onto $|\mu\rangle$.

   Then $D_a(g)AP = AD_b(g)P = 0$. Since $D_b(g)$ is an irreducible representation, $P$ must be the projector onto the entire vector space, which means $|\mu\rangle$ is an arbitrary vector in the whole vector space. Therefore, from $A|\mu\rangle = 0$ for an arbitrary $|\mu\rangle$, we conclude that $A = 0$.

2. If there does not exist such a non-zero $|\mu\rangle$, then from $D_a(g)A = AD_b(g)$, we get $D_a(g) = AD_b(g)A^{-1}$, which contradicts our assumption that $D_a(g)$ and $D_b(g)$ are inequivalent irreducible representations.

From (1) and (2), we conclude that if $D_a(g)A = AD_b(g)$ where $D_a(g)$ and $D_b(g)$ are inequivalent irreducible representations for all $g \in G$, then there always exists a non-zero $|\mu\rangle$ such that $A|\mu\rangle = 0$. Hence, $A = 0$.

Part II: Intertwining Maps Within the Same Irrep #

If $D(g)$ is an irreducible representation of $G$ and $A$ is a linear operator satisfying:

$$D(g)A = AD(g)$$

(that is, $[D(g), A] = 0$) for all $g \in G$, then $A$ is proportional to the identity operator: $A \propto I$.

Proof: #

1. If there exists a non-zero vector $|\mu\rangle$ such that $A|\mu\rangle = 0$, then by the same argument as in Part I, we conclude that $A = 0$, which is proportional to the identity.

2. Otherwise, since any operator $A$ has at least one eigenvector $|\mu\rangle$ with some eigenvalue $\lambda$, we have $(A - \lambda I)|\mu\rangle = 0$.

   From $D(g)A = AD(g)$, we get $D(g)(A - \lambda I) = (A - \lambda I)D(g)$. Since there exists $|\mu\rangle$ such that $(A - \lambda I)|\mu\rangle = 0$, by the argument in Part I, we must have $A - \lambda I = 0$, i.e., $A = \lambda I$.

From (1) and (2), we conclude that $A \propto I$.

Significance #

Schur’s lemma is a powerful tool in representation theory with applications in quantum mechanics, particularly in understanding symmetries and conservation laws. It provides a rigorous mathematical foundation for the orthogonality of matrix elements of irreducible representations and plays a crucial role in the development of character theory.