Proof of Schur's Lemma

Schur's lemma is the cornerstone of representation theory, controlling the structure of intertwining operators between irreducible representations. It underlies the orthogonality relations, the Wigner-Eckart theorem, and selection rules throughout physics.

Statement

Theorem8.3Schur's Lemma

Let $\rho_1: G \to \mathrm{GL}(V_1)$ and $\rho_2: G \to \mathrm{GL}(V_2)$ be irreducible representations of a group $G$ over $\mathbb{C}$ .

(Part I) If $T: V_1 \to V_2$ is a linear map satisfying $T\rho_1(g) = \rho_2(g)T$ for all $g \in G$ (an intertwining operator), then either $T = 0$ or $T$ is an isomorphism.

(Part II) If $V_1 = V_2 = V$ and $\rho_1 = \rho_2 = \rho$ , then any intertwining operator $T: V \to V$ satisfying $T\rho(g) = \rho(g)T$ for all $g \in G$ is a scalar multiple of the identity: $T = \lambda I$ for some $\lambda \in \mathbb{C}$ .

Proof

Part I.

Consider the kernel $\ker T = \{v \in V_1 : Tv = 0\}$ . If $v \in \ker T$ , then $T(\rho_1(g)v) = \rho_2(g)(Tv) = \rho_2(g)\cdot 0 = 0$ , so $\rho_1(g)v \in \ker T$ . Thus $\ker T$ is an invariant subspace of $V_1$ under $\rho_1$ .

Since $\rho_1$ is irreducible, the only invariant subspaces of $V_1$ are $\{0\}$ and $V_1$ . Therefore either $\ker T = V_1$ (i.e., $T = 0$ ) or $\ker T = \{0\}$ (i.e., $T$ is injective).

Consider the image $\mathrm{im}\,T = \{Tv : v \in V_1\} \subset V_2$ . If $w = Tv \in \mathrm{im}\,T$ , then $\rho_2(g)w = \rho_2(g)Tv = T\rho_1(g)v \in \mathrm{im}\,T$ . Thus $\mathrm{im}\,T$ is an invariant subspace of $V_2$ under $\rho_2$ .

Since $\rho_2$ is irreducible: either $\mathrm{im}\,T = \{0\}$ (i.e., $T = 0$ ) or $\mathrm{im}\,T = V_2$ (i.e., $T$ is surjective).

Combining: if $T \neq 0$ , then $T$ is both injective and surjective, hence an isomorphism $V_1 \cong V_2$ .

Part II.

Since we work over $\mathbb{C}$ and $V$ is finite-dimensional, $T$ has at least one eigenvalue $\lambda \in \mathbb{C}$ (by the fundamental theorem of algebra applied to the characteristic polynomial $\det(T - \lambda I) = 0$ ).

Consider $S = T - \lambda I$ . For any $g \in G$ : $S\rho(g) = (T - \lambda I)\rho(g) = T\rho(g) - \lambda\rho(g) = \rho(g)T - \lambda\rho(g) = \rho(g)(T-\lambda I) = \rho(g)S$

So $S$ intertwines $\rho$ with itself. By Part I, either $S = 0$ or $S$ is an isomorphism. But $\lambda$ is an eigenvalue of $T$ , so $\ker S = \ker(T-\lambda I) \neq \{0\}$ , which means $S$ is not an isomorphism. Therefore $S = 0$ , i.e., $T = \lambda I$ . $\square$

■

ExampleConsequences of Schur's Lemma

Orthogonality of characters. For compact groups, Schur's lemma combined with integration over the group yields: $\int_G \overline{\chi_\lambda(g)}\chi_\mu(g)\,d\mu(g) = \delta_{\lambda\mu}$ . This makes the character table an effective tool for decomposing representations. Commutant algebra. For a representation $\rho = \bigoplus_i n_i\rho_i$ , the algebra of all operators commuting with $\rho(g)$ for all $g$ is $\bigoplus_i M_{n_i}(\mathbb{C})$ (block diagonal matrices of size $n_i$ on each isotypic component). If $\rho$ is irreducible, the commutant is just $\mathbb{C}\cdot I$ .

RemarkSchur's Lemma over the Reals and for Infinite Dimensions

Over $\mathbb{R}$ , Part II generalizes: $T$ belongs to an associative division algebra over $\mathbb{R}$ , which by Frobenius' theorem is $\mathbb{R}$ , $\mathbb{C}$ , or $\mathbb{H}$ (the quaternions). This trichotomy classifies real irreducible representations into real type (commutant $\mathbb{R}$ ), complex type (commutant $\mathbb{C}$ ), and quaternionic type (commutant $\mathbb{H}$ ). Time-reversal symmetry in physics selects the type: Kramers degeneracy for half-integer spin arises from the quaternionic structure. For infinite-dimensional representations on Hilbert spaces, Part I holds for bounded intertwining operators between topologically irreducible representations, but Part II requires the additional assumption that $T$ is bounded.