Schur's Lemma

# Schur's Lemma

Lemma (Schur's Lemma): Let $G$ be a group and let $(V, \rho)$ be an irreducible group representation of $G$. Then every homomorphism $\varphi$ from $(V, \rho)$ to $(V, \rho)$ is multiplication by a scalar. |

**Proof:**Let $\lambda \in \mathbb{C}$ be an eigenvalue of $\rho$ and let $E_{\lambda}$ be the eigenspace. Then $E_{\lambda}$ is a subspace of $V$. Moreover, $E_{\lambda}$ is $G$-invariant since for all $g \in G$ and for all $v \in E_{\lambda}$:

\begin{align} \quad \varphi([\rho(g)](v)) = [\rho(g)](\varphi(v)) = [\rho(g)](\lambda v) = \lambda [\rho(g)](v) \end{align}

- Since $(V, \rho)$ is an irreducible group representation of $G$ we must have that $E_{\lambda} = V$. Therefore, for all $v \in V$ we have that:

\begin{align} \quad \varphi(v) = \lambda v \end{align}

- So the homomorphism $\varphi$ from $(V, \rho)$ to $(V, \rho)$ is multiplication by a scalar. $\blacksquare$

Corollary 2: Let $G$ be a group. If $(V_1, \rho_1)$ and $(V_2, \rho_2)$ are irreducible group representations of $G$ then $\mathrm{dim} (\mathrm{Hom}^G(V_1, V_2)) = \left\{\begin{matrix} 1 & \mathrm{if} \: V_1 \cong V_2 \\ 0 & \mathrm{otherwise} \end{matrix}\right.$. |

**Proof:**Let $\varphi \in \mathrm{Hom}^G(V_1, V_2)$. From the proposition on the Homomorphisms and Endomorphisms of Group Representations page we know that $\ker \varphi$ is a subrepresentation of $V_1$ and $\mathrm{range} \varphi$ is a subrepresentation of $V_2$. Since $(\rho_1, V_1)$ and $(\rho_2, V_2)$ are irreducible, this implies that $\ker \varphi = \{ 0 \}$ or $\ker \varphi = V_1$ and $\mathrm{range} \varphi = \{ 0 \}$ or $\mathrm{range} \varphi = V_2$.

- If $\ker \varphi = V_1$ then $\mathrm{range} \varphi = \{ 0 \}$. So $\varphi$ is the zero map.

- If $\ker \varphi = \{ 0 \}$ then $\varphi$ is injective. So $\mathrm{range} \varphi \neq \{ 0 \}$. Thus $\mathrm{range} \varphi = V_2$ so $\varphi$ is surjective and thus $\varphi$ is bijective. So $\varphi$ is an isomorphism from $V_1$ to $V_2$. $\blacksquare$

Corollary 3: Let $G$ be an abelian group. Then every irreducible group representation of $G$ is $1$-dimensional. |