----- > [!proposition] Proposition. ([[Schur's lemma for groups]]) > Let $G$ be a [[group]] and $(\rho, V), (\rho', V')$ two [[irreducible group representation|irreducible]] [[group representation|representations]] $\mathbb{F}$ of $G$ over a [[field]] $\mathbb{F}$. Let $T$ be a [[morphism of group representations|morphism]] $(\rho, V) \to (\rho', V')$. > > Then: > 1. Either $T=0$ or $T$ is an [[isomorphism]], > 2. If $\mathbb{F}$ is [[algebraically closed]] (e.g. $\mathbb{C}$) and $V=V'$, then in particular $T=\lambda I$ for some $\lambda \in \mathbb{F}$. Thus $\text{dim}_{\mathbb{C}}\text{Hom}_{G}(V, V)=1$. > [!note] Note. > Compare to [[Schur's Lemma for modules]] and [[Schur's Lemma for Lie algebras]]. Those also came later and so are cleaner. ^note > [!proof]- Proof. ([[Schur's lemma for groups]]) > **1.** We will show $\ker T = \{ 0 \}$ and $\im T = V$. > > - $\ker T$ is a $G$-[[group-invariant subspace|invariant]] [[linear subspace|subspace]] of $V'$ w.r.t. $\rho'$: let $v' \in \ker T$, then using $G$-equivariance of $T$ we get $T(\rho_{g}'v') = \rho_{g} (Tv') = \rho_{g}(0)=0$ implying $\rho _g'v' \in \ker T$. Since $\rho'$ is [[irreducible group representation|irreducible]], this means $\ker T = \{ 0 \}$ or $\ker T=V'$. If $\ker T = \{ 0 \}$ then $T$ [[linear map is injective iff kernel is trivial|is an]] [[injection]]. Else $\ker T = V'$, in which case $T=0$. > > - $\im T$ is a $G$-[[group-invariant subspace|invariant]] [[linear subspace|subspace]] of $V$ w.r.t. $\rho$: let $v = \im T$; $v = Tv'$ for some $v' \in V'$. Then using $G$-equivariance of $T$, we have $\rho_{g}v = \rho_{g}(Tv')=T(\rho_{g'}v') \in \im T.$ Since $\rho$ is [[irreducible group representation|irreducible]], either $\im T = V$ or $\im T = \{ 0 \}$. If $\im T = V$ then $T$ is a [[surjection]]. Else $\im T = \{ 0 \}$, in which case $T=0$. > > So either $T$ is a [[linear isomorphism]], or else $T=0$. > > **2.** Let $T \in \hom_{G}(V,V)$ have [[eigenvector]] $v$ with [[eigenvalue]] $\lambda$ (using that $\mathbb{F}$ is algebraically closed) Then $T-\lambda I$ must lie in $\hom_{G}(V,V)$ with nontrivial [[kernel of a linear map|kernel]]. Since $(\rho, V)$ is [[irreducible group representation|irreducible]] and by the above $K$ is $G$-[[group-equivariant map|equivariant]], we must have $V = \ker (T - \lambda I)$. Thus $T=\lambda I$. > > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` older version: Let $G$ be a finite [[group]] and $(\rho, V)$, $(\rho', V')$ two [[irreducible group representation|irreducible]] [[group representation|representations]] of $G$. Denote by $\hom_{G}(V',V) \subset \hom(V'V)$ the [[linear subspace|subspace]] of $G$-[[group-equivariant map|equivariant]] [[linear map|linear maps]] from $V'$ to $V$ w.r.t. $\rho, \rho'$. > > Let $T \in \hom_{G}(V',V)$. Then: > 1. Either $T$ is an [[morphism of group representations|isomorphism]], or else $T=0$. > 2. Suppose $(\rho',V')\cong (\rho, V)$. Then $T$ is multiplication by a scalar: $T=cI$. Thus, $\dim_{\mathbb{C}}(\hom_{G}(V_{},V_{}))=1.$