Schur's lemma

R. Virk

Department of Mathematics
University of Wisconsin
Madison, WI 53706

virk@math.wisc.edu

Lemma 0.1   Let $V$ be a countable dimensional vector space over $\mathbb{C}$. If $\varphi\in \mathrm{Hom}_{\mathbb{C}}(V,V)$, then there exists $c\in\mathbb{C}$ such that $T-c\cdot\mathrm{id}$ is not invertible on $V$.

Proof. Suppose that $\varphi-c\cdot\mathrm{id}$ is invertible for all $c\in\mathbb{C}$. Then $P(\varphi)$ is invertible for all non-zero polynomials $P$ in one variable. So, if $R = P/Q$ is a rational function with $P$ and $Q$ polynomials, then we can define $R(\varphi) = P(\varphi)(Q(\varphi))^{-1}$. This gives us a map $\mathbb{C}(x)\rightarrow\mathrm{Hom}_{\mathbb{C}}(V,V)$. If $v \in V$ is non-zero and $R(\varphi)$ is as above, then $R(\varphi) = 0$ only if $P(\varphi) = 0$, which implies that $P$ is the zero polynomial (as otherwise we can find an eigenvector for $\varphi$). Thus, the map $\mathbb{C}(x) \rightarrow V$ is injective. This is a contradiction since $\mathbb{C}(x)$ is of uncountable dimension over $\mathbb{C}$. $\qedsymbol$

Lemma 0.2 (Schur's lemma)   Suppose that $V$ is a countable dimension vector space over $\mathbb{C}$ and that $A$ is an algebra that acts irreducibly on $V$. If $\varphi\in \mathrm{Hom}_{\mathbb{C}}(V,V)$ commutes with the action of $A$, then $\varphi$ is a scalar multiple of the identity operator.

Proof. By the previous lemma, there exists $c\in\mathbb{C}$, such that $\varphi-c\cdot\mathrm{id}$ is not invertible. As $\ker(\varphi - c\cdot\mathrm{id})$ is a submodule of $V$ it is either 0 or all of $V$. If it is 0, then $\mathrm{im}(\varphi-c\cdot\mathrm{id})$ is all of $V$ and $\varphi-c\cdot\mathrm{id}$ is invertible, which is a contradiction. Thus, $\varphi-c\cdot\mathrm{id}= 0$ and $\varphi=c\cdot\mathrm{id}$. $\qedsymbol$