Statement and Proof of the Lemma

Theorem (Schur's Lemma): Let V and W be vector spaces; and let ${\displaystyle \rho _{V}}$ and ${\displaystyle \rho _{W}}$ be irreducible representations of G on V and W respectively.^[2]

1. If ${\displaystyle V}$ and ${\displaystyle W}$ are not isomorphic, then there are no nontrivial G-linear maps between them.
2. If ${\displaystyle V=W}$ finite-dimensional over an algebraically closed field (e.g. ${\displaystyle \mathbb {C} }$); and if ${\displaystyle \rho _{V}=\rho _{W}}$, then the only nontrivial G-linear maps are the identity, and scalar multiples of the identity. (A scalar multiple of the identity is sometimes called a homothety.)

Proof: Suppose ${\displaystyle f}$ is a nonzero G-linear map from ${\displaystyle V}$ to ${\displaystyle W}$. We will prove that ${\displaystyle V}$ and ${\displaystyle W}$ are isomorphic. Let ${\displaystyle V'}$ be the kernel, or null space, of ${\displaystyle f}$ in ${\displaystyle V}$, the subspace of all ${\displaystyle x}$ in ${\displaystyle V}$ for which ${\displaystyle f(x)=0}$. (It is easy to check that this is a subspace.) By the assumption that ${\displaystyle f}$ is G-linear, for every ${\displaystyle g}$ in ${\displaystyle G}$ and choice of ${\displaystyle x}$ in ${\displaystyle V'}$, ${\displaystyle f((\rho _{V}(g))(x))=(\rho _{W}(g))(f(x))=(\rho _{W}(g))(0)=0}$. But saying that ${\displaystyle f(\rho _{V}(g)(x))=0}$ is the same as saying that ${\displaystyle \rho _{V}(g)(x)}$ is in the null space of ${\displaystyle f:V\rightarrow W}$. So ${\displaystyle V'}$ is stable under the action of G; it is a subrepresentation. Since by assumption ${\displaystyle V}$ is irreducible, ${\displaystyle V'}$ must be zero; so ${\displaystyle f}$ is injective.

By an identical argument we will show ${\displaystyle f}$ is also surjective; since ${\displaystyle f((\rho _{V}(g))(x))=(\rho _{W}(g))(f(x))}$, we can conclude that for arbitrary choice of ${\displaystyle f(x)}$ in the image of ${\displaystyle f}$, ${\displaystyle \rho _{W}(g)}$ sends ${\displaystyle f(x)}$ somewhere else in the image of ${\displaystyle f}$; in particular it sends it to the image of ${\displaystyle \rho _{V}(g)x}$. So the image of ${\displaystyle f(x)}$ is a subspace ${\displaystyle W'}$ of ${\displaystyle W}$ stable under the action of ${\displaystyle G}$, so it is a subrepresentation and ${\displaystyle f}$ must be zero or surjective. By assumption it is not zero, so it is surjective, in which case it is an isomorphism.

In the event that ${\displaystyle V=W}$ finite-dimensional over an algebraically closed field and they have the same representation, let ${\displaystyle \lambda }$ be an eigenvalue of ${\displaystyle f}$. (An eigenvalue exists for every linear transformation on a finite-dimensional vector space over an algebraically closed field.) Let ${\displaystyle f'=f-\lambda I}$. Then if ${\displaystyle x}$ is an eigenvector of ${\displaystyle f}$ corresponding to ${\displaystyle \lambda ,f'(x)=0}$. It is clear that ${\displaystyle f'}$ is a G-linear map, because the sum or difference of G-linear maps is also G-linear. Then we return to the above argument, where we used the fact that a map was G-linear to conclude that the kernel is a subrepresentation, and is thus either zero or equal to all of ${\displaystyle V}$; because it is not zero (it contains ${\displaystyle x}$) it must be all of V and so ${\displaystyle f'}$ is trivial, so ${\displaystyle f=\lambda I}$.

Corollary of Schur's Lemma

An important corollary of Schur's lemma follows from the observation that we can often build explicitly ${\displaystyle G}$-linear maps between representations by "averaging" over the action of individual group elements on some fixed linear operator. In particular, given any irreducible representation, such objects will satisfy the assumptions of Schur's lemma, hence be scalar multiples of the identity. More precisely:

Corollary: Using the same notation from the previous theorem, let ${\displaystyle h}$ be a linear mapping of V into W, and set

$${\displaystyle h_{0}={\frac {1}{|G|}}\sum _{g\in G}(\rho _{W}(g))^{-1}h\rho _{V}(g).}$$

Then,

1. If ${\displaystyle V}$ and ${\displaystyle W}$ are not isomorphic, then ${\displaystyle h_{0}=0}$.
2. If ${\displaystyle V=W}$ is finite-dimensional over an algebraically closed field (e.g. ${\displaystyle \mathbb {C} }$); and if ${\displaystyle \rho _{V}=\rho _{W}}$, then ${\displaystyle h_{0}=I\,\mathrm {Tr} [h]/n}$, where n is the dimension of V. That is, ${\displaystyle h_{0}}$ is a homothety of ratio ${\displaystyle \mathrm {Tr} [h]/n}$.

Proof: Let us first show that ${\displaystyle h_{0}}$ is a G-linear map, i.e., ${\displaystyle \rho _{W}(g)\circ h_{0}=h_{0}\circ \rho _{V}(g)}$ for all ${\displaystyle g\in G}$. Indeed, consider that

${\displaystyle {\begin{aligned}(\rho _{W}(g'))^{-1}h_{0}\rho _{V}(g')&={\frac {1}{|G|}}\sum _{g\in G}(\rho _{W}(g'))^{-1}(\rho _{W}(g))^{-1}h\rho _{V}(g)\rho _{V}(g')\\&={\frac {1}{|G|}}\sum _{g\in G}(\rho _{W}(g\circ g'))^{-1}h\rho _{V}(g\circ g')\\&=h_{0}\end{aligned}}}$

Now applying the previous theorem, for case 1, it follows that ${\displaystyle h_{0}=0}$, and for case 2, it follows that ${\displaystyle h_{0}}$ is a scalar multiple of the identity matrix (i.e., ${\displaystyle h_{0}=\mu I}$). To determine the scalar multiple ${\displaystyle \mu }$, consider that

${\displaystyle \mathrm {Tr} [h_{0}]={\frac {1}{|G|}}\sum _{g\in G}\mathrm {Tr} [(\rho _{V}(g))^{-1}h\rho _{V}(g)]=\mathrm {Tr} [h]}$

It then follows that ${\displaystyle \mu =\mathrm {Tr} [h]/n}$.

This result has numerous applications. For example, in the context of quantum information science, it is used to derive results about complex projective t-designs.^[3]

Application to central characters

Definition: Let ${\displaystyle R}$ be a ${\displaystyle k}$-algebra. An ${\displaystyle R}$-module ${\displaystyle M}$ is said to have central character ${\displaystyle \chi :Z(R)\to k}$ (here, ${\displaystyle Z(R)}$ is the center of ${\displaystyle R}$) if for every ${\displaystyle m\in M,z\in Z(R)}$ there is ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle (z-\chi (z))^{n}m=0}$, i.e. if every ${\displaystyle m\in M}$ is a generalized eigenvector of ${\displaystyle z}$ with eigenvalue ${\displaystyle \chi (z)}$.

If ${\displaystyle \operatorname {End} _{R}(M)=k}$, say in the case sketched above, every element of ${\displaystyle Z(R)}$ acts on ${\displaystyle M}$ as an ${\displaystyle R}$-endomorphism and hence as a scalar. Thus, there is a ring homomorphism ${\displaystyle \chi :Z(R)\to k}$ such that ${\displaystyle (z-\chi (z))m=0}$ for all ${\displaystyle z\in Z(R),m\in M}$. In particular, ${\displaystyle M}$ has central character ${\displaystyle \chi }$.

If ${\displaystyle R=U({\mathfrak {g}}),k=\mathbb {C} }$ is the universal enveloping algebra of a Lie algebra, a central character is also referred to as an infinitesimal character and the previous considerations show that if ${\displaystyle {\mathfrak {g}}}$ is finite-dimensional (so that ${\displaystyle R=U({\mathfrak {g}})}$ is countable-dimensional), then every simple ${\displaystyle {\mathfrak {g}}}$-module has an infinitesimal character.

In the case where ${\displaystyle k=\mathbb {C} ,R=\mathbb {C} [G]}$ is the group algebra of a finite group ${\displaystyle G}$, the same conclusion follows. Here, the center of ${\displaystyle R}$ consists of elements of the shape ${\displaystyle \sum _{g\in G}a(g)g}$ where ${\displaystyle a:G\to \mathbb {C} }$ is a class function, i.e. invariant under conjugation. Since the set of class functions is spanned by the characters ${\displaystyle \chi _{\pi }}$ of the irreducible representations ${\displaystyle \pi \in {\hat {G}}}$, the central character is determined by what it maps ${\displaystyle u_{\pi }:={\frac {1}{\#G}}\sum _{g\in G}\chi _{\pi }(g)g}$ to (for all ${\displaystyle \pi \in {\hat {G}}}$). Since all ${\displaystyle u_{\pi }}$ are idempotent, they are each mapped either to 0 or to 1, and since ${\displaystyle u_{\pi }u_{\pi '}=0}$ for two different irreducible representations, only one ${\displaystyle u_{\pi }}$ can be mapped to 1: the one corresponding to the module ${\displaystyle M}$.

Application to the Casimir element

Suppose ${\displaystyle {\mathfrak {g}}}$ is a Lie algebra and ${\displaystyle U({\mathfrak {g}})}$ is the universal enveloping algebra of ${\displaystyle {\mathfrak {g}}}$. Let ${\displaystyle \pi$ :{\mathfrak {g}}\rightarrow \mathrm {End} (V)} be an irreducible representation of ${\displaystyle {\mathfrak {g}}}$ over an algebraically closed field. The universal property of the universal enveloping algebra ensures that ${\displaystyle \pi }$ extends to a representation of ${\displaystyle U({\mathfrak {g}})}$ acting on the same vector space. It follows from the second part of Schur's lemma that if ${\displaystyle x}$ belongs to the center of ${\displaystyle U({\mathfrak {g}})}$, then ${\displaystyle \pi (x)}$ must be a multiple of the identity operator. In the case when ${\displaystyle {\mathfrak {g}}}$ is a complex semisimple Lie algebra, an important example of the preceding construction is the one in which ${\displaystyle x}$ is the (quadratic) Casimir element ${\displaystyle C}$. In this case, ${\displaystyle \pi (C)=\lambda _{\pi }I}$, where ${\displaystyle \lambda _{\pi }}$ is a constant that can be computed explicitly in terms of the highest weight of ${\displaystyle \pi }$.^[8] The action of the Casimir element plays an important role in the proof of complete reducibility for finite-dimensional representations of semisimple Lie algebras.^[9]

See also Schur complement.