## § Schur's lemma

#### § Statement

if $r_v : G \rightarrow GL(V), r_w: G \rightarrow GL(W)$ are two irreducible representations of the group $G$, and $f: V \rightarrow W$ is an equivariant map (that is, $f\forall g \in G, \forall v \in V, (r_v(g)(v)) = r_w(g)(f(v))$), then we have that either $f = 0$ or $f$ is an isomorphism.
• Said differently, this implies that either $r_v$ and $r_w$ are equivalent, and $f$witnesses this isomorphism, or $V$ and $W$ are not isomorphic and $f$ is the zero map.

#### § Proof

• First, note that $ker(f)$ and $im(f)$ are invariant subspaces of $G$.
• Let $k \in ker(f)$. hence:
\begin{aligned} &r_w(g)(f(k)) = 0 \\ &f(r_v(g)(k) = r_w(g)(f(k)) = 0 \\ &r_v(g)(k) \in ker(f) \\ \end{aligned}
So if $k \in ker(f)$ then so does $r_v(g)(k)$ for all $g$. Hence, the kernel is an invariant subspace.
• Next, let $w \in im(f)$, such that $w = f(v)$ hence:
\begin{aligned} &f(v) = w \\ &r_w(g)(w) = r_w(g)(f(v)) = f(r_v(g)(v)) \\ &r_w(g)(w) \in im(f) \\ \end{aligned}
So if $w \in im(f)$ then $r_w(g)(w) \in im(f)$ for all $g$. Hence, image is an invariant subspace.
• Since $V$ is irreducible, we must have that either $ker(f) = 0$ or $ker(f) = V$.If this were not the case, then we could write $V = ker(f) \oplus ker(f)^\perp$non-trivially. This contradicts the irreducible nature of $V$. Thus, either $f$sends all of $V$ to $0$ (ie, $f$ is the zero map), or $f$ has trivial kernel(ie, $f$ is injective).
• Since $W$ is irreducible, we must have that either $im(f) = 0$ or $im(f) = W$by the exact same argument; $im(f)$ is an invariant subspace, and $W$ isirreducible thus has non non-trivial invariant subspaces. Thus either $im(f) = 0$($f$ is the zero map), or $im(f) = W$ ($f$ is surjective).
• Thus, either $f$ is the zero map, or $f$ is both injective and surjective; that is,it is bijective.
• The real star of the show is that (1) we choose irreducible representations,and (2) kernel and image are invariant subspaces for the chosen representations,thus we are forced to get trivial/full kernel/image.

#### § Strengthing the theorem: what is $f$?

We can additionally show that if $f$ is not the zero map, then $f$ is constant times the identity. That is, there exists a $\lambda$ such that $f = \lambda I$.
• $f$ cannot have two eigenvalues. If it did, the eigenspaces of $\lambda_1$ and$lambda_2$ would be different subspaces that are stabilized by $f$. This can'thappen because $V$ is irreducible. So, $f$ has a single eigenvalue $\lambda$.
• Thus, if $f$ has full spectrum, it's going to be $f = \lambda I$.
• $f$ has full spectrum since we tacitly assume the underlying field is $\mathbb C$and $f$ has full rank.