## § 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 $W$ is irreducible 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't happen 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.