§ Schur's lemma

§ Statement

if rv:G→GL(V),rw:G→GL(W)r_v : G \rightarrow GL(V), r_w: G \rightarrow GL(W) are two irreducible representations of the group GG, and f:V→Wf: V \rightarrow W is an equivariant map (that is, f∀g∈G,∀v∈V,(rv(g)(v))=rw(g)(f(v))f\forall g \in G, \forall v \in V, (r_v(g)(v)) = r_w(g)(f(v))), then we have that either f=0f = 0 or ff is an isomorphism.
  • Said differently, this implies that either rvr_v and rwr_w are equivalent, and ffwitnesses this isomorphism, or VV and WW are not isomorphic and ff is the zero map.

§ Proof

  • First, note that ker(f)ker(f) and im(f)im(f) are invariant subspaces of GG.
  • Let k∈ker(f)k \in ker(f). hence:
rw(g)(f(k))=0f(rv(g)(k)=rw(g)(f(k))=0rv(g)(k)∈ker(f) \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∈ker(f)k \in ker(f) then so does rv(g)(k)r_v(g)(k) for all gg. Hence, the kernel is an invariant subspace.
  • Next, let w∈im(f)w \in im(f), such that w=f(v)w = f(v) hence:
f(v)=wrw(g)(w)=rw(g)(f(v))=f(rv(g)(v))rw(g)(w)∈im(f) \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∈im(f)w \in im(f) then rw(g)(w)∈im(f)r_w(g)(w) \in im(f) for all gg. Hence, image is an invariant subspace.
  • Since VV is irreducible, we must have that either ker(f)=0ker(f) = 0 or ker(f)=Vker(f) = V. If this were not the case, then we could write V=ker(f)⊕ker(f)⊥V = ker(f) \oplus ker(f)^\perpnon-trivially. This contradicts the irreducible nature of VV. Thus, either ffsends all of VV to 00 (ie, ff is the zero map), or ff has trivial kernel (ie, ff is injective).
  • Since WW is irreducible, we must have that either im(f)=0im(f) = 0 or im(f)=Wim(f) = Wby the exact same argument; im(f)im(f) is an invariant subspace, and WW is irreducible thus has non non-trivial invariant subspaces. Thus either im(f)=0im(f) = 0( ff is the zero map), or im(f)=Wim(f) = W ( ff is surjective).
  • Thus, either ff is the zero map, or ff 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 ff?

We can additionally show that if ff is not the zero map, then ff is constant times the identity. That is, there exists a λ\lambda such that f=λIf = \lambda I.
  • ff cannot have two eigenvalues. If it did, the eigenspaces of λ1\lambda_1 and lambda2lambda_2 would be different subspaces that are stabilized by ff. This can't happen because VV is irreducible. So, ff has a single eigenvalue λ\lambda.
  • Thus, if ff has full spectrum, it's going to be f=λIf = \lambda I.
  • ff has full spectrum since we tacitly assume the underlying field is C\mathbb Cand ff has full rank.