§ 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 isirreducible 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 andlambda2lambda_2 would be different subspaces that are stabilized by ff. This can'thappen 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.