§ Schur's lemma
§ Statement
if rv:G→GL(V),rw:G→GL(W) are two irreducible
representations of the group G, and f:V→W is an equivariant map
(that is, f∀g∈G,∀v∈V,(rv(g)(v))=rw(g)(f(v))),
then we have that either f=0 or f is an isomorphism.
- Said differently, this implies that either rv and rw are equivalent, and fwitnesses 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∈ker(f). hence:
rw(g)(f(k))=0f(rv(g)(k)=rw(g)(f(k))=0rv(g)(k)∈ker(f)
So if k∈ker(f) then so does rv(g)(k) for all g. Hence, the kernel
is an invariant subspace.
- Next, let w∈im(f), such that w=f(v) hence:
f(v)=wrw(g)(w)=rw(g)(f(v))=f(rv(g)(v))rw(g)(w)∈im(f)
So if w∈im(f) then rw(g)(w)∈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)⊕ker(f)⊥non-trivially. This contradicts the irreducible nature of V. Thus, either fsends 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)=Wby 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 λ such that f=λI.
- f cannot have two eigenvalues. If it did, the eigenspaces of λ1 andlambda2 would be different subspaces that are stabilized by f. This can'thappen because V is irreducible. So, f has a single eigenvalue λ.
- Thus, if f has full spectrum, it's going to be f=λI.
- f has full spectrum since we tacitly assume the underlying field is Cand f has full rank.