Quantum Mechanics for Mathematicians

Theorem(Schur’s lemma).If a complex representation(π,V)is irreducible,
then the only linear mapsM:V →V commuting with all theπ(g)areλ 1 ,
multiplication by a scalarλ∈C.

Proof.Assume thatMcommutes with all theπ(g). We want to show that
(π,V) irreducible impliesM=λ 1. Since we are working over the fieldC(this
doesn’t work forR), we can always solve the eigenvalue equation

det(M−λ 1 ) = 0

to find the eigenvaluesλofM. The eigenspaces

Vλ={v∈V:Mv=λv}

are non-zero vector subspaces ofVand can also be described asker(M−λ 1 ),
the kernel of the operatorM−λ 1. Since this operator and all theπ(g) commute,
we have
v∈ker(M−λ 1 ) =⇒ π(g)v∈ker(M−λ 1 )

soker(M−λ 1 )⊂V is a representation ofG. IfV is irreducible, we must
have eitherker(M−λ 1 ) =Vorker(M−λ 1 ) = 0. Sinceλis an eigenvalue,
ker(M−λ 1 ) 6 = 0, soker(M−λ 1 ) =V and thusM=λ 1 as a linear operator
onV.

More concretely Schur’s lemma says that for an irreducible representation, if a
matrixMcommutes with all the representation matricesπ(g), thenMmust
be a scalar multiple of the unit matrix. Note that the proof crucially uses the
fact that eigenvalues exist. This will only be true in general if one works with
Cand thus with complex representations. For the theory of representations on
real vector spaces, Schur’s lemma is no longer true.
An important corollary of Schur’s lemma is the following characterization of
irreducible representations ofGwhenGis commutative.

Theorem 2.2. IfGis commutative, all of its irreducible representations are
one dimensional.

Proof.ForGcommutative,g∈G, any representation will satisfy

π(g)π(h) =π(h)π(g)

for allh∈G. Ifπis irreducible, Schur’s lemma implies that, since they commute
with all theπ(g), the matricesπ(h) are all scalar matrices, i.e.,π(h) =λh 1 for
someλh∈C.πis then irreducible when it is the one dimensional representation
given byπ(h) =λh.