Theorem 13.3.1 (Schur’s Lemma). Let be irreducible representations of , and a homomorphism of representations. Then:
Either or is an isomorphism.
Over : if is irreducible and is a homomorphism of representations (i.e., ), then for some .
Proof. (i) is a subrepresentation of . Since is irreducible, either () or ( injective). Similarly, is a subrepresentation of , so either () or ( surjective).
(ii) Over , has an eigenvalue . Then is also a -homomorphism with nontrivial kernel. By (i), .∎
Example 13.3.1 (Center acts by scalars). Let be an irreducible representation of , and . Since commutes with all , the map commutes with all , i.e., . By Schur, for some . So every central element acts as a scalar on every irreducible representation. In particular, irreducible representations of abelian groups are 1-dimensional (giving another proof of the earlier result).
13.3.1 Irreducible Representations of
Example 13.3.2 (The three irreducible representations of ).
The trivial representation : for all .
The alternating (sign) representation :.
The standard representation : start with the permutation representation (basis , with ). This has an invariant subspace . The complement has and is irreducible.
Proof (Irreducibility of ). Let and . Then acts on with eigenvalues and (since and ). If is a -eigenvector, then is a -eigenvector (since , using ). Since , and are linearly independent, so they span . Any invariant subspace of must contain an eigenvector of (since is diagonalizable on ), hence contains or , and then contains . So is irreducible. ∎
Proposition 13.3.1 (These are all irreducibles of ). The representations are the only irreducible representations of over , and any representation of decomposes as for unique .
Note. We will prove uniqueness (and a method to find ) using characters below. The key dimension check: , which we will show is a necessary condition.