Complete Reducibility and Schur’s Lemma — Studies in Algebra and Group Theory

31 Complete Reducibility and Schur’s Lemma

14 November 2025 · Denis Auroux

Representations

Recall.

Definition 31.1. A representation of a group is a vector space (usually finite-dimensional; we mostly consider ) + an action of on by linear operators, i.e. such that for all , is a linear map.

Equivalently: a homomorphism , the group of invertible linear operators .

Definition 31.2.

A subrepresentation is a subspace which is invariant under , i.e. for all .
A representation is irreducible if it has no nontrivial subrepresentations.

Example 31.3. If is a cyclic group, then a representation of is a vector space together with such that .

We need linear algebra to prove:

Lemma 31.4. Let be a finite-dimensional -vector space and of finite order, . Then is diagonalizable.

Returning to : invariant subspaces of are also invariant for , hence subrepresentations; in a basis which diagonalizes , each is an eigenvector for not only but also all , so is a 1-dimensional subrepresentation.

splits into a direct sum of 1-dimensional (irreducible) representations , for a basis of eigenvectors of , each given by a homomorphism There are distinct irreducible representations ( the complex -th roots of 1), and every complex representation of is a direct sum of some number of copies of each of these.

Now, if is a -representation of a finite abelian group , , with , the -action is equivalent to the data of such that and which pairwise commute, (then , with the generators of the factors ).

By the lemma each is diagonalizable, and by the homework, commuting diagonalizable operators are simultaneously diagonalizable. In fact: the eigenspaces of are invariant under all , and the restriction of to an eigenspace of is of finite order, hence diagonalizable by the lemma. Proceed by induction on .

This shows that splits into a of 1-dimensional subrepresentations. Those now correspond to homomorphisms .

Definition 31.5. For a finite abelian group, define its dual .

This is an abelian group using pointwise multiplication: if are homomorphisms, then so is (⚠ This uses the fact that is abelian .)

Concretely, for , as well, via , though there is no canonical map .

Similarly, the same ( is determined by the images of the generators of , which are roots of 1 in ).

This completes the classification of (complex) representations of finite abelian groups!

Homomorphisms of representations

Definition 31.6. Given two representations of , a homomorphism of representations is a linear map that is equivariant, i.e. compatible with the group actions:

We denote the set of homomorphisms of representations (-equivariant linear maps) by (as opposed to all linear maps, ).

New representations from old

We can make new representations out of old ones. In particular:

If are two representations of and , then and are preserved by , hence are subrepresentations of and . (, so .)
If is a subrepresentation, then is also a representation (since , maps cosets to cosets: ).
representations of is also a representation (), and so is (, extended by linearity).
(all linear maps) is also a -representation, but this requires care: given , what can we expect of ?

Answer: if , then . So: where the first is the action of on and is the action of on . (Check: ✓.)

Comparing with the above: given ,

Specializing to , where can be equipped with the trivial representation (every acts by ): the dual representation of is with , i.e. acts on by , the transpose of the action of on .

Then the isomorphism is an isomorphism of representations (i.e. a -equivariant isomorphism): does map .

Complete reducibility

Theorem 31.7. Let be any representation of a finite group (over , or of characteristic 0), and suppose is an invariant subspace (i.e., a subrepresentation). Then there exists another invariant subspace such that (as a direct sum of representations).

Corollary 31.8. Any finite-dimensional representation of a finite group decomposes into a direct sum of irreducibles.

Two proofs of the theorem. The first one uses:

Lemma 31.9. If is a -representation of a finite group , then there exists a positive definite Hermitian inner product on which is preserved by : i.e. all the linear operators are unitary.

Proof of the lemma. Let be any Hermitian inner product on , and use the averaging trick to set Then is still Hermitian and positive definite (hence an inner product), and . ∎

Proof of the theorem. Equip with a -invariant Hermitian inner product as in the lemma. Then if , unitary . So is a complementary invariant subspace. ∎

Proof (Alternative proof). Choose any complementary subspace such that . Let be the projection onto with kernel (, ). Define Then is a homomorphism of representations (i.e. -equivariant: for all ), so is an invariant subspace.

Since , is surjective and (dimension/rank formula and ). ∎

Remark (Remarks).

The proof fails if (more specifically, if divides ). This is one of the reasons that modular representations (= over fields of characteristic ) are more complicated.
It also fails if is infinite (and doesn’t carry a finite invariant measure), as we can’t use the averaging trick. (Averaging works for compact Lie groups such as , , ….)

Example 31.10. or acting on by . Then the first factor is invariant under , but there is no complementary invariant subspace.

Notes not reached in this lecture, covered at the start of {{LINK:L32}}

Schur’s lemma

Goal: given , find its irreducible representations, and describe how others decompose into irreducibles.

Lemma 31.11 (Schur’s Lemma).

If are irreducible representations of , and is any homomorphism of representations, then either or is an isomorphism.
Over : if is irreducible and is a homomorphism of representations, then is a multiple of the identity.

Proof.

Given , is an invariant subspace of , i.e. a subrepresentation. Since is irreducible, either ( injective) or (). Similarly, is invariant, hence either zero () or ( surjective). Hence either or is an isomorphism.
Over , any has an eigenvalue . Then is also equivariant and has nonzero kernel, hence by the above. Thus .

∎

Example 31.12. Let be an irreducible representation of , and , the center of ( commutes with all ). Then the action of satisfies for all : so is equivariant, i.e. by Schur’s lemma acts by a multiple of .

In particular, if is abelian and is irreducible, then every element of acts by a multiple of ; this gives another proof that irreducible representations of finite abelian groups are 1-dimensional.

(Next time: irreducible representations of .)