Schur’s Lemma (First and Second Forms)
Point of post: In this post we discuss Schur’s lemma, and the necessary prerequisite concepts, such as intertwinors.
Schur’s lemma will function as one of the main tools in proving theorems. In essence, Schur’s lemma says that the kinds of maps between two representation spaces of a finite group which “commute” with irreps (in a sense made precise below) are very limited. In essence, any such map is either the zero map or invertible. We shall use this basic concept to prove other “versions” of Schur’s lemma which will prove useful in different situations.
Let be a finite group and and two representations. Then a map is called an interwinor for and if for each the following diagram commutes:
Our first theorem regarding intertwinors will play a pivotal role in our proof of the first form of Schur’s lemma.
Theorem: Let be a finite group and and be two representations of . Then, if is an intertwinor for and then is invariant under and is invariant under .
Proof: Suppose that then for each one has that and thus . Since was arbitrary it follows that is invariant under .
Secondly, assume that then . The conclusion follows.
With just this simple fact about intertwinors we can prove the first form of Schur’s lemma:
Theorem(Schur’s Lemma First Form): Let be a finite group and and be irreps with . If is an intertwinor for and then either or is invertible.
Proof: Suppose that then is a proper -invariant subspace of and since is an irrep it follows that and thus is injective. Suppose then that isn’t surjective and , then is a proper -invariant subspace of and thus by assumption that is an irrep it follows that which contradicts that . It follows that and thus is surjective. Combining the two gives the desired result.
Corollary: If and are irreps and there exists a non-zero intertwinor then .
Proof: Indeed, by Schur’s lemma we know that must be an isomorphism such that , and thus by earlier theorem it follows that as desired.
We now prove a somewhat stronger version of Schur’s lemma when we are only interested in the case when and . Put more directly:
Theorem(Schur’s Lemma Second Form): Let and be an intertwinor for and itself. Then, for some .
Proof: Since is a complex vector space we know that admits at least one eigenvalue . Let . We note then that is non-trivial since if is an eigenvalue corresponding to then . This said we see that
and thus is an intertwinor, but since can’t be an isomorphism since it’s kernel is non-trivial it follows from Schur’s lemma that from where it follows that as desired.
1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.
2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print