## Schur’s Lemma (*-representation Form)

**Point of post: **In this post we discuss the notion of Schur’s lemma as it applies to irreducible -representations, and with this show that every irrep of an abelian group must have degree one.

*Motivation*

In our last post we saw how the existence of a non-zero intertwinor for an irrep guarantees the equivalence of the two irreps, since any such intertwinor must be an isomorphism. We also saw how if the two irreps and representation spaces coincided that the intertwinor must, in fact, be a multiple of the identity map. In this post we describe the notion of an *irreducible -representation* and then give a version of Schur’s second lemma for such representations. This will allow us to conclude, after some work, that every irrep of an abelian group must have degree one.

*Irreducible -representations and the -representation Form of Schur’s Lemma*

Let be a finite group and the group algebra. Then, if is a pre-Hilbert space and a -representation we call *-invariant *(or *invariant *when is evident) if is invariant under for each . We then, unsurprisingly, define a -representation to be *irreducible* if the only -invariant subspaces of are and .

Our first result result about irreducible -representations is a fundamental one. Namely:

**Theorem: ***Let be a finite group, the group algebra, and a pre-Hilbert space. Then if is an irreducible -representation then the induced representation is irreducible. Conversely, if is an irrep then the induced -representation*

**Proof: **Suppose first that is an irrep and is invariant under for each . We note then that for every one has

and thus is -invariant, but by assumption this implies that is either or . From where it follows that is -invariant implies that is either or and thus is an irrep.

Conversely, suppose that is an irrep and let be the induced irrep. Assume then that is invariant under then we have in particular that is invariant under for each and thus is -invariant and thus by assumption is either or , and by applying similar logic as last time we may conclude that is -invariant implies that is or and thus is irreducible.

With this we now have that third form of Schur’s lemma, the -representation form:

**Theorem(Schur’s Lemma Third Form--representation Form): ***Let be a finite group, the group algebra, and a pre-Hilbert spaces. Then if and are two irreducible -representations and such that for every then either or is an isomorphism. Moreover, if and then for some .*

**Proof: **We note first that since is irreducible the induced representations and $ must be irreps, but by definition from where it follows by the first form of Schur’s lemma that is either zero or an isomorphism.

Assume now that and then using the fact that the induced representation is irreducible and for each implies by the second form of Schur’s lemma that for some .

As a quick application of Schur’s Lemmas we prove the extremely interesting following theorem:

**Theorem: ***Let be a finite abelian group, a pre-Hilbert space, and an irreducible representation. Then, .*

**Proof: **Consider the induced -representation . Since is an irrep we know that is an irreducible representation. Note though that for any and any one has that

and thus by the third form of Schur’s lemma we may conclude that there exists such that for each . But, this clearly implies that every subspace of is invariant under every and thus -invariant. Thus, if we could choose with and by earlier remark this would be -invariant contradicting the irreducibility of . It follows that .

**References:**

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

[…] The necessity of this theorem was a previous theorem and so it suffices to show sufficiency. To do this we merely note that if each irrep has degree one […]

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