How does Schur's Lemma mean that the Dirac representation is reducible?

Question

In chapter 3 of Peskin and Schroeder, when they're talking about "Dirac Matrices and Dirac Field Bilinears," they introduce $\gamma^{5}$ and give some properties of it. One of the properties is $[\gamma^{5},S^{\mu\nu}]=0$. Then they say that this means the Dirac representation must be reducible, "since eigenvectors of $\gamma^{5}$ whose eigenvalues are different transform without mixing (this criterion for reducibility is known as Schur's Lemma)."

I've looked at the wikipedia page for Schur's Lemma, and at various math notes online about Schur's lemma, and I don't see the relevance here. I understand Schur's Lemma to be something like this: that if you have an irreducible representation of a algebra on a vector space, and a linear operator on that vector space commutes with that representation for every element in the algebra, then the linear operator is either 0 or invertible.

How does this reduce down to "since eigenvectors of $\gamma^{5}$ whose eigenvalues are different transform without mixing"?

Answer 1

The reasoning is supposed to go as follows:

1. $\gamma^5$ commutes with all algebra elements, hence with the whole image of the algebra representation.

2. $\gamma^5$ has at least two different eigenvalues, meaning it is not a scalar multiple of the identity.

3. If the representation of the $S^{\mu\nu}$ (that form the Lorentz algebra $\mathfrak{so}(1,3)$) were irreducible, $\gamma^5$ would be a scalar multiple of the identity by Schur's lemma, which would contradict 2.

4. Therefore, the representation of the $S^{\mu\nu}$ must be reducible.

Caveat: The Dirac representation is irreducible as the representation of the Clifford algebra, see e.g. this question and its answers.

Answer 2

@ACuriousMind 's answer is impeccable, but since it appears you are forgetting the routine usage of the term "Schur's lemma" in physics, and you are glomming onto the more recondite mathematical corners of it (the "second lemma"), instead of the straightforward criterion of reducibility invoked (the "first lemma"), I might as well remind you where you really have used it before, when you learned about spin.

Suppose somebody gave you a 5-dimensional representation J of SU(2), where the 3 generators given satisfy the standard Lie algebra of angular momentum, and you wanted to know whether it is the spin 2 irreducible representation, 5 , or else a reducible representation, say the direct sum of spin 1 and spin 1/2, that is 3 ⊕ 2. What do you do, without transforming the Js to bring them to a recognizable form?

You evaluate the quadratic Casimir, C = J·J, which, by dint of the algebra, commutes with all three Js. So, by Schur's lemma, if your rep is irreducible, the spin 2, it is proportional to the identity, the proportionality constant being 2(2+1)=6. If, by contrast, it is a diagonal matrix with different eigenvalues, three of them 2=1·(1+1), and two of them 3/4=1/2(1/2+1), then by this criterion of reducibility your rep must be reducible: your operator C can be similarity-transformed to 2 I(3) ⊕ 3/4 I(2) , and so then J = L ⊕ σ/2, a block matrix with spin 1 gens L in the upper block and Pauli matrices in the lower, disjoint block.

Edit per request on ref: Dimitry Vvedenski's (Imperial College) Group Theory book has a neat coverage of both Schur Lemmas.