[Physics] How does Schur’s Lemma mean that the Dirac representation is reducible

dirac-matricesrepresentation-theory

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"?

Best Answer

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.