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I have a question about an excerpt from Peskin & Schröder "Introduction to QFT" (see below). I understand the claim that I have marked as:

"Let $\{t^a\}$ be a basis of a Lie algebra such that there is an irreducible representation $d$ such that tr$[d(t^a) d(t^b)] = C(d) \delta^{ab}$ for some non-zero constant $C(d)$. Then for every other irrep. $d'$, there is a constant $C(d')$ such that tr$[d(t^a) d(t^b)] = C(d) \delta^{ab}$."

My first question is whether this interpretation is correct. My second is: If it is correct, then what is a nice argument to see that it is true? (A reference to some textbook containing a proof would also be very welcome).

Excerpt from Peskin Schröder

Qmechanic
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SPHerical
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1 Answers1

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I found an answer to my question in chapter 14.9 of "Symmetries, Lie Algebras and Representations" by Jürgen Fuchs an Christoph Schweigert. If anyone else ever stumbles over this SE page: The precise statement would be

Let $\mathfrak{g}$ be a simple Lie algebra and let $\{t^a\}$ be a basis of it such that there is a representation $d$ of $\mathfrak{g}$ such that tr$[d(t^a) d(t^b)] = C(d) \delta^{ab}$ for some non-zero constant $C(d)$. Then for every other representation $d'$, there is a (possibily zero) constant $C(d')$ such that tr$[d(t^a) d(t^b)] = C(d') \delta^{ab}$.

The proof is a beautifully simple argument: Given an arbitrary rep. $d$, we can define an ad invariant inner product on $\mathfrak{g}$ via tr$[d(t^a) d(t^b)]$. But for simple Lie algebras, the Killing form is the unique ad invariant inner product up to a constant prefactor. Thus all of these inner product are proportional to each other.

SPHerical
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