# Confusion about group representation terminology in physics

## Background

I see a lot of confusing group representation terminology in physics writing. Here is a typical example, taken from D. J. Griffiths' Introduction to Elementary Particles, talking about quark flavor combinations in baryons:

As for flavor, there are $$3^3 = 27$$ possibilities: $$uuu, uud, udu, udd, \dotsc, sss$$, which we reshuffle into symmetric, antisymmetric, and mixed combinations; they form irreducible representations of $$SU(3)$$, just as the analogous spin combinations form representations of $$SU(2)$$.

## Vantage point

I am very new to representation theory, but as I understand it, a representation is a homomorphism $$\rho : G \to GL(V)$$ from a group $$G$$ (of transformations) to the general linear group of a vector space $$V$$. Hence $$V$$ is where the spin (or isospin, or flavor, etc.) vectors "live", while the representation provides the matrices transforming elements of $$V$$. Now, some representations have a (proper nontrivial) invariant subspace $$V_1 \subset V$$, for which $$\rho(g)v \in V_1$$ if $$v \in V_1$$. Such a representation is called reducible (or more accurately decomposable or completely reducible, i believe) and may be block diagonalized, $$\rho_D(g) = U \rho(g) U^{-1} = \begin{pmatrix} \rho_1(g) & 0 \\ 0 & \rho_2(g) \end{pmatrix},$$ by some invertible matrix $$U$$. In other terms, $$\rho_1$$ and $$\rho_2$$ are representations of $$G$$ on $$V_1$$ and $$V_2$$ (where $$V_2 = V \setminus V_1$$ is also an invariant subspace), respectively, and we have $$V = V_1 \oplus V_2$$ and $$\rho_D = \rho_1 \oplus \rho_2$$. A representation that is not reducible is called an irreducible representation.

## Question

So then, if the above is correct, isn't the quoted terminology wrong? Griffiths says that the symmetric, antisymmetric, and mixed flavor combinations form irreducible representations of $$SU(3)$$, but as I understand it these combinations are not even part of the representation, but elements in a vector space on which the representation acts. Would not the correct statement be that these combinations form invariant subspaces under $$SU(3)$$, and that each such invariant subspace transforms according to some irreducible representation of $$SU(3)$$?

• The answer to your concluding rhetorical question is, of course, "yes". Your instructor did not explain to you that physicists informally conflate the vector spaces and subspaces with the matrices and block matrices that act on them in characterizing both as representations? – Cosmas Zachos Apr 26 at 15:48
• @CosmasZachos Well I would not call it a rhetorical question. Rather something that I was almost sure about, but for which I wanted confirmation by someone more familiar with the subject, so that I could put my worries to rest. But now with your answer I feel very confident, so thank you! Unfortunately this was brushed over pretty quickly in lectures, and what little understanding of representation theory I have, I had to read on the side. – ummg Apr 26 at 16:07
• @CosmasZachos If you post (some variation of) your comment as an answer, I will mark it as accepted. If not, in due time, I might write an answer myself, citing your comment as per the recommendations here. – ummg Apr 26 at 16:55
• Just to add that I had exactly the same confusion when I was learning this material and no one ever outright told me that physicists use the word "representation" in a lazy way. Your final statement in your question (ending with a question mark) is correct. (As I have often found to be the case... ultimately the physics way of talking about things is very convenient and efficient because it focuses on what is important for physics, but can make it difficult to learn because it is not precise) – Andrew Apr 26 at 17:17
• – Qmechanic Apr 26 at 23:04

that the symmetric, antisymmetric, and mixed flavor combinations [...] form invariant subspaces under $$SU(3)$$, and that each such invariant subspace transforms according to some irreducible representation of $$SU(3)$$