# Confusion about tensors in $SU(3)$

I have some confusion regarding the notion of tensors in $$SU(3)$$ (or some other matrix Lie group, but let's keep the discussion to $$SU(3)$$).

For concreteness, I will refer to Peskin and Schroeder's discussion on this topic (see chapter 17.1), but I think the discussion is similar in many sources. They represent quarks by $$q_i$$, transforming according to the fundamental representation, and antiquarks by $$\bar{q}^i$$, transforming according to the anti-fundamental representation. While not spelled out very explicitly, it seems they only put the matrix indices downstairs. Hence the transformation laws should look like $$q'_{i} = U_{ij}q_j$$ and $$\bar{q}'^i = U_{ij}^* \bar{q}^j = (U^\dagger)_{ji}\bar{q}^j.$$ First of all, is this a correct understanding of their notation (Q1)?

Moving on, they define $$\epsilon_{ijk}$$ as the totally antisymmetric combination of three fundamental representations. Thus, it transforms according to $$\epsilon'_{ijk} = U_{il}U_{jm}U_{kn}\epsilon_{lmn} = \mathrm{det}(U)\epsilon_{ijk} = \epsilon_{ijk},$$ by virtue of $$\mathrm{det}(U) = 1$$. They also use the symbol $$\epsilon^{ijk}$$ but do not explicitly define it. If the index rules are consistent, I reckon it should transform according to $$\epsilon'^{ijk} = U_{il}^* U_{jm}^* U_{kn}^* \epsilon^{lmn} = (U^\dagger)_{li} (U^\dagger)_{mj} (U^\dagger)_{nk} \epsilon^{lmn} = \mathrm{det}(U^\dagger)\epsilon^{ijk}=\epsilon^{ijk},$$ by virtue of $$\mathrm{det}(U^\dagger) = 1$$. Is this a correct understanding of the meaning of $$\epsilon^{ijk}$$ in their notation (Q2)?

Next, they state that the only allowed (simple) hadrons are $$\bar{q}^iq_i$$, $$\epsilon^{ijk}q_iq_jq_k$$ and $$\epsilon_{ijk}\bar{q}^i\bar{q}^j\bar{q}^k$$, from the requirement that hadrons be invariant under $$SU(3)$$. However, this statement somewhat surprises me given the transformation laws established above. For any $$SU(3)$$ tensor $$f^{ijk}$$ (not related to the structure constants), where the upstairs indices would transform just like they do for $$\epsilon^{ijk}$$, it seems to me that the object $$f^{ijk}q_iq_jq_k$$ would be invariant simply by virtue of being fully contracted. To whit: $$f^{ijk}q_iq_jq_k \to f'^{ijk}q'_i q'_j q'_k = U_{il}^* U_{jm}^* U_{kn}^{*}f^{lmn}U_{ir}q_rU_{js}q_sU_{kt}q_t = (U^\dagger)_{li}U_{ir}(U^\dagger)_{mj}U_{js}(U^\dagger)_{nk}U_{kt}f^{lmn}q_rq_sq_t = \delta^l_r\delta^m_s\delta^n_tf^{lmn}q_rq_sq_t=f^{lmn}q_lq_mq_n = f^{ijk}q_iq_jq_k.$$ Am I missing anything here (Q3)?

Thinking a bit more about this, perhaps the issue lies in how we would actually go about defining the object $$f^{ijk}$$ introduced above. In the case of the Levi-Civita, the invariance, in the sense that the components stay the same, means that we can sensibly define the tensor $$\epsilon^{ijk}$$. However, for an object like $$f^{ijk}$$, which may not be invariant, we would have to define its components in some preferred $$SU(3)$$ "frame" (for lack of a better word), which probably does not make much sense. Is this where the issue lies (Q4)?

I would like to add that I am well aware that one sometimes does not view the Levi-Civita $$\epsilon^{ijk}$$ and $$\epsilon_{ijk}$$ as tensors, but rather as index symbols, not subject to any $$SU(3)$$ transformations. To me, this is the more sensible viewpoint. In this case, the invariance of $$\epsilon^{ijk}q_iq_jq_k$$ does depend crucially on the properties of the Levi-Civita, because the transformation now looks like $$\epsilon^{ijk}q_iq_jq_k \to \epsilon^{ijk}q'_i q'_j q'_k = \epsilon^{ijk}U_{ir}q_rU_{js}q_sU_{kt}q_t = \mathrm{det}(U)\epsilon^{rst}q_rq_sq_t = \epsilon^{rst}q_rq_sq_t = \epsilon^{ijk}q_iq_jq_k,$$ by virtue of $$\mathrm{det}(U)=1$$. However, Peskin and Schroeder do indeed define $$\epsilon_{ijk}$$ and $$\epsilon^{ijk}$$ as tensors, so this argument does not apply.

• Related : Symmetry in terms of matrices. Commented Apr 17 at 10:22
• @Frobenius Thanks! Commented Apr 19 at 21:29

• (Q2) : As you point out later in your post, there are two distinct ways of thinking about what $$\epsilon_{ijk}$$ is. Indeed I agree that what you call the "index symbol" way of thinking about $$\epsilon_{ijk}$$ is easier to reason about. In this framework, what you have are quark representations $$q_i$$ (fundamental representation "$$3$$", labelled by $$\square$$ in Young Tableaux notation), and antiquark representations $$\overline{q}^i$$ (antifundamental representation "$$\overline{3}$$", labelled by two squares stacked on top of one another¹). You can then play the game of taking tensor products of these basic representations to try and find $$SU(3)$$-singlets. For example, taking one quark and one antiquark: $$3 \otimes \overline{3} = 1 + 8$$ and you find that there's a singlet and an octet. The two different representations are given by different ways of contracting the quark and antiquark indices together, for instance (as you point out in your post), you have mesonic operators $$\overline{q}^i q_i$$ corresponding to the singlet. The octet is given by $$\overline{q}^i {T^{Ai}}_j q_j$$ where $$T^A$$ are the Gell-mann matrices. Now, for the case of three quarks you find that: $$3 \otimes 3 \otimes 3 = 1 + 8 + 8 + 10, \qquad \text{but} \qquad \wedge^3(3) = 1$$ The naive tensor product of three-copies of the quark representation suggests you should get $$1 + 8 + 8 + 10$$, but since quarks are fermionic, you need to impose the anticommutativity, hence take the wedge product of the representations instead, where you find there is only a singlet. $$\epsilon^{ijk}$$ now is simply a tensor describing how to perform the contraction of the three quark representations into a singlet $$\epsilon^{ijk} q_i q_j q_k$$, and it itself does not transform under $$SU(3)$$. But now comes the question: I know from above that I should expect $$\epsilon^{ijk}$$ to exist, but how do I know how to write it down? We expect: $$\epsilon^{ijk} U_{ia} U_{jb} U_{kc} q_a q_b q_c = \epsilon^{ijk} q_i q_j q_k \qquad \implies \qquad \epsilon^{ijk} U_{ia} U_{jb} U_{kc} = \epsilon^{ijk}$$ Thus, if we acted on $$\epsilon^{ijk}$$ by pretending the upper indices were conjugate indices, it would have to transform as a singlet (which is what Peskin and Schroder write down). And we have an explicit example of such an object, the fully antisymmetric tensor.
• Thank you very much for your answer! In the tensor product $3\otimes 3\otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$, is it not the case that the singlet by definition is fully symmetrized, as we antisymmetrize over columns in Young diagrams? From this perspective, it is a bit unnatural to call the Levi-Civita a "tensor", as it just appears as a shorthand for writing down a certain linear combination. Commented Apr 16 at 20:05
• However, I still feel that I lack insight regarding Q3 and Q4. Just like P&S simply define $\epsilon_{ijk}$ to be the totally antisymmetric combination of three 3's, let me define $f_{ijk}$ to be the totally symmetric combination of three 3's. However, this object would transform like $f'_{ijk} = U_{il}U_{jm}U_{kn}f_{lmn}$ which is no longer totally symmetric. Thus, $f_{ijk}$ is ill-defined. It seems to me, then, that $\epsilon_{ijk} = U_{il}U_{jm}U_{kn}\epsilon_{lmn} = \epsilon_{ijk}$ shows that $\epsilon_{ijk}$ is invariant, yes, but more importantly that it is actually well-defined! Commented Apr 16 at 20:13