I am very confused about what are the branching rules for representations of $E6$ into a $SU(3)\times SU(3)\times SU(3)$ subgroup. At least in the physics literature, there seems to be a serious mismatch among various famous papers.

For concreteness, let us focus on the adjoint representation of $E6$, the one with dimension $78$. It seems that there are (at least) two ways to do this:

Case A) $78\to (8,1,1)\oplus (1,8,1)\oplus (1,1,8)\oplus (3,3,3)\oplus (\bar{3},\bar{3},\bar{3})$

Case B) $78\to (8,1,1)\oplus (1,8,1)\oplus (1,1,8)\oplus (3,\bar{3},\bar{3})\oplus (\bar{3},3,3)$

In reference 1), in page 110, it is claimed that the branching rule is given by A). Also reference 2), equation C.7, page 58, agrees about case A). Also, by using the Mathematica package "LieArt", one can compute that the branching rule is given by A). However, references 3) and 4) argue about the correctness of case B).

This is extremely confusing to me, especially since references 2), 3), 4) all consider the same physical setup. So even if there are multiple $SU(3)^3$ subgroups of $E6$ with different branching rules, those papers should agree on which subgroup is considered in the physical setup, and the branching rules therefore must agree.

So, my concrete questions are:

1) Are there multiple $SU(3)\times SU(3)\times SU(3)$ subgroups of $E6$? If so, in what sense they are different?

2) If there are many $SU(3)^3$ inside $E6$, what are the branching rules for all of them?

3) On the physical grounds, consider the superconformal index of the $E6$ Minahan-Nemeschansky theory. The full flavor symmetry is $E6$ nut only $SU(3)^3$ is manifest in the class-S construction. Which is the correct $SU(3)^3$ subgroup to be considered, in this case?

Furthermore, nothe that the same thing happens for the representation of dimension $27$. Various references give different branching rules.


Reference 1) http://citeseerx.ist.psu.edu/viewdoc/download?doi=

Reference 2) https://arxiv.org/pdf/1310.3841.pdf

Reference 3) http://inspirehep.net/record/1358241

Reference 4) http://inspirehep.net/record/849570

  • $\begingroup$ Of course, the simply connected compact group $G$ of type $E_6$ has infinitely many subgroups isomorphic to ${\rm SU}(3)\times {\rm SU}(3)\times{\rm SU}(3)$. Indeed, let $H$ be one of them. Then consider $g\cdot H\cdot g^{-1}$, where $g\in G$. $\endgroup$ – Mikhail Borovoi Jul 29 at 21:54
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    $\begingroup$ Moreover, let ${\bf H}={\rm SU}(3)\times {\rm SU}(3)\times{\rm SU}(3)$, and let $ \iota\colon {\bf H}\hookrightarrow G$ be an embedding. There exist a number of pairwise non-conjugate such embeddings. $\endgroup$ – Mikhail Borovoi Jul 29 at 22:02
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    $\begingroup$ Let $\sigma\colon {\rm SU}(3) \to {\rm SU}(3)$ denote the outer automorphism of complex conjugation. Consider the automorphism $\tau={\rm id} \times\sigma \times\sigma$ of $\bf H$. Consider the embedding $$\iota'=\iota\circ\tau\colon{\bf H}\to{\bf H}\hookrightarrow G.$$ I suspect that, passing from $\iota$ to $\iota'$, you get case B from case A. Thus A and B are equivalent. $\endgroup$ – Mikhail Borovoi Jul 29 at 22:12
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    $\begingroup$ Since the representations labeled $1$ ans $8$ are self-dual, the difference between decompositions A and B amounts to the exterior automorphism $(g_1,g_2,g_3)\mapsto (g_1,\bar{g_2},\bar{g_3})$ of $G={\rm SU(3)\times SU(3)\times SU(3)}$. So you just need to be careful about a specific embedding of $G$ into the ambient group $H$ (what you call "E6"), including the labeling of the factors. (Edit cross-posted with the later comments by Mikhail Borovoi.) $\endgroup$ – Victor Protsak Jul 29 at 22:17

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