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By the Young Tableaux construction, a triplet of $SU(2)$ (diagramatically, two boxes side by side) is supposed to be a two indices symmetric tensor.

However, one of the most known and minimal extensions of the standard model is that of adding a complex colorless scalar triplet in the $(\mathbf{1},\mathbf{3},1)$ representation of the SM group:

$$ \mathbf{\Phi} \equiv \begin{pmatrix} \frac{\Phi^0}{\sqrt{2}} & \Phi^{++} \\ \Phi^0 & -\frac{\Phi^0}{\sqrt{2}} \end{pmatrix}, $$

which is not symmetric. Why?

A lot of times I have seen the triplet appearing as $\epsilon\mathbf{\Phi}$, where $\epsilon$ is the rank-2 antisymmetric tensor, and this matrix product is indeed symmetric, but I don't know exactly what is happening.

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  • $\begingroup$ Linked, and also here. $\endgroup$ – Cosmas Zachos Jul 13 at 16:35
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    $\begingroup$ For a real triplet (adjoint!), this matrix would be Hermitian. But for two such, a complex triplet, it is manifestly not. $\endgroup$ – Cosmas Zachos Jul 13 at 16:50
  • $\begingroup$ @CosmasZachos Then I would guess that the symmetry is "spoiled" because of the $U(1)_Y$ factor of the representation. Is this correct? $\endgroup$ – GaloisFan Jul 13 at 16:56
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    $\begingroup$ Well, indirectly. You need a complex doublet for the U(1) to have something nontrivial to act on. The easiest way to see this is how the two juxtaposed real adjoints transform, and how each came out of the symmetric composition of its own doublet... The indices of the Pauli vector matrix are not the symmetrized indices going into the makeup of the adjoints... $\endgroup$ – Cosmas Zachos Jul 13 at 17:03
  • $\begingroup$ I meant "complex triplet" above! $\endgroup$ – Cosmas Zachos Jul 13 at 18:08

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