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Given Higgs fields, transforming as the representation $R$ of a group $G$. Does the physical fact that scalars=boson commute mean that we are only allowed to write symmetric tensor products of the representation in the Lagrangian?

Of course, every term in the Lagrangian must be invariant under $G$, but given a representation $R$ there are several possibilities to get an invariant.

Considering the cubic tensor product, we have in general

$$ R\otimes R \otimes R = (R_1 \oplus R \oplus \ldots)_{sym} \oplus (R_2 \oplus R \oplus \ldots)_{antisym} \oplus (R_3 \oplus R \oplus \ldots)_{mixedsym} $$

In general the representation $R$ appears in each part of this decomposition, i.e. in the symmetric, the antisymmetric and the mixed symmetric product and thus we can get something invariant under $G$ by multiplying these parts of the product with $R$. Nevertheless, the physical fact that Higgs field commute, means that we are only allowed to write the singlet from the tensor product with the symmetric cubic in the Lagrangian? I.e.

$$ ((R\otimes R \otimes R)_{sym}^{R} \otimes R )_1,$$

where $ ((R\otimes R \otimes R)_{sym}^{R}$ denotes $R$ projected from the symmetric part of the cubic product and the subscript $1$ at then end that we project $R \otimes R = 1 \oplus \ldots$ onto the singlet.

In contrast, $((R\otimes R \otimes R)_{antisym}^{R} \otimes R )_1$ and $((R\otimes R \otimes R)_{mixedsym}^{R} \otimes R )_1$ are forbidden, because of Bose-Einstein statistics?!

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