# What precisely and mathematically does it mean to have $W$ bosons carry electric charges?

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What precisely and mathematically does it mean to have $$W$$ bosons carry electric charges?

We know from Wikipedia that experiments say that $$W$$ bosons carry electric charges: $$W^\pm$$ carry $$+$$ and $$-$$ of electric charges.

However, the gauge fields do not sit at any representation of the gauge group, but the gauge fields transform under the adjoint action of the gauge group. See eg https://physics.stackexchange.com/a/191982/42982.

So how can $$W^\pm$$ carry $$+$$ and $$-$$ of electric charges, which are charge 1 and -1 representation of $$U(1)$$ electromagnetic gauge group? But gauge fields do not sit at any representation of the gauge group?

Some may say, "well it is easy, we have lagrangian terms (which need to be neutral) like $$\bar{\nu}_e W^{+ \mu} \partial_{\mu} e^-.$$ $$\bar{\nu} W^{- \mu} \partial_{\mu} e^+.$$ So these explain why $$W^\pm$$ carry $$+$$ and $$-$$ of electric charges."

But again gauge fields do not sit at any representation of the gauge group (say $$U(1)$$?), precisely and mathematically.

This post imported from StackExchange Physics at 2020-12-13 12:44 (UTC), posted by SE-user annie marie heart
What exactly do you mean by saying that the gauge fields do not sit at any representation of the gauge group but they transform under the adjoint representation?

This post imported from StackExchange Physics at 2020-12-13 12:44 (UTC), posted by SE-user Oбжорoв
The adjoint action is a representation of the group (cf. en.wikipedia.org/wiki/Adjoint_representation). The post of mine you link is about the difference between the global symmetry group $G$ and the group of gauge transformations and does not seem relevant here. I don't understand the question.
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