Weak interactions seem the most universal after gravitation. A very few particles avoid them, only the photon and gluon, plus right leptons. The photon, however, is a part of the electroweak interactions and is a superposition of the B and W3 bosons. This leaves the gluon as one of a few particles immune to the electroweak interactions.
What is the reason for gluon not having a weak charge? Is it because the gluon is a gauge boson of a different type of interactions and different gauge bosons cannot share their charges with each other? However the Higgs boson does have the weak charge. Is this because the Higgs boson is not a gauge boson and this fact allows it to have the weak charge?
It would be great if someone could clarify the intuitive meaning of this setup and hopefully not just by formulas alone without an interpretation of their physical meaning.
As far as I can tell, all three answers are correct, just use different approaches. I am upvoting all three. The answer of @CosmasZachos is detailed, much appreciated. The answer of @annav is the closest to the question asked:
SU(3) has the gluon as the gauge boson, and it has, by construction, no weak vertices.
While to some this point may seem obvious enough to just skip and move straight to higher orders, it is worth pointing out to the rest of us. Checking Anna's answer as correct for being the closest to the actual question asked, although again, all three answers are very helpful.
To summarize, gluons do not directly interact weakly, because they are the gauge bosons of the SU(3) symmetry, which is defined independently of the SU(2) symmetry of the weak interactions. The reason for this definition is observation, but once the definition is in place, it becomes a theory prohibiting gluons from direct weak interactions. The higher order interactions (like those via virtual quarks in Anna's diagram) are not prohibited, but (as far as I understand) have not been observed.