Explanation for non-abelian gauge symmetry leading to charged gauge bosons Is there a (somewhat) intuitive explanation for why a non-abelian gauge symmetry leads to its gauge bosons requiring boson-boson interactions (being charged)?
Any QFT text derives this result but in tens of pages of math and usually just end with concluding that for example yes, the weak interaction leds to 3 and 4 boson vertices..  I'm not allergic to the math per se but it feels as a quite important consequence and should have a condensed, intuitive explanation.
 A: In non-abelian gauge theories, two generators of infinitesimal gauge transformations don't commute:
$$
[t^a,t^b]=f^{abc}t^c~,\tag{1}
$$
where we don't distinguish upper and lower Lie algebra indices.
Now take a general definition of a field strength,
$$
F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu - [A_\mu,A_\nu]~.\tag{2}
$$
For an abelian gauge theory the last term (commutator) vanishes, because there is only one generator and it obviously commutes with itself. For a non-abelian gauge theory
$$F_{\mu\nu}=F^a_{\mu\nu}~t^a~,~A_\mu=A^a_\mu~t^a~.$$
Plugging this into $(2)$ you can see that since generators don't commute $\rightarrow(1)$, the last term doesn't vanish and $~F_{\mu\nu}F^{\mu\nu}~$ leads to trilinear and quartic terms which are absent in abelian gauge theories.
A: The field strength tensor $\mathbf{F} \equiv F_{\mu\nu}^a T^a$ of the gauge field transforms in the adjoint representation. Under a gauge transformation $V$, the field strength tensor transforms explicitly as $\mathbf{F} \to V \mathbf{F} V^{-1}$. In the U(1) case, everything commutes, so $\mathbf{F}$ doesn't transform, and we say that the gauge boson is uncharged. However, in the general non-abelian case, the transformation is non-trivial, which is why we say that the gauge boson is charged.
A: From the other way around it is perhaps easier to ask why boson-self interactions require a non-abelian gauge symmetry. If you write down the most general 3-particle amplitude for boson self-interaction you find that it violate bose-symmetry unless the coupling is anti-symmetric.
Taking it a step further you can glue together two 3-particle amplitudes to find a 4-particle amplitude. One then finds that the only way to preserve Unitarity and locality is if the coupling constant satisfy the Jacobi identity, i.e if it is a non-abelian theory.
So in summary boson-self interactions leads to a non-abelian theory due to locality, unitarity and lorentz invariance.  This is discussed in great detail in Schwartz & Elvang. 
