# Tag Info

6

That's because you are forgetting that $A$ has a Yang-Mills index. You better write this in components, which reads $\epsilon^{\mu\nu\rho} g_{IJ} \Big( A^I_\mu \partial_\nu A_\rho^J + \frac{1}{3} f^J{}_{KL} A^I_\mu A^K_\nu A^L_\rho \Big)$ This component notation also answers your second question.

4

The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...

4

No, the gauge group of electromagnetism is still $\mathrm{U }(1)$. This is because there is a difference between the gauge group $G$ of a physical theory on a manifold $M$ and the group of gauge transformations $\mathcal{G}$ consisting of local functions $M\to G$ and the algebra of infinitesimal gauge transformations consisting of local functions ...

4

Yes, you missed the whole point of gauge invariance which clearly has to be a symmetry – the Hamiltonian has to be and is invariant under these transformations. In the case of a global symmetry, a Hamiltonian may afford "not to be symmetric" under it. But once a group of transformations is said to be a gauge symmetry, the failure of $H$ to be invariant would ...

2

I think what they mean is FI-term is not gauge invariant under the full gauge symmetry of the theory, but under this remaining gauge freedom after WZ gauge, which is $U(1)$.

2

I believe this is a valid gauge condition. You can use the hint given by AccidentalFourierTransform in a comment and for arbitrary $A^\mu$ find such $\Omega$ that $(A_\mu+\Omega_{,\mu})(A^\mu+\Omega^{,\mu})=0$, as the latter equation can be resolved with respect to $\Omega_{,0}$, so a Cauchy problem can be posed. Of course, the usual mathematical issues of ...

1

A massless vector is the gauge field $A_\mu$ and the massless symmetric traceless tensor is the metric perturbation $h_{\mu\nu}$. They can couple only to conserved currents, namely the $U(1)$ vector current $j_\mu$ and the stress tensor $T_{\mu\nu}$ respectively. The requirement for this arises from gauge invariance. Coupling a gauge field to a current ...

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