# Tag Info

2

Leibniz rule holds for covariant derivatives, both in gauge theories and gravity. Mathematically, a derivation is one for which the Leibniz rule holds. How does it work for non-abelian covariant derivatives. I will give you an example. Let $\Phi^\dagger \Phi$ be invariant under local non-abelian gauge transformations. Then $$\partial_\mu (\Phi^\dagger ... 0 The value of lambda must be fixed if you want gauge invariance, written in differential forms is 1/3,$$S_{CS}=\frac{k}{4\pi}\int \langle A\wedge dA+\frac{2}{3}A\wedge A\wedge A \rangle\,, where $A=A_\mu dx^\mu$ and $\langle \rangle$ is an invariant quadratic trace (so you will normally work with particular types of Lie algebras, see the killing ...

1

I can't be sure what the source meant without seeing the context, however I suspect the author meant the following. A $U(1)$ gauge transformation acting on a charged scalar field gives: $$\phi (x) \rightarrow e ^{ i \alpha (x) } \phi (x)$$ Under such a transformation the normalization is invariant since $\phi$ simply gains a ...

2

I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian. II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point ...

Top 50 recent answers are included