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If we have a general gauge group whose action is $$ \Phi(x) \rightarrow g(x)\Phi(x), $$ with $g\in G$.

Then introducing the gauge covariant derivative $$ D_{\mu}\Phi(x) = (\partial_{\mu}+A_{\mu})\Phi(x).$$

My notes state the gauge potential $A_{\mu} \in L(G)$, $L(G)$ being the Lie Algebra of the group $G$.

What's the connection between the Lie Algebra of the group and the gauge potential?

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The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation $g:\Sigma\to G$ (on a spacetime $\Sigma$) connected to the identity may be written as $\mathrm{e}^{\mathrm{i}\chi(x)}$ for some Lie algebra valued $\chi: \Sigma\to\mathfrak{g}$. The derivative of a transformed field is $$ \partial_\mu(g\phi) = \partial_\mu(g)\phi + g\partial_\mu\phi = g(g^{-1}(\partial_\mu g) + \partial_\mu)\phi$$ and it is the $g^{-1}(\partial_\mu g) = \partial_\mu\chi$ that we want to cancel here by adding the gauge field so that $D_\mu(g\phi) = gD_\mu\phi$. Since $\partial_\mu\chi$ is Lie algebra valued, so must the gauge field $A$ we add, and it has to transform as $$ A\overset{g(x)}{\mapsto} gAg^{-1} - g^{-1}\mathrm{d} g$$ to cancel the terms we want to cancel.

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  • $\begingroup$ what is $\mathrm{d}$ again? Total derivative? $\endgroup$ – SuperCiocia May 3 '15 at 22:03
  • $\begingroup$ @SuperCiocia: The exterior derivative, a component-free notation for the spacetime vector $\partial_\mu g$ in this case (note that I didn't write indices on either side of the transformation). $\endgroup$ – ACuriousMind May 3 '15 at 22:17
  • $\begingroup$ from what I'm reading on Wikipedia it looks like $\mathrm{d}$ is a differential and not a derivative...? As in, it's a $\mathrm{d}f$ term instead of a $\frac{\partial}{\partial x}f$ $\endgroup$ – SuperCiocia May 3 '15 at 22:33
  • $\begingroup$ @SuperCiocia: What do you mean "It's a differential, not a derivative"? The exterior derivative maps $k$-forms to $k+1$-forms, and since $A$ is a $1$-form (the dual of a vector field), but $g$ is a $0$-form on spacetime, both summands in the transformation law need to be $1$-forms, and so the $\mathrm{d}$ there is an exterior derivative. You may as well write it in coordinates if you are uncomfortable with my notation: $A_\mu\mapsto gA_\mu g^{-1} - g^{-1}\partial_\mu g$. $\endgroup$ – ACuriousMind May 3 '15 at 22:39
  • $\begingroup$ sorry, I didn't mean to doubt you. Thanks very much for the answer. Since we are here already, I have always wondered by the commutator of the gauge covariant derivatives is equal to the field strength, i.e. $[D_{\mu}, D_{\nu}] \propto F_{\mu\nu}$... Does it have something do to with the Lie Algebra in this question? Or should I ask a different question? $\endgroup$ – SuperCiocia May 4 '15 at 9:59

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