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Understanding gauge fields as connections on a principal G-bundle

+ 4 like - 0 dislike
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I have read that gauge theories may be formulated in the language of differential geometry though the use of Ehresmann connections on a principal $G$ bundle, where $G$ is the Lie group of the theory. While I understand the various equivalent definitions of the connection as either a $\mathfrak{g}$-valued 1-form or a decomposition of the tangent bundle into vertical and horizontal subspaces, I have yet to understand how the connection may be identified with a gauge field or how it's curvature may be identified with the field strength.

On an abstract level, I see how there could be a correspondence between these ideas. It makes sense that a gauge transformation could be thought of as a type of parallel transport and hence could be described via a connection, but I have not seen an explicit mathematical formalism that describes this correspondence. Could someone perhaps provide me with some insight?

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user Jackson Burzynski
asked Nov 30, 2016 in Theoretical Physics by Jackson Burzynski (20 points) [ no revision ]
The gauge field is the connection on the principal bundle. Is that what you are asking? I'm not completely sure I'm understanding you

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user coconut
@coconut The way I have seen the gauge field introduced in particle physics textbooks is as a correction term by demanding that a global symmetry hold locally. My question I suppose is how is this realized geometrically in terms of a connection on the principal bundle.

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user Jackson Burzynski

2 Answers

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The correspondence between the connection and the gauge field is as usual in quantum field theory -- this geometrical object is interpreted as an integration variable in the Feynman integral. In absolutely the same way sections of associated bundles are interpreted as matter fields, metric on the manifold as the gravitational field, maps from the manifold to some target space as some fields in $\sigma$-models and so on. There is an enlightening discussion of the subject in Ch. 8 of this beautiful book on Mirror Symmetry.

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user Andrey Feldman
answered Nov 30, 2016 by Andrey Feldman (600 points) [ no revision ]
+ 2 like - 0 dislike

A gauge field is a connection on a principal bundle. I'll try to show here roughly how this formulation is related to the common construction in physics of the gauge fields in the process of making locally invariant a theory that only has global invariance.

Let $G$ be a Lie group, $G\times V\rightarrow V:(g,v)\mapsto gv$ a representation of $G$, $M$ the space-time manifold, $E\rightarrow M$ a principal $G$-bundle and $F\overset{\pi}{\rightarrow} M$ an associated bundle with fiber isomorphic to $V$. The space of gauge transformations $\Gamma(E)$ is equipped with a canonical action over the space of fields $\Gamma(F)$ as $(g,\phi)\mapsto g\phi$ where $(g\phi)(x)=g(x)\phi(x)$.

Observe that for a constant $g$, derivatives have the nice property that $\partial_\mu(g\phi)=g\partial_\mu\phi$. This isn't true for a general gauge transformation, so we want to construct an object similar to $\partial_\mu$ that satisfies the previous equation for any $g$. By adding a connection $A$ of the principal bundle $G$ (and using the Lie algebra representation $(T,v)\mapsto Tv$ corresponding to the representation of $G$ over $V$) we can define the covariant derivative $D_\mu$ to be $D_\mu\phi(x)=\partial_\mu\phi(x)+A_\mu\phi(x)$. Now we have our object $D_\mu$ for which the equation $D_\mu(g\phi)=gD_\mu\phi$ holds for any $g$.

The action functional is a map $S:\Gamma(F)\rightarrow\mathbb{R}$. It is said to have global invariance under $G$ if $S[\phi]=S[g\phi]$ for every constant gauge transformation $g$. If the condition $S[\phi]=S[g\phi]$ holds for any $g\in\Gamma(E)$ the action is said to be locally (gauge) invariant.

Suppose that we want to construct locally invariant actions from globally invariant ones. Usually, global invariance will depend on the fact that $\partial_\mu(g\phi)=g\partial_\mu\phi$, so an action with global invariance will not have in general local invariance. However, by replacing every $\partial_\mu$ by a $D_\mu$ in the expression for the action one obtains a new one which extends invariance to every gauge transformation because now $D_\mu(g\phi)=gD_\mu\phi$ is satisfied for all $g$.

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user coconut
answered Nov 30, 2016 by coconut (45 points) [ no revision ]
Thank you very much, this helped clear things up for the most part.

This post imported from StackExchange Physics at 2017-02-14 10:48 (UTC), posted by SE-user Jackson Burzynski

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