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How does defining the curvature in Yang-Mills as $F = d_A A$ make sense?

+ 3 like - 0 dislike

In Yang-Mills theory, the covariant differential is defined as

$d_A \phi = d\phi+[A,\phi].$

The curvature is defined as

$F = dA + \frac{1}{2}[A,A],$

note the factor of $\frac{1}{2}$. Yet, I often see authors define the curvature as

$F = d_A A.$

How does this definition make sense, given the factor of $\frac{1}{2}$ difference?

asked Mar 18 in Theoretical Physics by Zoe (15 points) [ no revision ]
reshown Mar 18 by Dilaton

Notice that this factor is mostly a matter of convention. By simply rescaling the structure constants in the Lie algebra (which may always be done) it may be changed at will

1 Answer

+ 4 like - 0 dislike

This corresponds to the case that the G-bundle over which you consider the connection 1-form $A$ and hence the curvature $F$ is non-abelian. That is, in the case you consider, say a $U(1)$-bundle the curvature on the associated bundle will be just 

$$ F = dA$$

The symbol $d_A$ means you are dealing with the non-abelian case and hence you get the extra commutator. The factor of $1/2$ is there due to antisymmetry. Try to expand the brackets by picking a trivilization over frame and you will see that the factor is needed so that all is good. 

answered Mar 18 by conformal_gk (3,505 points) [ no revision ]

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