Right, so in Yang-Mills theory, the vector potential is modified from:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]F = dA[/tex]

To:

[tex]F = dA + A\wedge A[/tex]

However, it is my understanding that the exterior/wedge product is anticommutitive, so that for a given exterior algebra over a vector space, V:

[tex]\omega \wedge \omega = 0, \forall \omega \epsilon \Lambda(V) [/tex]

Why then is the second term in the curvature, F, not non-zero? I assume I'm missing something, could someone fill me in?

(Sorry, this probably fits better in Topology & Geometry section, but the question technically is a question about multilinear algebra)

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# A question about the Exterior Product in Yang-Mills theory

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