3
$\begingroup$

I'm confused about application of the Hodge star operation within Yang Mills theory. Using differential forms, and given the connection $A$, the curvature in Yang-Mills theory is $F=dA+A \wedge A$. The Bianchi identities and the field equations are then $d_{A}F=0$ and $*d_{A}*F=0$, where $d_{A}$ is the covariant derivative $d_{A}F=dF+[A,F]$. Now the commutator of a $P$-form and $Q$-form is $[P,Q]=P\wedge Q-Q\wedge P$ if either or both of $P$, $Q$ are even, and $[P,Q]=P\wedge Q+Q\wedge P$ if both $P$ and $Q$ are odd. I want to express the Bianci identity and the YM equations in full in terms of $A$ and the wedge product. In full, the Bianchi identity in terms of connections and wedges is \begin{equation} d_{A}F=dA\wedge A-A\wedge dA + A\wedge dA + A\wedge A\wedge A - dA\wedge A - A\wedge A \wedge A= 0. \end{equation} But what is the corresponding full expression in terms of $A$ for the YM equations $*d_{A}*F$? And/or how is it derived directly from the action $\int *F \wedge F$? It is the correct use of the Hodge star operation that I'm messing up on. (Can't find any notes/texts that answers this in detail.)

$\endgroup$
1
$\begingroup$

The Yang-Mills action is $$ S = \int F \wedge \ast F $$ Its variation is \begin{align} \delta S &= 2 \int \delta F \wedge \ast F = 2 \int ( d \delta A + \delta A \wedge A + A \wedge \delta A ) \wedge \ast F \\ &= 2 \int \delta A \wedge ( d \ast F + A \wedge \ast F - \ast F \wedge A ) \\&= 0 \end{align} The equations are then $$ d \ast F + A \wedge \ast F - \ast F \wedge A = d_A \ast F = 0 ~. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.