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  Mistake or Rewriting of Yang-Mills in Nakahara

+ 0 like - 0 dislike

I am familiar with Yang-Mills equation of motion E.O.M. (without matter or source fields) in differential form.

$$ D * F =0 $$ and Bianchi identity $$ D F=0 $$ where $F= dA + A \wedge A$ and $D=d + [A, ]$ as the covariant derivative version of exterior derivative $d$.

However, in Nakahar book Geometry, Topology and Physics, Second Edition ,

we can compare E.O.M. to his (1.269) below,

and Bianchi identity to his (1.266) below.

My question is that: Did Nakahara make any mistake? Or are his equations the rewriting of my Yang-Mills Equations above? If so, how do we convert to make the rewriting precise?

enter image description here

This post imported from StackExchange Physics at 2020-11-09 09:44 (UTC), posted by SE-user annie marie heart
asked May 19, 2019 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
Note that the indices in Nakahara's formulae are contracted and have only one free index, while yours are equations for 3-forms and would have 3 free indices. Now think about what transforms a 3-form to a 1-form...

This post imported from StackExchange Physics at 2020-11-09 09:44 (UTC), posted by SE-user ACuriousMind
...in 4 spacetime dimensions

This post imported from StackExchange Physics at 2020-11-09 09:44 (UTC), posted by SE-user Kosm
You gotta remember that to get a "normal" contraction you always have to throw in a Hodge star. For example, the action contains $F \wedge \star F$, not $F \wedge F$ even though the $F$'s are just contracted in index notation. You can show this by just expanding everything in components.

This post imported from StackExchange Physics at 2020-11-09 09:44 (UTC), posted by SE-user knzhou

1 Answer

+ 2 like - 0 dislike

Let's take abelian gauge theory and Minkowski metric for simplicity.

Apply Hodge star to equations of motion $d*F=0$: $$ *d*F=*(\frac{1}{2}\partial_\mu\tilde{F}_{\nu\rho}~dx^\mu\wedge dx^\nu\wedge dx^\rho)=\frac{1}{2}\partial_\mu \tilde{F}_{\nu\rho}~dx^\sigma \eta_{\sigma\omega}\epsilon^{\omega\mu\nu\rho}=\\ =-\partial_\mu F^{\omega\mu}\eta_{\omega\sigma}dx^\sigma=0~\rightarrow~ \partial_\mu F^{\omega\mu}=0, $$ where I used $\tilde{F}_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}$ and Levi-Civita contractions. Bianchi identities can be "derived" in the same way (by expressing $F_{\mu\nu}$ in terms of $\tilde{F}_{\mu\nu}$).

Or you can simply convince yourself that the following expressions of Bianchi identities are equivalent $$ \partial_\mu F_{\nu\rho}+\partial_\rho F_{\mu\nu}+\partial_\nu F_{\rho\mu}=0~\Leftrightarrow~\epsilon^{\sigma\mu\nu\rho} \partial_\mu F_{\nu\rho}=0. $$ Substitute $\tilde{F}$ above for equations of motion.

This post imported from StackExchange Physics at 2020-11-09 09:44 (UTC), posted by SE-user Kosm
answered May 19, 2019 by Kosm (55 points) [ no revision ]

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