0
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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... $\endgroup$ – ACuriousMind May 19 at 9:38
  • $\begingroup$ ...in 4 spacetime dimensions $\endgroup$ – Kosm May 19 at 9:45
  • $\begingroup$ 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. $\endgroup$ – knzhou May 19 at 12:43
2
$\begingroup$

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.

$\endgroup$

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.