4
$\begingroup$

I am aware of the differential form language proof of the fact that for arbitrary gauge transformations the Chern-Simons' term shifts by a WZW term (on the boundary).

But I am getting confused if in local coordinates I try to prove that under infinitesimal gauge transformations the Chern-Simons' term is invariant (may be upto total derivatives)

So I took my infinitesimal gauge transformation as $\delta A^a_\mu = \partial _\mu \epsilon ^a + f^{abc}A_\mu^b \epsilon ^c$ where my structure constants $f^{abc}$ are totally antisymmetric and cyclic. Then I look at the two terms that I have in the CS form, $\epsilon^{\mu \nu \lambda}A^a_\mu \partial_\nu A^a_\lambda$ and $\epsilon^{\mu \nu \lambda}f^{abc}A^a_\mu A^b_\nu A^c_\lambda$ and I vary them infinitesimally to get,

$\delta (\epsilon^{\mu \nu \lambda}A^a_\mu \partial_\nu A^a_\lambda ) = 2\epsilon^{\mu \nu \lambda}( \partial_\mu \epsilon ^a \partial _\nu A^a_\lambda + f^{abc}A_\mu ^b \partial _\nu A_\lambda ^a \epsilon ^c)$

$\delta (\epsilon^{\mu \nu \lambda}f^{abc}A^a_\mu A^b_\nu A^c_\lambda) = 3\epsilon^{\mu \nu \lambda}f^{abc} ( \partial _\mu \epsilon^a A_\nu ^b A_\lambda ^c + f^{adf} A_\mu ^d A_\nu ^b A_\lambda ^c \epsilon ^f) $

  • Now I can't see how any linear combination of the above two terms can make their combined variation 0 (or a total derivative). I would like to know as to what am I missing here.
$\endgroup$
1
$\begingroup$

Use the Jacobi identity $f^{abc}f^{adf}+f^{adb}f^{acf}+f^{acd}f^{abf}=0$ to the last term.

$\endgroup$

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.