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.

1 Answer 1


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


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