I have to prove the (non abelian) gauge invariance of the following lagrangian (for a certain value of $\lambda$): $$\mathcal L= -\frac14 F^{\mu\nu}_aF_{\mu\nu}^a + \frac{k}{4\pi}\epsilon^{\mu\nu\rho}\operatorname{tr}[A_{\rho}\partial_{\mu}A_{\nu} + \lambda A_{\mu}A_{\nu}A_{\rho}]$$ Is there an easier way than expliciting every component of A, going to first order gauge transformation, ... ? Cause it seems really ugly to me.


3 Answers 3


There is a nice reason for this, which Witten often explains. Imagine that your three dimensional space is the boundary of a four-dimensional space, for example, you can imagine that space is the surface z=0 of regular four dimensional space x,y,z,t. Further, you can imagine that space is closed into a sphere, which doesn't affect things except for some boundary conditions at infinity (the physics shouldn't care about such things, also note that this is implicitly Euclidean). If you close the three dimensional space-time into a sphere, the interior of the sphere is like the rest of the values of z for the plane case.

You can extend any 3 dimensional gauge field configuration to the imaginary fourth dimension arbitrarily, so that any gauge field on the surface of the sphere can be extended to many different gauge fields on the interior.

On the interior, you can construct the manifestly gauge invariant operator:

$$ \epsilon_{\mu\nu\lambda\sigma} F^{\mu\nu}F^{\lambda\sigma} = F\tilde{F}$$

It is important to note that this quantity is a perfect divergence:

$$ F\tilde{F} = \partial_\mu J^\mu_\mathrm{CS} $$

where J is the Chern-Simons current in 4-dimensions. Using Stokes theorem, for any four-dimensional gauge field configuration

$$ \int F\tilde{F} = \int d(*J) = \int_\partial *J $$

Where the last equality is Stoke's theorem, and the previous equality is writing the diverence of a current as the Poincare dual of a three-form.

So the manifestly Gauge invariant $F\tilde{F}$ integral on any gauge field on the interior of the sphere is equal to the integral of the three form *J on the boundary of the sphere. So the integral of *J must be gauge invariant. I didn't work out the actual form of *J, but it is the quantity you are trying to prove gauge invariant.

Although Witten's argument is conceptually illuminating, so it is the correct argument, verifying gauge invariance explicitly is not much more difficult than understanding all parts of the argument. Still, it is good to know the conceptual reason, because the reason the Chern-Simons style things are important is exactly because they are the boundary terms of integrals of those gauge invariant field tensor combinations which are perfect derivatives.

  • 3
    $\begingroup$ This nice argument of course requires that the 3-manifold we are interested in is "co-bounding" (the boundary of a 4-manifold) in the first place. Depending on which extra structure one has (e.g. spin structure etc.) this is not always the case. But the good message is that essentially this argument goes through fully generally if one refines it "locally" as expressed in the language of stacks. An exposition of this is at the beginning of this article: ncatlab.org/schreiber/show/… $\endgroup$ Aug 10, 2013 at 12:44

Usually when you're doing this sort of thing, you try to group the operators into combinations which have simple gauge transformation properties. For instance, I would suggest rearranging the trace factor into $\operatorname{tr}[A_\rho(\partial_\mu + \lambda A_\mu)A_\nu]$, and see what you can figure out about the derivative operator $\partial_\mu + \lambda A_\mu$.

But when it comes down to it, the tricks take you only so far. Proving gauge invariance "from scratch" always entails a certain amount of tedious algebra.

  • $\begingroup$ Ok I will try with this tip. $\endgroup$
    – toot
    Jan 17, 2012 at 1:46

The value of lambda must be fixed if you want gauge invariance, written in differential forms is $1/3$, $$S_{CS}=\frac{k}{4\pi}\int \langle A\wedge dA+\frac{2}{3}A\wedge A\wedge A \rangle\,,$$ where $A=A_\mu dx^\mu$ and $\langle \rangle$ is an invariant quadratic trace (so you will normally work with particular types of Lie algebras, see the killing symmetric tensor for instance). You can see that $dL_{CS} \sim \langle FF \rangle$ where $F=dA+A\wedge A$ and $d\wedge A=\partial_\mu dx^\mu \wedge A$. The volume measure is $d^4x=dx^0 dx^1 dx^2=\frac{1}{3!}\epsilon_{\mu\nu\rho}dx^\mu\wedge dx^\nu\wedge dx^\rho$. You can either work with covariant indices $\mu, \nu, \cdots$ or in terms of differential forms. A gauge transformation is given by $$A \longrightarrow A'=g^{-1}Ag+g^{-1}dg\,,$$ where $g=\exp{G}$ and $G=a^I T_I$ lives in the Lie algebra $[T_I,T_J]=C^K{}_{IJ}T_K$. The 'parameter' of the transformation is a 0-form $a^I=a^I(x)$. The infinitesimal form of the gauge transformations is $$A'=[A,G]+dG=DG\,,$$ where $DG$ is the covariant derivative acting on the zero form $G$.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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