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Homework Statement
##(a)## Consider the following action in three space-time dimensions
##S_{CS}=\displaystyle \int d^3x\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho##
It is known as the Chern-Simons action.
Demonstrate that it is gauge-invariant.
Show that any solution to the equations of motion for the above action is gauge-equivalent to a trivial solution.
##(b)## Find the equations of motion for the following Maxwell-Chern-Simons action:
##S=-\frac{1}{4e^2}\displaystyle \int d^3x F_{\mu\nu}F^{\mu\nu} + \displaystyle \int d^3x\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho##
Show that each component of ##F_{\mu\nu}## satisfies the Klein-Gordon equation.
Homework Equations
The Attempt at a Solution
[/B]
##(a)## The action is gauge invariant if it is invariant under the transformation ##A_\mu \rightarrow A_\mu + \partial_\mu \Lambda##, where ##\Lambda## is a function of the space-time coordinates. Under this transformation, ##S_{CS} \rightarrow S_{CS}' = k \displaystyle \int d^3x \epsilon^{\mu\nu\rho} (A_\mu + \partial_\mu \Lambda) \partial_\nu (A_\rho + \partial_\rho \Lambda)##
##= k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}(A_\mu \partial_\nu A_\rho + A_\mu\partial_\nu \partial_rho \Lambda + \partial_\mu \Lambda\partial_\nu A_\rho + \partial_\mu \Lambda \partial_\nu \partial_\rho \Lambda)##
## =S_{CS} + k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}(A_\mu\partial_\nu \partial_rho \Lambda + \partial_\mu \Lambda\partial_\nu A_\rho + \partial_\mu \Lambda \partial_\nu \partial_\rho \Lambda)##
##(*)## Note that ##\epsilon^{\mu\nu\rho}A_\mu\partial_\nu \partial_\rho \Lambda = - \epsilon^{\mu\rho\nu}A_\mu\partial_\nu \partial_\rho \Lambda## by anti-symmetry of ##\epsilon##
## =- \epsilon^{\mu\nu\rho}A_\mu\partial_\rho \partial_\nu \Lambda## after renaming indicies
## = - \epsilon^{\mu\nu\rho}A_\mu\partial_\nu \partial_\rho \Lambda## due to the symmetry ##\partial_\nu \partial_\rho=\partial_\rho \partial_\nu##
##=0##.
Same argument holds for the last term in the integrand and we are left with ##S_{CS}'=S_{CS} + k \displaystyle \int d^3x \epsilon^{\mu\nu\rho}\partial_\mu \Lambda \partial_\nu A_\rho##
Applying the product rule backwards and using the same argument as above in ##(*)## we find that ## \displaystyle \int d^3x \epsilon^{\mu\nu\rho}\partial_\mu \Lambda \partial_\nu A_\rho = \displaystyle \int d^3x \partial_\mu (\epsilon^{\mu\nu\rho} \Lambda \partial_\nu A_\rho)##
Applying the divergence theorem this becomes ##\displaystyle \int dS n_\mu \epsilon^{\mu\nu\rho} \Lambda \partial_\nu A_\rho##, where ##n_\mu## is a normal vector pointing outward from the surface, which is a surface integral equal to ##0## under the assumption that the fields fall off sufficiently fast at infinity. Hence the action is gauge-invariant.
The Lagrangian for this action is ##\mathcal{L}_{CS}=\epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho##. I'm not going to write out the full procedure for getting the equations of motion, but we find that they are
##k\epsilon^{\rho\mu\nu}F_{\mu\nu}=0##
I am unsure what the phrase "gauge equivalent to a trivial solution" means. I know that the equations of motion are gauge-invariant so nothing will change, but I don't really know what it's asking me to do.##(b)## Again, not going through finding the e.o.m, but we find that they are ##\partial_\mu F^{\mu\nu} + ke^2\epsilon^{\nu\alpha\beta}F_{\alpha\beta}=0##
The Klein-Gordon equation is ##(\partial_\mu \partial^\mu + m^2)\phi=0##. I was reading a paper on this theory and it said this can be done by writing ##\tilde{F^\mu}=\epsilon^{\mu\nu\rho}F_{\nu\rho}## and then the e.o.m. can be written as ##(\partial^\nu \partial_\nu + (ke^2)^2)\tilde{F^\mu}=0##. I have no idea why this is justified.Any help on the two issues above would be very appreciated. Thank you.