8
$\begingroup$

I am reading papers about topological massive gravity (TMG) in 3-dimensional spacetime. I come across two kinds of formalism to describe TMG. In the first kind, the gravitational Chern-Simons (CS) term is constructed by Christoffel connection $\Gamma$ which reads \begin{equation} S_1=\int \Gamma \wedge d\Gamma +\frac23\Gamma\wedge\Gamma\wedge\Gamma \end{equation} The other involves spin-connection $\omega$ \begin{equation} S_2=\int \omega\wedge d \omega+\frac23\omega\wedge\omega\wedge\omega \end{equation}

My questions are:

  1. Are those two kinds of formalism are equal? By equal I mean $S_1=S_2$ up to some boundary term?

  2. What the differences between those two formalisms? When should we use the first one or the second?

$\endgroup$
4
$\begingroup$

Topologically massive gravity formulated in terms of $\Gamma$ or $\omega$ is the same; for example in Maloney's paper on the geometric microstates of the three-dimensional black hole, he uses the $\Gamma$ formulation, and Witten's talk on revisiting $2+1$-dimensional gravity includes the Chern-Simons term with the spin connection $\omega$, but they are both describing the same theory.

One subtlety I'd like to point out, when performing reformulations of fields, is it can modify it in a non-trivial way. In particular, gravity in $2+1$ dimensions is considered trivial, but when formulated in terms of a gauge field out of the connection and vierbein,

$$A = \begin{pmatrix} \omega & e\\ -e & 0 \end{pmatrix}$$

allowing a non-invertible $e$ and thus non-classical configurations. It turns out this makes the Einstein-Hilbert action ill-defined unless the coupling is quantised.

Since you seem interested in topological massive gravity, and since it wasn't mentioned in the papers you listed, I would like to point out it has been conjectured by Witten and others that if the Frenkel-Lepowsky-Meurman conjecture is true, it may be that the dual $k=1$ CFT is the theory whose symmetry is described by the Monster group - I don't know if this avenue is of interest to you.


To explicitly show the equivalence, it seems rather messy. As a starting point, note the relation between the spin connection $\omega$ and the affine connection $\Gamma$ as,

$$\Gamma^\lambda_{\mu\nu} = e^\lambda_A \partial_\mu e^A_\nu + e^\lambda_A e^B_\nu {\omega_\mu}^A_B$$

where capital Roman indices denote the orthonormal basis and Greek indices the coordinate basis. We can substitute this into the Chern-Simons term for $\Gamma$, which is explicitly in index notation,

$$\mathcal L \sim \epsilon^{\lambda\mu\nu} \Gamma^\rho_{\lambda \sigma} \left( \partial_\mu \Gamma^\sigma_{\rho\nu} + \frac{2}{3} \Gamma^\sigma_{\mu\kappa} \Gamma^\kappa_{\nu\rho}\right),$$

using the definition of the wedge product and exterior derivative. Using the relation between the connections, we thus have,

$$\mathcal L \sim \epsilon^{\lambda\mu\sigma}\left(e^\rho_A \partial_\lambda e^A_\nu + e^\rho_A e^B_\sigma {\omega_\lambda}^A_B \right) \bigg[ \partial_\mu \left( e^\sigma_A \partial_\rho e^A_\nu + e^\sigma_A e^B_\nu {\omega_\rho}^A_B\right)+ \bigg.\\ + \bigg. \frac{2}{3} \left(e^\sigma_A \partial_\mu e^A_\kappa + e^\sigma_A e^B_\kappa {\omega_\mu}^A_B \right) \left( e^\kappa_A \partial_\nu e^A_\rho + e^\kappa_A e^B_\rho {\omega_\nu}^A_B\right)\bigg].$$

At this point it is a matter of tedious manipulation, which should hopefully show an equivalence to the theory in terms of the spin connection.

$\endgroup$
  • $\begingroup$ Thank you for your answer very much! It is a good starting point. $\endgroup$ – phys_student Nov 26 '16 at 1:15
  • $\begingroup$ @phys I've updated my answer with confirmation. $\endgroup$ – JamalS Dec 1 '16 at 22:20

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.