# Equation of motion from $D=3$ Lorentz Chern-Simons action

In three dimensions, the well known Lorentz Chern-Simons action is $$S_{\text{CS}}=\int\text{d}^3x\varepsilon^{\mu\nu\rho}\bigg(\omega_{\mu}{}^{ab}R_{\nu\rho ab}+\frac{2}{3}\omega_{\mu a}{}^{b}\omega_{\nu b}{}^{c}\omega_{\rho c}{}^{a}\bigg) \tag{1}$$ where $\omega_{\mu ab }$ is the Lorentz spin connection and $R_{\mu\nu ab}$ is its corresponding field strength, $$R_{\mu\nu ab}=\partial_{\mu}\omega_{\nu ab}-\partial_{\nu}\omega_{\mu ab}+\omega_{\mu a}{}^{f}\omega_{\nu fb}-\omega_{\nu a}{}^{f}\omega_{\mu fb}. \tag{2}$$ I would like to obtain the equation of motion corresponding to an arbitrary variation in the vielbein $e_{a}{}^{\mu}$, which is related to the spin connection via the zero torsion condition (or equivalently the vielbein compatibility condition). Following the instructions on the foot note of page 438 in [1] (page 30 from the title page), I vary (1) with respect to the spin connection to obtain, $$\delta[S_{\text{CS}}]=\int\text{d}^3x\varepsilon^{\mu\nu\rho}R_{\nu\rho}{}^{ab}\delta\omega_{\mu ab}.\tag{3}$$ Now by varying the vielbein compatibility condition, we can obtain $\delta\omega_{\mu ab}$ in terms of a variation in the vielbein, \begin{align} 0=&\nabla_{\mu}e_{\nu}{}_a=\partial_{\mu}e_{\nu a}+\omega_{\mu ab}e_{\nu}{}^{b}-\Gamma_{\mu\nu}^{\rho}e_{\rho a}\implies \delta\omega_{\mu ab}=e_{b}{}^{\nu}\big(\delta\Gamma_{\mu\nu}^{\rho}e_{\rho a}-\nabla_{\mu}\delta e_{\nu a}\big). \tag{4} \end{align} After we insert (4) into (3), the second term from (4) does not contribute because after integrating by parts, we have a term of the form $\varepsilon^{\mu\nu\rho}(\nabla_{\mu}R_{\nu\rho}{}^{ab})e_{b}{}^{\sigma}\delta e_{\sigma a}$ which vanishes by virtue of the second Bianchi identity on $R$. Therefore the total variation is $$\delta[S_{\text{CS}}]=\int\text{d}^3x\varepsilon^{\mu\nu\rho}R_{\nu\rho\alpha}{}^{\sigma}\delta\Gamma_{\mu\sigma}^{\alpha}\tag{5}$$ which can now be completely expressed in terms of the variation of the metric. Everything I have done until now has followed the instructions of the aforementioned footnote. The authors now state that by expressing $\delta\Gamma_{\mu\sigma}{}^{\alpha}$ in terms of $\delta g_{\mu\nu}$ one can show that the result is $$\delta[S_{\text{CS}}]=\int\text{d}^3xC^{\mu\nu}\delta g_{\mu\nu}\tag{6}$$ where $C^{\mu\nu}$ is the Cotton tensor defined by $$C^{\mu\nu}=\varepsilon^{\mu\alpha\beta}\nabla_{\alpha}\widetilde{R}_{\beta}{}^{\nu},\qquad \widetilde{R}_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{4}g_{\alpha\beta}R, \tag{7}$$ ($\widetilde{R}$ is the Schouten tensor). However, despite the authors comments I am not able to arrive at (6) from (5). The reason is the following. Upon applying the well known formula $$\delta\Gamma_{\mu\sigma}^{\alpha}=\frac{1}{2}g^{\alpha\beta}\big(\nabla_{\mu}\delta g_{\beta\sigma}+\nabla_{\sigma}\delta g_{\beta\mu}-\nabla_{\beta}\delta g_{\mu\sigma}\big) \tag{8}$$ to (5) and integrating all three terms by parts, the first vanishes due to the Bianchi identity again, whilst the second is equal to the third. This leads me to the following result, $$\delta[S_{\text{CS}}]=\int\text{d}^3x\varepsilon^{\mu\nu\rho}\nabla^{\beta}R_{\nu\rho\beta}{}^{\sigma}\delta g_{\mu\sigma}\tag{9}$$ which, after using the identity $$\nabla^{\beta}R_{\nu\rho\beta\sigma}=\nabla_{\nu}R_{\sigma\rho}-\nabla_{\rho}R_{\sigma\nu},\tag{10}$$ reduces to $$\delta[S_{\text{CS}}]=2\int\text{d}^3x\varepsilon^{\mu\nu\rho}\nabla_{\nu}R_{\rho}{}^{\sigma}\delta g_{\mu\sigma}\tag{11}.$$ One would have also arrived at the same conclusion had they used the observation that in D=3, $$R_{\alpha\beta\gamma\delta}=g_{\alpha\gamma}\widetilde{R}_{\beta\delta}+g_{\beta\delta}\widetilde{R}_{\alpha\gamma}-g_{\alpha\delta}\widetilde{R}_{\beta\gamma}-g_{\beta\gamma}\widetilde{R}_{\alpha\delta}. \tag{12}$$ The integrand in (11) clearly is not equivalent to that of (6), the terms proportional to the Ricci scalar are not apparent. What went wrong?

[1]: S. Deser, R. Jackiw, S. Templeton; Topologically Massive Gauge Theories (1982).

Note that the term that is missing is $$-\frac14\varepsilon^{\mu\alpha\beta}\nabla_\alpha(\delta^\nu_\beta R)\delta g_{\mu\nu}=-\frac14\varepsilon^{\mu\alpha\nu}\nabla_\alpha R\,\delta g_{\mu\nu}$$ which vanishes due to $\varepsilon^{\mu\cdot\nu}\delta g_{\mu\nu}\equiv 0$, inasmuch as the first tensor is skew-symmetric and the second one is symmetric. Thus, the result in the OP and the one in the paper are identical.
• No: if $\delta S=\int A^{\mu\nu}\delta g_{\mu\nu}$ the equations of motion are not $A^{\mu\nu}=0$, but $A^{(\mu\nu)}=0$. The EoM are insensitive to the skew part of the variation; only the symmetric part is set to zero. You can add whichever skew tensor you like to $A^{\mu\nu}$, and this won't affect the EoM. – AccidentalFourierTransform Jun 13 '18 at 14:28
• But the equation of motion is supposed to be $C^{\mu\nu}=0$, which can only be obtained by adding the antisymmetric tensor in your answer to the integrand. Implying that the EoM is sensitive to the addition of an antisymmetric tensor? – NormalsNotFar Jun 13 '18 at 14:51
• Nope. In general terms, when you perform the variation of the action, you end up with $\delta S=\int A^{\mu\nu}\delta g_{\mu\nu}$, for some tensor $A^{\mu\nu}$. The equations of motion are, then, $A^{(\mu\nu)}=0$. You can add any skew tensor to $A$, and the EoM remain the same, $A^{(\mu\nu)}=0$. In your case, you don't need to "add" an antisymmetric tensor to obtain $C^{\mu\nu}$. Instead, you have to symmetrise whatever tensor you end up with. This is precisely what the authors did (note that the Cotton tensor is symmetric!). – AccidentalFourierTransform Jun 13 '18 at 15:00