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).
