Q1: What is the difference of boson and fermions for their Gravitational Chern-Simons theory?
I suppose in general if the metric is not flat, we have vierbein ${e_{\hat{b}}}^{\nu}$, with $$ g_{\mu\nu} {e_{\hat{a}}}^{\mu} {e_{\hat{b}}}^{\nu} = \eta_{\hat{a}\hat{b}}, $$ where $g_{\mu\nu}$ is curved and $\eta_{\hat{a}\hat{b}}$ is Lorentzian flat. With the spin-connection is $$ \omega^{\hat{c}}_{\hat{b}\mu} = {e^{\hat{c}}}_{\nu}\partial_\mu {e_{\hat{b}}}^{\nu} + {\Gamma^\nu}_{\sigma\mu} {e^{\hat{c}}}_{\nu} {e_{\hat{b}}}^{\sigma}, $$ where ${\Gamma^\nu}_{\sigma\mu}$ are the Christoffel symbols.
SET-UP: Now let us imagine there are some matter fields bosons $\phi$ or fermions $\psi$ coupling to the spacetime metric, and we integrate out the bosons $\phi$ or fermions $\psi$ to get the effective actions involving gravitational Chern-Simons action in 2+1D.
So a 2+1D gravitational Chern-Simons action can be (spin connection $\omega$): $$ S=\int\omega\wedge\mathrm{d}\omega + \frac{2}{3}\omega\wedge\omega\wedge\omega \tag{1} $$
I am sure this works for fermions.
Q2: however, if bosons have spin 0 or spin 1, do we still have spin-connection? I suppose still yes?
Q3: or, do we have a 2+1D gravitational Chern-Simons action for bosons to be ( the connection $\Gamma$): $$ S=\int\Gamma\wedge\mathrm{d}\Gamma + \frac{2}{3}\Gamma\wedge\Gamma\wedge\Gamma \tag{2} $$
Q4: what are the differences between Eq.1 and Eq.2 in terms of background matter fields? (suppose the metric $g_{\mu\nu}$ is coupled to matters, either bosons or fermions.)
Please feel free give Reference. And make sure that your answer is to my point Q1-4.