# 2D Chern Simons action by integrating out fermions

In Qi, Hughes, and Zhang's paper (https://arxiv.org/abs/0802.3537), they show how the Chern number appears as a coefficient of response function. Given the Hamiltonian (49) of a (2+1) or (4+1)D insulator coupled to $U(1)$ gauge field $$H[A]=\sum_{m,n}(c^\dagger_{m \alpha} h^{\alpha \beta}_{mn} e^{iA_{mn}} \,c_{n \beta}+h.c.) + \sum_m A_{0m}c^\dagger_{m\alpha} c_{m \alpha}, \tag{49}$$ the effective action of gauge field $A^\mu$ can be obtained by usual coherent state path integral $$e^{i S_{eff} [A]} = \int Dc D\bar{c} \exp \{i \int dt [\sum_m \bar{c}_{m \alpha}\,(i\partial_t)c_{m \alpha} - H[A]]\}$$ where $c$ and $\bar{c}$ are Grassmann variables. Then, the Gaussian integral gives the determinant: $$e^{i S_{eff} [A]} = \det \left[ (i\partial_t - A_{0m}) \delta^{\alpha \beta}_{mn}-h^{\alpha \beta}_{mn}e^{A_{mn}}\,\right].\tag{50}$$ I don't quite understand the last step. Where does the $\int dt$ part go when we perform the Gaussian integral? Also, the determinant contains the term like $\prod_m (i\partial_t - A_{0m})$. What does this term mean since it's not even a number?

In the paragraph below Eq. (50), they mention "In the case of the (2 + 1)D insulators, the effective action $S_{eff}$ contains a Chern-Simons term $(C_1/4\pi)A_\mu \epsilon^{\mu \nu \tau}\partial_\nu A_\tau$." Can we get this CS term from Eq. (50) easily?

In Bernevig's book, there is an exercise in the end chapter 13: Explicitly integrate out the fermions in the 2d Dirac Hamiltonian $H=\sum_k c^\dagger_k[(k_i+A_i)\sigma_i+m \sigma_z]c_k$, where i = 1, 2, to obtain the Chern-Simons form $\epsilon_{ij}A_i \partial_0 A_j$ with coefficient $\frac{1}{2}\text{sign}(m)$. I couldn't figure out this. Can anyone give some hints on this? Thanks!