I have a question re: equation (6.6) in Altland and Simons, which claims that the effective action of an interacting electron gas takes the form $$S\propto \sum_q \phi_q\left( \frac{\mathbf{q}^2}{4\pi}-e^2\Pi_{q}\right)\phi_{-q}.\tag{6.6}$$ What I don't understand is where the negative sign in front of the $e^2\Pi_q$ term comes from. If I follow their derivation, starting with equation (6.4), I have
$$S[\phi] = \frac{\beta}{8\pi L^d}\sum_q\phi_q\mathbf{q}^2\phi_{-q} - \text{tr}\ln\left[-i\omega + \frac{\mathbf{p}^2}{2m}-\mu+\frac{ie}{L^d}\phi\right] \tag{6.4}$$ $$=\frac{\beta}{8\pi L^d}\sum_q\phi_q\mathbf{q}^2\phi_{-q} + \text{tr}\ln\left[G_0^{-1}(\phi)-\frac{ie}{L^d}\phi\right]$$ Now, expanding the logarithm in small powers of $\phi$, we have
$$\text{tr}\ln\left[G_0^{-1}(\phi)-\frac{ie}{L^d}\phi\right] \approx \text{tr}\ln G^{-1}_0 + \frac{ie}{L^d}\text{tr}[G_0\phi] + \frac{1}{2}\frac{e^2}{L^2d}\text{tr}[G_0\phi G_0\phi]$$
where the quadratic term is positive because of the factor of $i^2$ that multiplied out the $-1$ in the Taylor expansion of $\ln(1+x)$. NOTE: here I think there should actually be a $-$ sign in front of the second term because $\ln(1+x)\approx -x -\frac{x^2}{2}$ but whatever, because we don't care about this term. Anyways. We can show that $$\text{tr}[G_0 \phi G_0 \phi] = \frac{L^d}{2T}\sum_q\Pi_q \phi_q\phi_{-q}$$ Putting this back together with equation 6.4, shouldn't we have $$S\propto \sum_q \phi_q\left( \frac{\mathbf{q}^2}{4\pi}+e^2\Pi_{q}\right)\phi_{-q}$$ I am not sure where the negative sign came from in this derivation.
