Assuming that the density of the universe is dominated by matter, so that $\rho = \rho_m$, how can the deceleration parameter today be shown to be $q_0 = \frac{\Omega_m}{2} - \Omega_{\Lambda}$
The acceleration equation; $\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho + \frac{3p}{c^2}) + \frac{\Lambda}{3}$,
the continuity/fluid equation; $\dot{\rho}=-3H(\rho+\frac{p}{c^2})$,
and the Friedmann equation $H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}$ are all needed as far as I can see, as well as $q=\frac{-\ddot{a}a}{\dot{a}^2}$.
$\rho_\Lambda = \frac{\Lambda}{8\pi G}$, and $p=-\rho_\Lambda c^2$ as matter does not contribute to $p$. I just can't manage it.
Thanks!