# Hubble Parameter in terms of Density Parameters

So I have the equation (derived from the Friedman equations with $$a(t_0)=1$$ and the powers of $$a(t)$$ quoted)$$H^2 = H_0^2\left(\frac{\Omega_{R_0}}{a^4(t)}+\frac{\Omega_{M_0}}{a^3(t)}+\Omega_{\Lambda_0}+\frac{1-\Omega_0}{a^2(t)}\right)$$

Where the "$$0$$" subscript denotes the value of the quantities at the initial time $$t_0$$, the subscripts $$R_0$$, $$M_0$$, $$\Lambda_0$$ denote the density parameter of radiation, matter and dark energy respectively and $$\Omega$$ is the total density parameter.

It says that for $$\Omega_{M_0}=1$$, it follows that $$\Omega_{R_0}=\Omega_{\Lambda_0}=0$$.

How exactly does this follow? I think that I might be missing something here because I'm not entirely sure I understand how this is so straight forward.

They're imposing spatial flatness. $$\Omega_0 = \Omega_{R_0} + \Omega_{M_0} + \Omega_{\Lambda_0}$$. The curvature parameter $$\Omega_k = 1 - \Omega_0$$. Space is flat when $$\Omega_k=0$$, thus $$\Omega_{R_0} + \Omega_{M_0} + \Omega_{\Lambda_0} = 1$$. You can relax the flatness requirement, though, in the more general case, though all observations, so far, are consistent with $$\Omega_k=0$$.