Singularity in Robertson Walker metric with flat spatial slices In Sean Carroll's GR book, pg. 76, a special case of the Robertson-Walker metric, where the spatial slices are flat is given by
$$ds^2=-dt^2+a^2(t)[dx^2+dy^2+dz^2].$$
It was said that $t=0 $ represents a true singularity of the geometry (the 'Big Bang') and should be excluded from the manifold. The range of the $t$ coordinate is therefore $0<t<\infty$.
Why is $t=0$ a singularity? What is infinite or undefined when $t=0$ ?
 A: If you're looking for some physical quantities that blow up as $t \rightarrow 0$ in FRW cosmology, you can note that the Hubble factor $H = \frac{\dot{a}(t)}{a(t)}$ diverges at this point. Similarly, to pin home the fact this is a proper singularity, if you calculate the curvature invariants, e.g. $R$, $R^{\mu \nu}R_{\mu \nu}$ you find they're all inversely proportional to some power of $t$ and also diverge as $t \rightarrow 0$. Therefore we have a curvature singularity.
A: Carroll mentions that in the solution under consideration, the scale factor $a\rightarrow 0$ as $t\rightarrow 0$.  Since the metric must always be non-degenerate, this represents a singular point. Operationally, if the metric is degenerate than the dual metric $g^{\mu\nu}$ becomes undefined.
A: Remeber that the solution to the Friedman equations for the scale factor $$a(t) = a_0 t^{\lambda}$$ where $\lambda$ is a constant. This is obviously zero at $t=0$. At this point the spatial part of the metric $$ds^2=-dt^2+a^2(t)[dx^2+dy^2+dz^2]$$ vanishes.
