Consider a flat Robertson-Walker metric.
When we say that there is a singularity at $t=0$, clearly it is a coordinate dependent statement. So it is a "candidate" singularity.
In principle there is "another coordinate system" in which the corresponding metric has no singularity as we approach that point in the manifold.
However, we know that Big Bang is "a true" singularity, but how should we test that?
Is it intuitively self-evident, or should we check rigorously all scalars based on the Ricci tensor? If so "which order of scalar" goes to infinity at that point called Big Bang?
Answer
The singularity comes from the scale factor $a(t)$:
$$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$
By solving the Friedmann equations for the scale factor we know that:
$$a(t) = a_0 t^{\lambda}$$
where $\lambda$ is some positive number that depends on the matter-radiation ratio of the universe. At $t=0$ the scale factor becomes $a(0)=0$. So at $t=0$ the spacial part of the metric becomes zero. You can check that scalars will blow up by showing that the volume element $\sqrt{-g} ~d^4x \,$ gives nonsense. At $t=0,$ $g=det(g_{\mu \nu}) = 0,$ meaning the volume element is zero. This is not anything that can be fixed by a coordinate transformation.
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