Is there any quick way of proving that for a conformal metric of the form:


where the $\eta^{}_{\mu\nu}$ is the usual Minkowski metric, the Weyl tensor vanishes in the whole space? Knowing that its given by:


Can this fact help us in proving that any 2D Riemannian manifold is conformally planar? If so, how?

  • $\begingroup$ Can you just use the fact that the Weyl tensor is conformally invariant? $\endgroup$ – Javier Oct 4 at 0:13
  • $\begingroup$ Actually they asked me to prove that one necessary and suficient condition for the metric to be conformal as above would be for the Weyl tensor to vanish everywhere, so the only quick way i found was to plug the metric in and by index manipulation conclude that all the terms would cancel each other out, Unfortunatly they force me to go this way sorry. $\endgroup$ – MicrosoftBruh Oct 4 at 9:37

The metrics, $g_{ab}$ and $\bar{g}_{ab}$ are conformally equivalent if we can write $$ \bar{g}_{ab} = \Omega^2g_{ab}\,, $$ where $\Omega\equiv \Omega(x)$ is a non-zero differentiable function of the space-time coordinates.

In this case, their Weyl tensors are equivalent, i.e., $\bar{C}^{a}_{\,\,bcd} = {C}^{a}_{\,\,bcd}$.

Setting $g_{ab}$ equal to the Minkowski space-time metric (the metric of flat space-time in special relativity), $\eta_{ab}$, we have that all the components of the Riemann tensor and the Ricci tensor are equal to zero and that the Ricci scalar is zero.

This means that the Weyl tensor for Minkowski space-time vanishes. This follows from the expression of the Weyl tensor in terms of the Riemann tensor, the Ricci tensor, the Ricci scalar and the metric tensor that is given in the opening post. Since

$$ {C}^{a}_{\,\,bcd} \equiv 0 $$

for $\eta_{ab}$ it follows that $\bar{C}^{a}_{\,\,bcd} \equiv 0$ for $\bar{g}_{ab}\,.$

Without doing all the maths that's the shortest proof I can give.

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