Weyl Tensor Theorems

Question

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

$g^{}_{\mu\nu}=\Omega^2\eta^{}_{\mu\nu}$

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

$C^{}_{ijkl}=R^{}_{ijkl}-\frac{1}{n-2}(R^{}_{ik}g^{}_{jl}-R^{}_{il}g^{}_{jk}+R^{}_{jl}g^{}_{ik}-R^{}_{jk}g^{}_{il})+\frac{R}{(n-1)(n-2)}(g^{}_{ik}g^{}_{jl}-g^{}_{il}g^{}_{jk})$

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

Answer 1

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.