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Can someone please explain the difference between the information that the Ricci curvature tensor gives us and that which the Laplacian offers?

Here is my shot at it:

My concept of the Ricci curvature tensor is that it gives us information about the local curvature (how much it diverges from a flat plane) which can then be compared to any other local curvature. Universal comparisons of curvature can be made based on it.

The Laplacian also gives us information about curvature, but it does not do so in the "absolute" way that the Ricci tensor does. Universal comparisons of curvature cannot be made based on it.

Thanks

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1 Answer

up vote 1 down vote accepted

OP wrote (v1):

The Laplacian also gives us information about curvature, but it does not do so in the "absolute" way that the Ricci tensor does.

It is in some sense the other way around. From the Beltrami-Laplacian $\Delta$, it is possible to reconstruct the metric tensor. Sketched Proof: If $U \subseteq M$ is an open coordinate neighborhood with coordinates $x^i$, then the metric

$$ds^2~=~g_{ij}~dx^i dx^j$$

can be deduced from the fact that

$$ 2g^{ij} ~=~ \Delta (x^ix^j)- x^i\Delta(x^j)- x^j\Delta(x^i). $$

Once one knows the metric tensor, one has full information to calculate the (Levi-Civita) Riemann curvature tensor, the Ricci tensor, etc., in the standard way.

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