- #1

Joker93

- 504

- 36

I was thinking about the Riemann curvature tensor(and the torsion tensor) and the way they are defined and it seems to me that they just need a connection(not Levi-Civita) to be defined. They don't need a metric. So, in reality, we can talk about the Riemann curvature tensor of smooth manifolds without even going to Riemannian manifolds(giving the smooth manifolds a metric).

Now, for the sake of clarity, consider the following example:

Say we have the smooth manifold ##S^2##. If we give it an appropriate metric, we can "turn it" into a round sphere or an ellipsoid. They have different metric and have different Levi-Civita connections. But, say we don't assign to them the Levi-Civita connection, but give them the same (generally not torsion-free) connection. This means that they have the same Riemann curvature tensor(and torsion tensor)!.

This, doesn't make much sense to me, geometrically.

In my mind, a connection gives a way to parallel transport and has nothing to do with the actual shape of the manifold(which makes sense since without a assigning a metric, we can't think of a smooth manifold as having a "concrete" shape). But, the metric does. So, how can the ellipsoid and the round sphere, with the same connections, have the same curvature?

Lastly, since we don't need a metric to define the curvature tensor(or the Ricci tensor) but we do need a metric to define the Ricci scalar, it seems to me that the Riemann curvaure tensor is not the most important quantity to describe/characterize the curvature of a Riemannian manifold. [But, it is the most important quantity to describe the curvature of a smooth manifold.] But, this does not make much sense either, because I can't imagine that one scalar can carry all the information about the curvature of an n-dimensional manifold that can be very complicated in shape.

**Attempt at explaining some of the above points:**

For the example where we have the ellipsoid and round sphere with the same connections:

I was always thinking about curvature in terms of distance, but if I think of it in terms of parallel transporting vectors around closed curves and having a difference in the angle between the vector at the start of the loop and the vector at the end of the loop, then I can make sense of it as follows:

The ellipsoid and the sphere might look different, but given the same connection, when parallel transporting a vector along the same loop it will change its angle in the same way.

For the question on the importance of the curvature tensor and Ricci scalar in the characterization of the curvature:

Since a lot of the characteristics of a Riemannian manifold are given to it by its structures that it has as a smooth manifold, it might mean that a large portion of the curvature of a Riemannian manifold are completely determined without any reference to a metric. But, if we also give a metric, then we gain some extra information for the curvature through the Ricci scalar**.

**Note that I am aware that there exist other tensors too. For example, I would not be surprised if the Weyl tensor contains some more information about the curvature that cannot be obtained through the Riemann curvature tensor and the Ricci scalar.

Thank you in advance for any help you provide!