On the dependence of the curvature tensor on the metric

[Joker93]

Hello!
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!

 

[Orodruin]

While you can put the connection to be the same for both the sphere and the ellipsoid, it will only be compatible with one of the metrics and not the other. Generally, when you have a metric, you can have connections that are not metric compatible. Note that metric compatibility is not the same thing as imposing a Levi-Civita connection. The Levi-Civita connection is the unique torsion free and metric compatible connection.

In other words, when you do impose a metric (which is the only thing that is going to separate your ellipsoid from the sphere), then you will not have both $\nabla g_S = 0$ and $\nabla g_E = 0$ (where $g_S$ and $g_E$ are the metrics on the sphere and ellipsoid, respectively).

Edit: Note: The metric is what allows you to define angles and lengths of tangent vectors and so metric compatibility is what is required for parallel transport to preserve angles and magnitudes.

 

[Joker93]

While you can put the connection to be the same for both the sphere and the ellipsoid, it will only be compatible with one of the metrics and not the other. Generally, when you have a metric, you can have connections that are not metric compatible. Note that metric compatibility is not the same thing as imposing a Levi-Civita connection. The Levi-Civita connection is the unique torsion free and metric compatible connection.

In other words, when you do impose a metric (which is the only thing that is going to separate your ellipsoid from the sphere), then you will not have both $\nabla g_S = 0$ and $\nabla g_E = 0$ (where $g_S$ and $g_E$ are the metrics on the sphere and ellipsoid, respectively).

Edit: Note: The metric is what allows you to define angles and lengths of tangent vectors and so metric compatibility is what is required for parallel transport to preserve angles and magnitudes.

I don't understand why do you emphasize metric compatibility.
Since the curvature tensor is the same for the smooth manifold $S^2$ and does not change if we give it two different metrics, then where does metric compatibility play any role in this?

Also, on your note, why do all the people explain the effects of curvature via the change in the angle of a vector when it is parallel transported along a closed loop(as you say we need a metric to describe this) since the curvature does not need a metric to be defined?

 

[Orodruin]

I don't understand why do you emphasize metric compatibility.
Since the curvature tensor is the same for the smooth manifold $S^2$ and does not change if we give it two different metrics, then where does metric compatibility play any role in this?

Because the assertion that angles and lengths are preserved by the parallel transport is tied to metric compatibility. Also, without the metric, there is no difference between your manifolds. You said it yourself in post #1. As far as manifolds go, the sphere and the ellipsoid are diffeomorphic. The only thing that separates them is the metric. Without the metric, they are the same manifold. The curvature tensor by definition is related to the connection, not to the metric. It tells you how a vector changes when you parallel transport it around an infinitesimal loop. It should therefore come as no surprise that the same connection gives the same curvature tensor.

Also, on your note, why do all the people explain the effects of curvature via the change in the angle of a vector when it is parallel transported along a closed loop(as you say we need a metric to describe this) since the curvature does not need a metric to be defined?

In the case of a 2D-manifold with a metric, you can relate the change in the angle to the integral of the Ricci curvature over the enclosed area (this also holds globally - not only infinitesimally - as long as the manifold is simply connected). It is not so much of an explanation as it is an example intended to bring some sort of intuition.

 

[Joker93]

In the case of a 2D-manifold with a metric, you can relate the change in the angle to the integral of the Ricci curvature over the enclosed area (this also holds globally - not only infinitesimally - as long as the manifold is simply connected). It is not so much of an explanation as it is an example intended to bring some sort of intuition.

So, that intuitive picture is not there to describe what fundamentally means for a manifold to have non-zero components for its Riemann curvature tensor, right? Those pictures are of Riemannian manifolds. So, how can I understand intuitively and geometrically what does it mean for a smooth manifold to have curvature, without the intuitive picture of parallel transport along a Riemannian manifold? In other words, if I would told you to give a graphical explanation of curvature for a smooth manifold, how would you describe/show it?

 

[Joker93]

@Orodruin Also, something related that I think contributes to my confusion.
If you watch 30 seconds from 5:45 in the lecture:
I am a bit confused about which structure gives "shape" to our manifold. The lecturer says that it is the connection that does so, while my intuition always pointed to me that it was the metric that did so. Might it be that the lecturer secretly talks about the Levi-Civita connection, and thus he secretly assigns a metric to each smooth manifold that he is talking about?
Thanks for your patience.

 

[Orodruin]

Those pictures are of Riemannian manifolds. So, how can I understand intuitively and geometrically what does it mean for a smooth manifold to have curvature, without the intuitive picture of parallel transport along a Riemannian manifold?

Essentially that the parallel transport of a vector from one point to the other generally depend on the path (and not only the homotopy of the path).

@Orodruin Also, something related that I think contributes to my confusion.
If you watch 30 seconds from 5:45 in the lecture:
I am a bit confused about which structure gives "shape" to our manifold. The lecturer says that it is the connection that does so, while my intuition always pointed to me that it was the metric that did so. Might it be that the lecturer secretly talks about the Levi-Civita connection, and thus he secretly assigns a metric to each smooth manifold that he is talking about?
Thanks for your patience.

The "very particular choice of nabla" he is talking about is the Levi-Civita connection of the typical metric on the sphere.