Metric & One Forms: A R2 Confusion

[Silviu]

Hello! I am a bit confused about how the metric transforms vector into one forms. If we have a 2-sphere and we take a point on its surface, we have a tangent plane there on which we define vectors at that point. A one form at that point is associated to a vector at that point through the metric on the sphere i.e. $\omega_\mu = g_{\mu \nu} A^\nu$. However, if I understood this correctly, the tangent space is $R^2$, in which the metric is $diag(1,1)$. So if both the vectors and the one-forms (or tensors in general) are defined at a point, so in the tangent space at that point, why are they different than the ones in $R^2$ i.e. if in $R^2$ we would use the $diag(1,1)$ metric to go from vectors to one forms, why in the tangent space of a point on a sphere, which is also $R^2$, we use the metric of the sphere and not $diag(1,1)$? Thank you!

 

[Orodruin]

You cannot say that the metric in $R^2$ is ${\rm diag}(1,1)$. First of all, this is a coordinate dependent (or, more generally, a basis dependent) statement. Second, a tensor is not equal to a matrix. However, you can represent the components of a tensor (given a basis) in matrix form.

To answer your question, a one-form is a linear map from the tangent space to scalars. In the case of the metric, $\omega_V(X) = g(X,V)$ for some fixed vector $V$ is such a linear map and therefore $\omega_V$ is a one-form.

 

[Silviu]

You cannot say that the metric in $R^2$ is ${\rm diag}(1,1)$. First of all, this is a coordinate dependent (or, more generally, a basis dependent) statement. Second, a tensor is not equal to a matrix. However, you can represent the components of a tensor (given a basis) in matrix form.

To answer your question, a one-form is a linear map from the tangent space to scalars. In the case of the metric, $\omega_V(X) = g(X,V)$ for some fixed vector $V$ is such a linear map and therefore $\omega_V$ is a one-form.

Ok, I understand the first part. However for the second one, I still don't understand why do we use the metric defined on the sphere rather than the metric of $R^2$, if the vectors and one-forms are defined in $R^2$?

 

[Orodruin]

The vectors and tensors are defined on the sphere, they are not tensors/vectors in $R^2$. That the tangent space at each point is isomorphic to $R^2$ is a completely different story.

 

[Silviu]

The vectors and tensors are defined on the sphere, they are not tensors/vectors in $R^2$. That the tangent space at each point is isomorphic to $R^2$ is a completely different story.

Wait, isn't the tangent space on the sphere $R^2$, i.e. a vector in there can't tell the difference between the tangent space and actual $R^2$? So if want to calculate the distance between 2 points in that tangent space, don't we use the metric of $R^2$?

 

[Orodruin]

Wait, isn't the tangent space on the sphere $R^2$, i.e. a vector in there can't tell the difference between the tangent space and actual $R^2$?

Yes, that is what I said.

So if want to calculate the distance between 2 points in that tangent space, don't we use the metric of $R^2$?

There is no metric on the tangent space because there is no notion of distance within the tangent space. The notion of distance is a notion on the manifold itself. This distance notion is given by the metric, which is a (0,2)-tensor on the manifold, meaning it maps two tangent vectors to a scalar.

 

[PeterDonis]

The vectors and tensors are defined on the sphere

Strictly speaking, they're defined on the tangent space at each point, correct?

 

[Orodruin]

Strictly speaking, they're defined on the tangent space at each point, correct?

The tensors and vectors are certainly elements of the appropriate products of tangent (and cotangent) spaces. I would call a vector "on the sphere" a vector that is in $T_p S^2$, where $p$ is some point on the sphere. When you say "metric on the sphere" you are clearly referring to a section of $TS^2$ (with certain properties). Since we are only talking a single point and not necessarily about vector fields, it would probably make more sense to talk about the point in question. In that respect, the metric at the given point of the sphere is the metric that determines lengths of vectors in the tangent space. The problem of the OP is somehow to think that there is a separate metric for the tangent space itself. I think this should have been resolved by the first part of #2.

 

[Daverz]

I think the confusion is caused by early training about vectors in the plane having a magnitude and angle. The vector space ℝ² doesn't include these notions (is this why some authors prefer the term "linear space" instead of "vector space"?) The metric is induced on the tangent space by the embedding of the sphere in ℝ³. This is worked out in many elementary differential geometry books.

 

[Orodruin]

I think the confusion is caused by early training about vectors in the plane having a magnitude and angle.

Very likely. That and the fact that people like to use non-holonomic orthonormal bases in curvilinear coordinates. This gives the impression that the inner product is always the sum of the products of the components.

The metric is induced on the tangent space by the embedding of the sphere in ℝ³. This is worked out in many elementary differential geometry books.

This is the standard metric on the sphere. Technically there are other metrics you can impose as well (of course, they do not correspond to the one induced by the embedding). Regardless, I think it is fair to point out that one can always find a set of coordinates such that the metric does become diag(1,1).