Curvature Tensor for Dual Vectors

[Phinrich]

TL;DR Summary

Is it possible to define a curvature tensor for dual vectors and what form would that take?

Good day all.

Given that in Sean Carroll`s Lectures on GR he states that when calculating the covariant derivative of a 1-Form the Christoffel symbols have a negative sign as opposed to for the covariant derivative of a vector, would it be naive to think that, given the usual equation for the Riemann Curvature Tensor of a vector, we can simply multiply all Christoffel symbols by -1 to obtain the Curvature Tensor for a 1-Form? Here I am interpreting "curvature tensor" as the result of a parallel transport of a 1-form around a closed path in spacetime. I also realize that, if I am correct, the resulting geometric object may well be different to the curvature tensor for a vector. If you do the substitution you basically end up swapping the first and second terms of the Riemann Tensor for a vector. Whether I am correct or wrong another interesting point is, can we define the Bianchi Identity for whatever geometric object results?

Thanks

Paul

 

[Orodruin]

There is no such thing as the "Riemann curvature tensor of a vector". There is only the Riemann curvature tensor. That this tensor is related to the parallel transport of a vector around a small loop is something else (although this is usually how it is introduced). The parallel transport of a one-form around a small loop can also be expressed in terms of the same Riemann curvature tensor, but the curvature tensor itself is the same geometrical object.

 

[Phinrich]

Thank you very much for your reply which is very clear.

Paul

 

[Ibix]

Presuming that what you want to do is parallel transport a one-form around a small loop, I'd be wary of that approach. Some of the Christoffel symbols involve the last two indices of the Riemann tensor, which define the loop you are moving your object around. I think you would want those to be defined in terms of vectors - although the same information is in the one form equivalents of those vectors, so maybe it doesn't matter.

I think that the right thing to do is lower the first index and raise the second - i.e. calculate $g_{ia}g^{jb}R^a{}_{bkl}$. Then you can grind through the algebra and see if it matches your guess.

 

[Phinrich]

Thank you for your suggestion.

I will try that and see what I get.

Paul

 

[Orodruin]

I think that the right thing to do is lower the first index and raise the second - i.e. calculate $g_{ia}g^{jb}R^a{}_{bkl}$. Then you can grind through the algebra and see if it matches your guess.

This would surprise me. The curvature tensor (or the parallel transport of a vector or one-form) a priori has no connection to the metric unless you impose the Levi-Civita connection.

 

[Ibix]

This would surprise me. The curvature tensor (or the parallel transport of a vector or one-form) a priori has no connection to the metric unless you impose the Levi-Civita connection.

As we're in the GR forum I assumed that we were using that connection. In that context, I think that what I proposed is just raising the index on the one form, applying the Riemann tensor (given some loop) to the resulting vector, then lowering the index on the result. Does that not work for some reason?

 

[Phinrich]

Thank you for your suggestion. Yes the question was posed in the context of GR. I will try what you suggest. I am no mathematical expert but a 'hobbyist'. Forgive me if I am wrong but if a vector has a single raised index and a dual vector a single lowered index then doesn't raising the index on the 1-form make it a vector ? Please corect my understanding if I am wrong here ?

 

[haushofer]

Thank you for your suggestion. Yes the question was posed in the context of GR. I will try what you suggest. I am no mathematical expert but a 'hobbyist'. Forgive me if I am wrong but if a vector has a single raised index and a dual vector a single lowered index then doesn't raising the index on the 1-form make it a vector ? Please corect my understanding if I am wrong here ?

Yes. The metric maps every dual vector to a 'normal' vector. This mapping is one to one. That's why an inner product is unambiguous. ;)

 

[Phinrich]

Thank you.

That makes sense to me.

Paul

 

[Ibix]

Forgive me if I am wrong but if a vector has a single raised index and a dual vector a single lowered index then doesn't raising the index on the 1-form make it a vector ?

Yes. That's why I'm not quite sure what @Orodruin's point is here. If he's saying that I stated a result without being clear on the limits of its applicability then I agree he has a point. If that's not what he means then, given our respective knowledge levels, I expect there's a learning opportunity coming my way...

 

[Phinrich]

Heres a WILD thought !

If we say that a Gradient is a 1-Form we can picture a representation of a Gradient in a vector space. If it was possible to take this representation and parallel transport it around a closed loop in the vector space we could define a curvature tensor which will likely be the same beast as if we had parallel transported a vector around the closed loop. If on the other hand we keep the dual vector in the dual space and could somehow parallel transport it around a closed loop in the dual space (if we can define such) would we be able to define a "curvature tensor" which would NOT be the same beast as in the first case ? I am sure that my lack of mathematical insight must be SCREAMING at you in this question.

 

[vanhees71]

A tensor is a tensor! It's and invariant object.

In a general differentiable manifold you have (tangent) vectors and co-vectors, i.e., at each point there's a tangent-vector space and of course also its dual space (i.e., the vector space of linear 1-forms). Though both spaces have the same dimension, there's no "canonical" (i.e., basis-independent) one-to-one mapping between vectors and co-vectors.

However in general relativity you have pseudo-Riemannian manifold, i.e., there's a fundamental non-degenerate symmetric bilinear form defined, which physicists call "the metric" though they should rather name it "the pseudo-metric", because the bilinear form is not positive definite but of the signature (1,3) (or (3,1) depending on the convention you use). Then there's indeed a canonical, i.e., basis-independent mapping between vectors and dual vectors and thus you can indentify vectors and dual vectors in a basis-independent way: That's because if you have a linear form $L^*$ defined on the vector space, and because the pseudo-metric is not degenerate there's a uniquely defined vector $L$ such that for any vector $V$ you have $L^*(V)=g(L,V)$. This provides a unique mapping between the linear form $L^*$ and the vector $L$.

That's why it also makes sense to talk about covariant and contravariant components of a vector and that's the reason, why you "draw indices up and down" by the components of the pseudo-metric.

 

[PeterDonis]

That's why I'm not quite sure what @Orodruin's point is here.

His point is that there is no such thing as "the curvature tensor for 1-forms". There is just the curvature tensor. Rearranging the indexes on the components so that you can compute the change in a 1-form when parallel transported around a closed loop, instead of a vector, does not make it "the curvature tensor of a 1-form" instead of "the curvature tensor of a vector". It just means you're computing a different quantity using the same curvature tensor.

 

[Phinrich]

Thank you for the clarification. It makes sense to me now.

 

[PeterDonis]

If we say that a Gradient is a 1-Form we can picture a representation of a Gradient in a vector space.

No, there is no such thing. If you have a metric, you can find a vector that corresponds to that 1-form (by using the metric to raise the index), and that will be an element of the space of vectors. But the gradient itself isn't.

 

[Orodruin]

Yes. That's why I'm not quite sure what @Orodruin's point is here. If he's saying that I stated a result without being clear on the limits of its applicability then I agree he has a point. If that's not what he means then, given our respective knowledge levels, I expect there's a learning opportunity coming my way...

The point was that the curvature is a statement about the connection. As such, even if you parallel transport the one-form around, the corresponding expression can only contain information pertinent to the curvature tensor itself. Now, due to the symmetries of the curvature tensor for the Levi-Civita connection, what you described $g_{ia} g^{jb} R^a_{bcd} = - R^j_{icd}$ and there is no actual information related to the metric here, just the curvature tensor itself. The general expression for the parallel transport of a one-form around a small loop, for an arbitrary connection, will be involving the same curvature tensor as the parallel transport of a tangent vector - regardless of the metric, which does not necessarily even exist or - if it does - may not be metric compatible.

 

[Ibix]

Thanks @PeterDonis and @Orodruin. So, if I'm understanding correctly, if you want to parallel transport a one-form $\omega_b$ around a loop then $g_{ai}g^{bj}R^i{}_{jkl}\omega_b$ would be the correct expression to use, at least in the context of vanilla GR where we can assume the existence of a metric and the Levi-Civita connection. Such a parallel transport may be more complicated to express, or even impossible, in a more general differential geometry context.

In the context of this thread, which was specifically discussing a "different curvature tensor", I should have made clear that $g_{ai}g^{bj}R^i{}_{jkl}$ is not a different tensor from $R^i{}_{jkl}$, just as $g_{ab}$ and $g^{ab}$ aren't different tensors. They're just different expressions of the same tensor. And I should have guessed that symmetries of the Riemann tensor meant that there'd be a simpler form for what I was trying to express.

Or am I still off track?

 

[Orodruin]

Such a parallel transport may be more complicated to express, or even impossible, in a more general differential geometry context.

No, this was not my point. My point was that your expression with the metrics inserted is actually equivalent to a more general expression that does not require the existence of a metric - nor that the connection is metric compatible if it does exist. The parallel transport of a one-form should relatively easy to express - in fact not much harder than that of a tangent vector. My point was that by just doing the parallel transpirt, you should arrive at an expression that does not involve the metric - only the curvature.

Technically, the metric and the inverse metric are very different tensors. They do not even belong to the same tensor space as the metric is type (0,2) and the inverse type (2,0). They are however mapped into each other by the natural bijection between those spaces that exists once you put the metric in place.

 

[strangerep]

@Phinrich : Going back to your original question,... it's simpler to work with a commutator of covariant derivatives, as is shown in most GR textbooks. Acting on a vector, we get the usual expression for the Riemann tensor (plus possibly some more stuff if torsion is nonzero).

To compute a similar case for that commutator's action on a covector, the simplest approach (imho) is to compute its action on a scalar $f := u^\alpha w_\alpha$, where both $u,w$ are arbitrary. Using the formula for its action on the vector $u^\alpha$ one can deduce the corresponding action on $w_\alpha$ .

 

[Phinrich]

As the one who started this thread I would like to thank all who contributed, namely PeterDonis, strangerep, Ibix and Orodruin and anyone else who I may have missed, for your valuable comments. I have learned a LOT from this discussion. Strangerep, I still need to process your comment above to fully understand it. In fact there are a few responses which I still need to digest and understand. Ibix your last comment about the metric and inverse metric being different beasts is what I also believe. Saying that doing the parallel transport of a 1-form and of a vector should arrive at an expression involving only the curvature was, I believe, my starting point for this discussion. My question was would that "curvature" be the same as the curvature arrived at by transporting a vector. Understanding that vectors and 1-forms exist in different spaces, if we could represent a 1-form in the same space as a vector and do the transport we would be transporting in the space and should get the same curvature but if we purely transport a 1-form in the dual space would we get a curvature different from that of transport of a vector in a space as opposed to the dual space. Hope this is clear and does not cinfuse the issue more.

 

[PeterDonis]

if we purely transport a 1-form in the dual space would we get a curvature different from that of transport of a vector in a space as opposed to the dual space

If you don't have a metric, there is no given relationship between 1-forms and vectors, so it doesn't even make sense to compare the curvature you get from transporting some vector in the vector space, to the curvature you get from transporting some 1-form in the dual space.

If you do have a metric, then the curvature you get from transporting a 1-form in the dual space has to be the same (except for raising/lowering indexes as necessary) as the curvature you get from transporting the corresponding vector (the vector you get by raising the 1-form's index with the inverse metric) in the vector space.

 

[Phinrich]

Thank you. That is how I understand it.

 

[Orodruin]

The easy way of arriving at the actual expression for the change in a 1-form $\omega$ after parallel transport around a loop based on already knowing the vector expression is the following:

Consider the arbitrary 1-form $\omega$ and an arbitrary tangent vector $V$. Both are parallel transported around an infinitesimal loop spanned by the tangent vectors $X$ and $Y$. The differences after parallel transport we can denote $\delta\omega$ and $\delta V$ and they need to be a 1-form and a tangent vector, respectively. Because of linearity, we know that
$$
\delta V^a = R^a_{bcd} V^b X^c Y^d \qquad \mbox{and} \qquad
\delta \omega_a = C^b_{acd} \omega_b X^c Y^d,
$$
where we know that $R^a_{bcd}$ are the components of the curvature tensor and $C^b_{acd}$ are some coefficients that we need to determine. Since $\omega$ and $V$ are parallel transported, $\omega(V) = \omega_a V^a$ is constant and therefore the same before and after parallel transport. This means that
$$
\omega(V) = (\omega + \delta\omega)(V+\delta V) = \omega(V) + \omega(\delta V) + \delta \omega(V)
$$
to linear order. In other words, $\delta \omega(V) = -\omega(\delta V)$. Expressing this in component form, we find
$$
C^b_{acd} \omega_b X^c Y^d V^a = - \omega_b R^b_{acd} V^a X^c Y^d.
$$
Since the 1-form $\omega$ and the vectors are all arbitrary, we must have $C^b_{acd} = - R^b_{acd}$. In other words, the parallel transport of the 1-form can also be expressed in terms of the Riemann curvature tensor only. In no part of the above did we have to refer to a metric and the result is general and holds irrespective of the connection that is used.

 

[Phinrich]

OK thank you for this detailed post. Referring to your statement that the two curvature tensors are related by ä minus sign. I started this post referring to a quote from Sean Carrols Book on GR and that he said the directional derivative for a 1-form was the same as for a vector except that each Christoffel symbol would be multiplied by -1. There are four terms in the curvature tensor, each involving some combination of Christoffel symbols. The third and 4th terms each involve the product of Christoffels so they should not change. Only the 1st ans 2nd terms should swop places in the expression for the Curvature tensor for the 1-Form. The 3rd and 4th terms also should swop signs. This would seem to agree with your analysis above.

 

[Orodruin]

Referring to your statement that the two curvature tensors are related by ä minus sign.

I have not said this. There is only one curvature tensor.

 

[Phinrich]

No you did not. Sorry.