Interpretation of torsion vs riemann tensorby ianhoolihan Tags: interpretation, riemann, tensor, torsion 

#1
May1412, 06:34 PM

P: 145

Hi all,
I am working through Visser's notes http://msor.victoria.ac.nz/twiki/pub...s4642011.pdf section 3.5 onward. I am trying to differentiate between the torsion and the Riemann curvature tensor in a heuristic manner. It appears from "Geometric interpretation 1" (page 73) that the torsion tensor is related to going along both paths of a parallelogram to get to opposite corners, and then finding the difference...how the infinitesimal parallelogram does not close. The Riemann tensor is introduced simply through the noncommutativity of the the covariant derivative, though looking around on the internet, I have also found references to infinitesimal parallelograms. Could someone explain, in the same sense as above, what motivates the calculation of the Riemann tensor (i.e. commutator of covariant derivatives)? I have heard references to the "twist" of frames along a geodesic for the torsion as opposed to the "roll" for the Riemann curvature  if you could explain these, that would also be appreciated. Cheers 



#2
May1412, 09:12 PM

P: 145

Hmmm, is it related to the fact that the torsion is to do with the "position not commuting" in the sense that for a parallelogram ABCD, the position reached by going from A> B > C is different to that reached by going from A > D > C ?
Whereas for the Riemann curvature tensor, we parallel transport a vector round an infinitesimal loop, but we return to the same position (with 0 torsion), and it is the change in direction of the vector we are concerned about? Cheers 



#3
May1412, 09:31 PM

P: 145

OK, more looking on my part at the wikipedia article in the first post: http://en.wikipedia.org/wiki/Riemann...trical_meaning. Does anyone know how to show the equation
[tex]\left.\frac{d}{ds}\frac{d}{dt}\tau_{sX}^{1}\tau_{tY}^{1}\tau_{sX}\tau_{tY}Z\right_{s=t=0} = (\nabla_X\nabla_Y  \nabla_Y\nabla_X)Z \ ?[/tex] Cheers 



#4
May1512, 01:15 AM

P: 2,045

Interpretation of torsion vs riemann tensor 



#5
May1512, 04:53 AM

P: 145

In saying that, while I agree with your construction in post#4, I always worry that such "head to tail" diagrams imply a distance, when in this case, I believe it is more about the direction. In your diagram, what I think would be the case with torsion is that [itex]\nabla_X Y  \nabla_Y X[/itex] is the direction composed of that of [itex][X,Y][/itex] as in your post, plus the direction due to the parallelogram not quite closing in terms of positions, as in Visser's notes, which is due to torison. (I.e. torsion is related to differences in position, not direction.) That's not clear, but look at Visser pg 7374, and you'll see what I mean. To quote him "in a manifold with torsion there are no infinitesimal parallelograms". Ben Niehoff's answer about torsion being a twist was helpful, though I'd need to think about it more to write the equation he talked of. As for the "rolling" associated with the Riemann tensor, I'm happy to hear thoughts. As for my question, I am now more interested in how to prove [tex]\left.\frac{d}{ds}\frac{d}{dt}\tau_{sX}^{1}\tau_{tY}^{1}\tau_{sX}\tau_{tY}Z\right_{s=t=0} = (\nabla_X\nabla_Y  \nabla_Y\nabla_X)Z [/tex] as I think I can do the interpretation once I've covered this. Any thoughts? I don't see why it should the product of propagators becomes a product and subtraction of covariant derivatives, i.e. I'd have thought it'd be [itex]\nabla_X+\nabla_Y  \nabla_Y\nabla_X[/itex] or [itex]\nabla_X\nabla_Y (\nabla_Y)(\nabla_X)=\nabla_X\nabla_Y\nabla_Y\nabla_X[/itex], instead of a mixture of both. This reminds me of how the commutator of a lie algebra comes from the commutator of the lie group. Still not sure. Ideas? Cheers 



#6
May1512, 05:07 AM

P: 145

Ahah, Figure 1 in this may help explain what I was talking about:
http://arxiv.org/PS_cache/arxiv/pdf/...711.1535v1.pdf 



#7
May1512, 09:00 AM

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I am confused as to what [itex]\tau_{sX}^{1}\tau_{tY}^{1}\tau_{sX}\tau_{tY}Z[/itex] even means. Assuming Z belongs to TpM, why would [itex]\tau_{sX}^{1}\tau_{tY}^{1}\tau_{sX}\tau_{tY}Z[/itex] belong to TpM (so that differentiation makes sense)?




#8
May1512, 02:48 PM

P: 2,045

I am not familiar with the equation you put up, so I probably can't help you there. I've seen many other ways to describe the curvature, the most geometric of which is parallel transport around a closed loop. However, I am used to using covariant derivatives rather than propagators. 



#9
May1512, 03:56 PM

P: 145





#10
May1512, 04:09 PM

P: 145

Covariant derivatives and parallel propagators go hand in hand. As in the post above, the equation I am referring to is parallel propagation around a "parallelogram" (I am not sure if it closes with torsion etc). As for the argument as why [itex][\nabla_X , \nabla_Y]Z[/itex] is important for curvature, I only half understand it. You are looking at how the vector [itex]Z[/itex] changes in the [itex]X[/itex] direction, and seeing how this changes in the [itex]Y[/itex] direction, and then subtracting the reverse. It doesn't exactly seem intuitive as to why this corresponds in some way to curvature  at least to me. Two points as an aside: 1) Is it propagation "around" a parallelogram (i.e. back to the start) or taking the two different routes from diagonal corners, and subtracting (as [itex][\nabla_X , \nabla_Y]Z[/itex] and your diagram imply)? This is part of my problem with the formula I am trying to work out: the LHS appears to be around the parallelogram, the RHS subtracting diagonal routes. 2) It seems like torsion is first order in covariant derivatives, and the riemann tensor second order. Do we have a third order? I.e. using a parallelepiped with three vectors? Cheers 



#11
May1512, 08:23 PM

P: 2,045

The idea is that in the presence of curvature, a vector which is parallel transported around a closed loop does not return to itself. We use that as a DEFINITION of curvature.
In a flat space, parallel transport around a closed loop will obviously return the vector to itself. Perhaps, imagine a 2sphere imbedded in 3space. The vectors must always be tangent to this 2 sphere. I can move the vectors around, but there is no universal definition of parallel present. See the figure in this article: http://en.wikipedia.org/wiki/Parallel_transport You can see the effects of curvature on the parallel translated vectors there. 



#12
May1512, 09:29 PM

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I feel like your formula can be proved by first showing that [tex]\nabla_XZ=\left.\frac{d}{dt}\right_{t=0}\tau_{tX}^{1}Z[/tex] and then using the chain rule on [itex]f(s,t,s,t)=\tau_{sX}^{1}\tau_{tY}^{1}\tau_{sX}\tau_{tY}Z[/itex]. (For the first identity, write Z in terms of a parallel frame along an integral curve of X.) 



#13
May1512, 10:00 PM

P: 145

Thanks Matterwave, I am well aware of all that.
I've been reading MTW etc, and I think I understand how torsion etc fits in. Like in your diagram, [itex][U,V][/itex] always closes the infinitesimal parallelogram, without any use of parallel propagation (and I found a mathematical proof for it in MTW). I believe the difference with the torsion is that [itex]T(U,V)[/itex] closes the parallelogram formed by the geodesics of [itex]U,V,U\parallel,V\parallel[/itex]. That is, the torsion depends on the choice of connection. (That is obviously true, given the formula). I am unsure if in your picture it would close the inner quadrilateral, as I believe you have to go along geodesics of the given curves. Then again, I'm still not sure. I also wonder if incorporating "twist" into your diagram when there is torsion may help...i.e. [itex]U\parallel[/itex] and [itex]V\parallel[/itex] have components in and out of the page due to their twisting when parallel propagated. What'd I'd like to see is an example of a connection with torsion, and a picture like yours drawn on "graph paper" as such. As for your example of the 2sphere which is always used to illustrate parallel propagation...well, if, in the usual way, parallel propagation is to "stay at a fixed angle with respect to the coordinates" then isn't a vector parallel propagated around a parallelogram (i.e. defined by latitude A to B, longitude C to D) unchanged? That would imply zero curvature, which is wrong... 



#14
May1512, 11:23 PM

P: 2,045

If you take, in my picture, U and V to be the tangent vectors to geodesic curves (w.r.t. a metric), then introducing torsion to that picture would mean the parallel propagated vectors can twist into or out of the page, but not in any other direction (or else the connection would no longer be metriccompatible).
You can take parallel transport on the 2sphere to be "move the vector like you would in flat 3D Euclidean space, and then project the resulting vector onto the tangent space of the sphere". That description matches with the LeviCivita connection. That is, in fact, what the picture shows, and you can clearly see that the resulting vector is different than the initial vector. 



#15
May1612, 01:21 AM

P: 145

In terms of your suggestion, it is actually [tex] \nabla_XZ=\left.\frac{d}{dt}\right_{t=0}\tau_{tX}Z [/tex] which I understand. I will think more about solving my problem, but I still don't see how. Cheers 



#16
May1612, 01:56 AM

P: 145

[tex]2{T^a}_{bc}X^b X^c \lambda_1 \lambda_2 + O(\lambda^3). [/tex] I am not sure if taking the limit of small [itex]\lambda[/itex] is the same as covariant derivatives, as you also need to get the geodesic equation in. As for the second point, I've discovered that since lines of constant latitude are not geodesics, a vector will rotate as it is transported along them. Now I just need to convince myself how that works. I do not understand your description however  do you mean to move the vector along the path on the 2sphere as you would in 3d Euclidean space? Ah, and then project it...I am going to get a ball and a toothpick to check with that lines of latitude cause precession, but I think I see what you mean. Now, all that remains is the equation... 



#17
May1612, 03:33 AM

P: 2,045

I'm going to leave it to quasar to answer the equation question...




#18
May1612, 08:24 AM

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[tex]\nabla_{\dot{x}_0}Y=\lim_{t\rightarrow 0}\frac{\tau_{x_t}^{1}Y_tY_0}{t}[/tex] And this is also what my book on riemannian geometry tells me. 


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