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## Interpretation of torsion vs riemann tensor

 Quote by ianhoolihan Yes, I also realised they must have assumed the vectors commute. In terms of your suggestion, it is actually $$\nabla_XZ=\left.\frac{d}{dt}\right|_{t=0}\tau_{tX}Z$$ which I understand. I will think more about solving my problem, but I still don't see how. Cheers
That's odd. Have you done this computation yourself? Because as far as I can tell, the formula in wiki is wrong by a sign. I.e., I find that

$$\nabla_{\dot{x}_0}Y=\lim_{t\rightarrow 0}\frac{\tau_{x_t}^{-1}Y_t-Y_0}{t}$$

And this is also what my book on riemannian geometry tells me.

 Recognitions: Gold Member Homework Help Science Advisor But anyway, this detail aside, the idea I was hinting to is the following. Set $f(x_1,x_2,x_3,x_4):=\tau_{x_1X}^{-1}\tau_{x_2Y}^{-1}\tau_{x_3X}\tau_{x_4Y}Z$ Notice that $\tau_{sX}^{-1}=\tau_{-sX}$ and $\tau_{tX}^{-1}=\tau_{-tX}$. We want to compute d²/dsdt of f(s,t,s,t). So by the chain rule, $$\left.\frac{d}{dt}\right|_{t=0}f(s,t,s,t)=\frac{ \partial f }{\partial x_2}(s,0,s,0)+\frac{\partial f}{\partial x_4}(s,0,s,0)=\tau_{-sX}(-\nabla_Y(\tau_{sX}Z))+\nabla_YZ$$ and $$\left.\frac{d^2}{ds dt}\right|_{t=0}f(s,t,s,t)=\frac{\partial^2 f}{\partial x_1\partial x_2}f(0,0,0,0) + \frac{\partial^2f}{\partial x_3\partial x_2}f(0,0,0,0)+0+0=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ$$

 Quote by quasar987 That's odd. Have you done this computation yourself? Because as far as I can tell, the formula in wiki is wrong by a sign. I.e., I find that $$\nabla_{\dot{x}_0}Y=\lim_{t\rightarrow 0}\frac{\tau_{x_t}^{-1}Y_t-Y_0}{t}$$ And this is also what my book on riemannian geometry tells me.
I guess it depends on how you define the parallel propagator. In the (different) notation I was using the numerator was $\Gamma(t \to 0)V_{\gamma{t}} - V_{\gamma(0)}$. You're right though --- after looking at their notation again, $\tau_{x_t} \leftrightarrow \Gamma(0 \to t)$, so it should be $\tau_{x_t}^{-1}$

 Quote by quasar987 But anyway, this detail aside, the idea I was hinting to is the following. Set $f(x_1,x_2,x_3,x_4):=\tau_{x_1X}^{-1}\tau_{x_2Y}^{-1}\tau_{x_3X}\tau_{x_4Y}Z$ Notice that $\tau_{sX}^{-1}=\tau_{-sX}$ and $\tau_{tX}^{-1}=\tau_{-tX}$. We want to compute d²/dsdt of f(s,t,s,t). So by the chain rule, $$\left.\frac{d}{dt}\right|_{t=0}f(s,t,s,t)=\frac{ \partial f }{\partial x_2}(s,0,s,0)+\frac{\partial f}{\partial x_4}(s,0,s,0)=\tau_{-sX}(-\nabla_Y(\tau_{sX}Z))+\nabla_YZ$$ and $$\left.\frac{d^2}{ds dt}\right|_{t=0}f(s,t,s,t)=\frac{\partial^2 f}{\partial x_1\partial x_2}f(0,0,0,0) + \frac{\partial^2f}{\partial x_3\partial x_2}f(0,0,0,0)+0+0=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ$$
Excellent. I was getting confused with how one does
$$\frac{d}{ds}f(s,t,s,t)$$
given there are two $s$ slots. Your solution of letting $x_1=s,x_2=t,x_3=s,x_4=t$ is neat.

Thanks a bunch quasar987, that has helped a lot. Now I'll head off and do the additional bit where the vector fields do not commute.

 Quote by Matterwave You can take parallel transport on the 2-sphere to be "move the vector like you would in flat 3-D Euclidean space, and then project the resulting vector onto the tangent space of the sphere". That description matches with the Levi-Civita connection. That is, in fact, what the picture shows, and you can clearly see that the resulting vector is different than the initial vector.
Matterwave, can you clarify what you mean by this. If you mean move the vector in 3-space with the base' of the vector moving along the curve, and the direction staying constant (in three space), and projecting on to the tangent space...well this does not work, as since the direction does not change, once it returns to the start, it is still pointing in the same direction, so has the same projection...

I found something on wikipedia that was useful:
 A more appropriate parallel transportation system exploits the symmetry of the sphere under rotation. Given a vector at the north pole, one can transport this vector along a curve by rotating the sphere in such a way that the north pole moves along the curve without axial rolling. This latter means of parallel transport is the Levi-Civita connection on the sphere.
This makes sense in the triangle' case, but also why vectors precess around a curve of constant latitude (not the equator). I can't put it into symbols, but I just have a feeling that if you want an axis of the sphere to track a line of constant latitude, there must be some axial rolling', which means precession??

The idea of the south pointing chariot is also helpful.

 You project at each infinitesimal movement so that the vector always stays in the tangent surfaces, you don't project at the end. Maybe that was your confusion? With this scheme you can imagine that parallel translating along a line of constant latitude on a sphere is the same as parallel transporting along the lip of a cone you put onto this sphere (like a hat). The angular displacement around a closed circle is exactly the angular defect of the cone (how much of an angle you cut off from a flat sheet of paper to make that cone). Now you can see that the "cone" that fits onto the equator (as opposed to any other latitude) is not a cone, but a cylinder. A cylinder is not intrinsically curved (only extrinsically) and so parallel transport around a closed circle on a cylinder returns the vector to its original position.

 Quote by Matterwave You project at each infinitesimal movement so that the vector always stays in the tangent surfaces, you don't project at the end. Maybe that was your confusion?
Hmmm, maybe. Say you have a vector point (tangentially) North on the sphere, and you wish to send it East along a line of constant latitude. You first move it dx along the curve, and then project it down. Since the tangent plane is at an angle to the original tangent plane, the new projected vector now has a component pointing slightly west(?). This becomes your new vector, and on you go. Correct?

 Yes, if I have it visualized correctly, that sounds right to me.

 Quote by ianhoolihan I found something on wikipedia that was useful: A more appropriate parallel transportation system exploits the symmetry of the sphere under rotation. Given a vector at the north pole, one can transport this vector along a curve by rotating the sphere in such a way that the north pole moves along the curve without axial rolling. This latter means of parallel transport is the Levi-Civita connection on the sphere.
 Quote by ianhoolihan Hmmm, maybe. Say you have a vector point (tangentially) North on the sphere, and you wish to send it East along a line of constant latitude.
OK, having thought about it more, does anyone know what axial rolling' is? Otherwise, for the vector at the fixed latitude above, you could just rotate it about the North/South pole axis, and it would come back to where it started...