# I Understanding Parallel Transport

Tags:
1. May 20, 2017

### Arcturus7

I'm currently in a GR class and have come across the notion of parallel transport, and I've searched and searched the last few days to try and understand it but I just can't seem to wrap my head around it, so I'm hoping someone here can clarify for me.

The way I picture parallel transport is thus; if I have a manifold M equipped with a metric, then I can construct a covariant derivative. I do this by using the Levi-Civita connection, which is characterised by having zero torsion and also preserving the metric. The upshot of this is simply that, a) the covariant derivative of the metric is zero, and b) because the connection is torsion free, then we have that the covariant derivatives w.r.t two different coordinates commute. Parallel transport is the act of, if you like, "picking up" a vector which is a member of the tangent space of a point p, and then moving it along come curve C whilst always keeping its orientation the same as it was at p.

This is what confuses me; if we keep it "pointing in the same direction" by using the covariant derivative to vary its components in conjunction with the varying basis vectors, then surely at any given point along the line the vector will not generally be in the tangent space of that point. I was under the impression that it only makes sense to describe vectors as being in the tangent plane of the manifold at a point.

The example I keep coming back to is that of parallel transport along a given latitude of a sphere (not a geodesic). I understand mathematically (i.e I can see why the equations show) that a vector will not be parallely transported along this line because it isn't a geodesic, however by naively looking at a sphere and imagining moving a vector around a fixed latitude I cannot see why the transported vector does not coincide with the original one. I understand also the argument of using a cone tangential to the circle, and this makes sense to me, but surely we should be able to arrive at that conclusion without resorting to the cone?

So I guess my first question is this; when we parallel transport a vector along a curve, does the vector have to remain in the tangent space of each point, or does it have to stay parallel to itself?

2. May 20, 2017

### Staff: Mentor

3. May 20, 2017

### Staff: Mentor

Here's a video on it because seeing is believing:

4. May 20, 2017

### Arcturus7

Thanks for the responses. I saw that video and thought it made sense - the guy just sort of "slides" the tangent plane (and this the tangent vectors) along the surface of the sphere. I totally get why that loop leads to two vectors that aren't parallel. However I also watched a video of Leonard Susskind trying to explain the topic and he seemed to suggest what I said in my post, that after each step we have to keep the vector being transported with the same orientation as before.

Maybe it's because I'm imagining embedding the manifold into Euclidean space. If the manifold weren't embedded, surely it wouldn't make any sense at all for a vector to NOT be in the tangent plane?

Edit 1: I also looked through the quotes thread and again, some part s made sense and others didn't. It's really troubling me that it seems I don't have s grasp on what's even happening here.

5. May 20, 2017

### Staff: Mentor

Yes. In general in the fiber of each point.
Yes. But this is a misleading image. What does it mean? It remains unchanged in direction and length, but for "parallel to itself" you needed some outer space, which you don't have. I found the picture of a gun on a tank in the thread I quoted funny but telling.

6. May 21, 2017

### Orodruin

Staff Emeritus
This. To point "in the same direction" is a heuristic that you should pay no attention to if it confuses you. If you want to use it, you must remember that the affine connection and the curve define what "in the same direction" means for different tangent spaces. A vector parallel transported along different curves between the same points generally will not result in the same parallel transported vector. This is the entire point of curvature.

Note that you would also typically say that the Levi-Civita connection is metric compatible (not that it preserves the metric, at least I have never encountered this nomenclature), which is equivalent to the metric being a parallel field, i.e., $\nabla_X g=0$ for all $X$.

7. May 21, 2017

### Arcturus7

I also liked this analogy but I'm not sure how much it helped me. In the case of circling a given latitude, if the tank barrel begins pointing towards north, then halfway round the Earth, where is the barrel pointing? Either; the barrel is pointing in "the same direction" as it was before, which would be into outer space somewhere with a component perpendicular to the earth which it didn't have originally (since by our own admission it was pointing north); or it is still pointing north because during the parallel transport we forced the vector to remain in the tangent space at each point. Again, I think the issue may be because I'm embedding the sphere into Euclidean space. If it wasn't embedded then I would definitely say the answer was for latter, but in that case I don't understand why circling a latitude would not result in the same vector.

*Edit* reading your comment again, I see that what happens IS the second case, since this is precisely keeping the gun-barrel-vector in the tangent space. That much feels clear, so thank you again!

In response to Orodruin, I suspected as much - that I was holding on to a sort of intuitive interpretation of something that just wasn't helping. Thanks for your help :)

8. May 22, 2017

### zwierz

that is wrong. when the torsion is equal to zero then the Riemann curvature tensor is responsible for for result of $\nabla_i\nabla_j-\nabla_j\nabla_i$
it is better to think the parallel transport as a mapping from one tangent space to another one that is from $T_pM$ to $T_qM$ where $p,q\in M$ are the end points of the curve. Scalar product of each two vectors is conserved under the parallel transport.

9. May 22, 2017

### Orodruin

Staff Emeritus
True. The correct statement is that $\nabla_X Y - \nabla_Y X$ is equal to the corresponding Lie bracket.

This is a special case. The connection might not be metric compatible, which would mean the scalar product would generally change, or there might not even be a metric, which would render the notion of a scalar product of two elements in the tangent space obsolete. What would be conserved regardless of the connection is the product of the parallel transports of a tangent vector in $T_pM$ and a 1-form in $T^*_pM$.

Last edited: May 22, 2017
10. May 22, 2017

### Arcturus7

AH yes of course you're correct zwierz, I don't know what I was thinking when I typed that!