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Imagining a non-embedded manifold that forces a reformulation of the tangent space at a point as partial derivatives of any arbitrary smooth functions on the manifold along a parameterized curve is mind-blowing. However, the idea of a derivative along a path integral being akin to a velocity or a tangent is helpful.

Unfortunately, when the need to compare vectors in different points on the manifold (hence, different tangent spaces) comes up, my imagination hits a second obstacle.

One easy and unsatisfactory perspective is to just think about smoothness and differentiability again. Unfortunately, all presentations on parallel transport and the covariant derivative rely on arrows on the manifold.

What would be a good mental caricature of these partial derivatives belonging to different tangent vectors and being transported?

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  • $\begingroup$ Can you elaborate on how all presentations of parallel transport rely on "arrows"? $\endgroup$
    – J. Murray
    Commented Apr 24, 2023 at 4:11
  • $\begingroup$ @J.Murray Vectors as arrows. Diagrams, lectures and presentations accessible online. Typically two arrows and the problem of how to subtract them when they don't belong to the same vector space. $\endgroup$
    – JAP
    Commented Apr 24, 2023 at 4:14
  • $\begingroup$ Example: youtu.be/AU2Hp2BDNi8 $\endgroup$
    – JAP
    Commented Apr 24, 2023 at 4:17
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    $\begingroup$ For the record, my intent was not to shame - I meant to point out that if $p\neq q$ are different points and $X_p$ and $Y_q$ are directional derivatives defined at $p$ and $q$ respectively, then the action of $X_p$ on a smooth function $f$ is to differentiate $f$ and evaluate the result at $p$, and likewise with $Y_q$. If you try to add $X_p+Y_q$ to obtain some vector $Z$, then where would $Z$ live? at what point would $Z$ evaluate a function after differentiation? $\endgroup$
    – J. Murray
    Commented Apr 24, 2023 at 17:00
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    $\begingroup$ The reason that you don't need a connection to differentiate a function along a curve is because in the definition of the difference quotient, you subtract $f\big(\gamma(t+h)\big)-f\big(\gamma(t)\big)$ - a real number from a real number. If you try the same with a vector field $\vec V$, you find $\vec V \big(\gamma(t+h)\big) - \vec V(\gamma(t)\big)$ - but the former of these terms is defined in the tangent space at $\gamma(t+h)$ while the latter is defined in the tangent space at $\gamma(t)$, and there is no a priori way to subtract them from one another. $\endgroup$
    – J. Murray
    Commented Apr 26, 2023 at 20:08

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You should feel good about being confused at this point, because it means that you're thinking deeply and trying to gain intuition about one of the most important parts of differential geometry — namely, how to compare vectors (and more generally tensors) at different points of a manifold. But there is no single answer to your questions, precisely because this ambiguity of how to "connect" the tangent space at one point to the tangent space at another is always present.

There are three main interesting approaches to the problem: the covariant derivative (or more general connection), the Lie derivative, and the exterior derivative. Even within those three, there are many choices to be made, rather than any canonically obvious choice for the "correct" derivative.

The one you use will be dictated by some other feature of your problem. For example, in physics, we usually assume that a metric exists with some meaning independent of coordinates. That, and a few more assumptions, are enough to basically specify one particular covariant derivative. Or there may be some special vector field (perhaps describing a physical symmetry), which will pick out a Lie derivative to use. Or maybe you only need to use differential forms (usually if you're integrating over the manifold or a surface in it), in which case you'll want to use the exterior derivative.

Now, as to parallel transport, this is defined in terms of a connection. In general, a connection is just exactly a function that tells you how much a given vector field changes as you move in a given direction. This is exactly the question you are pondering: how do you decide what the relationship is between vectors in two different tangent spaces? The connection tells you exactly how.

But connections are pretty arbitrary. They have to be linear, and they have to satisfy the Leibniz rule, but otherwise a general connection has basically no restrictions; you can make up your own. A vector field is said to be "parallel transported" along some curve if the field's covariant derivative along that curve is zero — which depends on your choice of connection. In fact, you could even just set up a frame (a set of vectors that span the tangent space) at each point, and define your connection so that they don't change from point to point. (At least, for some neighborhood of a given point.) Frankel points out that the natural connection for surveyors is exactly of this type, though it is not "metric compatible" and actually has torsion (and he references this paper for more details).

Having said that, if you have a metric (which we usually do in physics), then there is one common choice for the connection: the Levi-Civita connection — not the only choice, but a common one. It's special because it it has no torsion, and the metric is constant with respect to this derivative. It also has the nice property that the dot product between two vectors that are parallel-transported with respect to this connection is preserved.

So, the moral of the story is: parallel transport means more-or-less whatever you want it to. You decide what it means by choosing a connection. Therefore, if you want intuition about parallel transport, work on your intuition about connections.

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  • $\begingroup$ Thank you. I have a vague idea of these approaches and it's good to reinforce. However, my question is how to picture schematically, informally this problem of "moving as parallel as possible a vector between two points" after getting rid of vectors as arrows jutting out of the manifold, and replacing them by partial derivatives. $\endgroup$
    – JAP
    Commented Apr 24, 2023 at 10:38
  • $\begingroup$ I guess my point is that there is no single correct answer to this problem, so there can be no single correct picture. I've expanded on this in my answer. $\endgroup$
    – Mike
    Commented Apr 24, 2023 at 19:40

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