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I have a general question about the nature of parallel transport. As far as I understand it, given a tensor at point $p$ and a smooth curve going through $p$, the parallel transport equations determine a tensor at any other point $q$ along the curve.

My question is, does the parallel transport equation simply associate, for each point $q$ along the curve, a tensor in the vector space at point $p$ with the point $q$, or does it actually map the tangent space at $p$ into the tangent space at $q$.

Perhaps more fundamentally, given a chart and two points $p$ and $q$ in said chart, for any tensor in $V_p$ is there a tensor in $V_q$ with the same components, and if so, are the tangent spaces only fundamentally different because they are based at different points on the manifold?

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    $\begingroup$ If no answers appear in a few days, perhaps you could consider posting this at Mathematics. $\endgroup$
    – Kyle Kanos
    Commented Dec 8, 2017 at 10:58
  • $\begingroup$ Parallel transport doesn't typically preserve the components. $\endgroup$
    – Javier
    Commented Dec 8, 2017 at 13:33
  • $\begingroup$ On a semi-Riemannian manifold, the natural coordinate vector fields are parallel and so are their restirctions to a curve. Hence parallel translation from $p$ to $q$ along any curve is a canonical isomorphism $v_{p} \rightarrow v_{q}$. $\endgroup$ Commented Aug 19, 2019 at 3:33
  • $\begingroup$ You might find the wiki page on holonomies useful. $\endgroup$ Commented Dec 24, 2021 at 15:24

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For the manifolds met in relativity, the tangent spaces at all points are vector spaces of the same dimension. Furthermore, their (pseudo) metric structure is the same, being of the same signature. The tangent spaces are therefore all the same as Minkowski spacetime (or, in a more general setting, a fixed dimension vector space with a fixed pseudo metric), so in answer to your last paragraph:

Perhaps more fundamentally, given a chart and two points $p$ and $q$ in said chart, for any tensor in $V_p$ is there a tensor in $V_q$ with the same components, and if so, are the tangent spaces only fundamentally different because they are based at different points on the manifold?

Yes, they only differ by the points they are tangent to. Indeed one way to think of the tangent bundle to the manifold is simply as the disjoint union of all the tangent spaces; each being the same and distinguished only by being associated with different points.

Your understanding of parallel transport along a curve as a special kind of linear, bijective map between the tangent spaces at either end of the curve is correct.

This I believe answers your question, but for completeness parallel transport is a specialized general linear map between these tangent spaces insofar that its differential version is a covariant derivative along the tangent to the curve defined by a connection. In its most abstract form, a connection is simply a linear map that (1) assigns to a smooth section $\Gamma(T\,M)$ of the tangent bundle $T\,M$ (i.e. a vector field on the manifold $M$) another smooth section $\Gamma(T\,M \otimes T^* M)$ of the tangent bundle tensored with the cotangent bundle $T^* M$ and (2) fulfills the Leibniz rule when we multiply our smooth section by any smooth function. That $T\,M \otimes T^* M$ looks fearsome, but all we are saying is that we begin with a vector field $X$ and our function returns a vector-field-valued linear function of $\nabla_Y\,X$ of a second vector field $Y$ - the second argument is simply where we put the "direction" we want to differentiate along. When we're calculating parallel transports along a system of curves, the vector field $Y$ is the system of tangents to these curves (or, contrariwise, the curves are the flow of the vector field $Y$) and we put $Y$ as the argument into our covariant derivative $\nabla_Y X$ to work out the directional derivative of $X$ along $Y$. The Leibnitz rule condition is simply that $\nabla_Y (f\,X) = \nabla_Y X + \nabla_Y f \otimes X$ for any smooth function $f:M\to\mathbb{R}$ defined on the manifold.

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