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In most of the textbooks that I have been reading and many of the answers here, it has been an important point that we cannot compare vectors in the tangent spaces of two different points. That is, given two points on a manifold $p, q\in M$, there is no natural notion of parallel transport that can allow us to take a vector from $T_pM$ and add/subtract it to one in $T_qM$.

At the same time, we can define a vector field and then define when this function $M\to TM$ is smooth. To me, this implies that we are doing calculus with the vectors in the vector field, which would require some way to compare vectors based at different points, contradicting the first notion. In fact, in order to define a smooth vector field as a smooth section of the bundle, we need to define a structure on $TM$ to make it a smooth $2n$ dimensional manifold. In doing so, we obtain diffeomorphisms between any neighborhood in $TM$ and an open subset in $\mathbb{R}^n\times\mathbb{R}^n$. Then we can start talking about addition, subtraction, limits, and derivatives in the tangent bundle. Therefore, we have created an isomorphism between the tangent spaces and created a way to compare vectors at any two points in the neighborhood, correct?

I want to confirm that these ideas are compatible. We can exploit the local triviality of the bundle to develop a local method of parallel transport, right? And this allows us to create a local definition for a smooth vector field, circumnavigating the problem I mentioned in the first paragraph?

Noah M
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  • You could of course do that, but the important takeaway is that your local diffeomorphism of $TM$ and $\mathbb{R}^n\times\mathbb{R}^n$ is not intrinsic to the manifold. The way it is usually done it depends on the charts (i.e. the coordinates) you choose. So yes, you could to this, but what is it good for? Does it reveal any interesting properties of the geometry of your manifold? Do you need it do study some differential equations on your manifold? You have to give some context here. – Maik Pickl Nov 18 '19 at 16:06
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    Isn't this just what is done when you topologize the bundle by transferring the topology of $(\varphi(U)\times \mathbb R^n)$ to $TU$? – Matematleta Nov 18 '19 at 16:16
  • Related question? https://math.stackexchange.com/questions/3422771 – Ekhin Taylor R. Wilson Nov 18 '19 at 16:42

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Every local chart provides a way to identify tangent vectors at one point with tangent vectors at another point. It thereby provides notions of addition, subtraction, differentiability, etc. But different local chart, related to the first one by a smooth change of coordinates, will provide a different identification, different addition, different subtraction, but the same notion of differentiability. Thus, addition and subtraction are not well-defined just on the basis of the manifold structure; they vary according to which of the many available charts you use. But all charts (from the atlas defining your manifold) give the same notion of differentiability of vector fields, so that notion is well-defined just by the manifold, with no further arbitrary choices.

Andreas Blass
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    Related question? https://math.stackexchange.com/questions/3422771 – Ekhin Taylor R. Wilson Nov 18 '19 at 16:42
  • Thank you Andreas! That did clarify things for me greatly. I was missing the fact that the coordinate transform diffeomorphisms need not preserve any of the vector space operations. However, they do preserve the limits that are used to define derivatives. That's a really cool thought!

    So, because the charts cover the manifold and the smoothness of a section is chart-independent, we obtain a natural definition for a smooth section that can be applied to each point.

     – Noah M Nov 18 '19 at 21:11
  • @EkhinTaylorR.Wilson, thanks for the link! I hadn't considered the question that you posed, that's also really interesting and touches on some of the same ideas! – Noah M Nov 18 '19 at 21:18
  • @Andreas Blass, I had another question. I was reading this document about connections, and they stressed the point that on an arbitrary manifold there is no natural operation of differentiating vector fields. However, if we have a smooth structure on $M$ And $TM$, we should be able to start taking derivatives of maps between them, yes? And these smooth structures come directly from the definition of $M$, from its atlas. Is there something further that I'm missing? – Noah M Nov 19 '19 at 11:19