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I have an elementary understanding of differential geometry, and I know that the concept of a vector on a manifold can be defined in several different ways. Perhaps the easiest to understand, is in terms of equivalence classes of curves $\gamma:M\rightarrow\mathbb{R}^{n}$ on a manifold $M$. In this case, vectors are defined at each point $p\in M$, as tangent vectors to curves at that point, "living" in the tangent space $T_{p}M$ to that point. A tangent vector at a point $p\in M$ is then an equivalence class of curves, mutually tangent, at that point. Another way is to construct the notion of a vector using derivations.

My question is (and apologies if it's a silly one), can there exist vectors $v$ in a given tangent space $T_{p}M$ that are not tangent to a curve passing through $p$ (essentially, are there cases where $v^{i}\neq\frac{\mathrm{d}x^{i}}{\mathrm{d}t}$)?

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  • $\begingroup$ If you define the tangent vectors via equivalence classes of curves, it should be obvious that there cannot be such a vector, since you need at least one curve in the equivalence class which defines this vector. $\endgroup$ – M. Winter Feb 20 '18 at 15:00
  • $\begingroup$ @M.Winter Sorry, I realise what I wrote sounds a little stupid. My initial reason for asking the question was that I thought (briefly) that it was a particular quirk to the definition in terms of equivalence classes of curves, but I now realise that it would be inconsistent if this weren't also true for the other definitions. $\endgroup$ – user35305 Feb 20 '18 at 15:18
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No. I mean, if you define the tangent space to consist of equivalence classes of curves, then each tangent vector is a collection of curves, all of which have the same derivative (in local coordinates) at the given point.

If you define it by derivations, then there's some work to do (involving the chain rule, and some other stuff) showing that for each derivation, there's a collection of curves whose associated tangent vector (in definition 1) really does correspond to this derivation ... but that turns out to be true, too.

And if your manifold is smoothly-enough embedded in euclidean space, then there's yet another (more geometric) notion of tangent plane/tangent vector, and it, too, corresponds to the first two.

Spivak's Comprehensive Intro to Diff'l Geometry (vol 1) has about 6 definitions of tangent vector/tangent space, and spends a good deal of time showing that they're all equivalent.

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  • $\begingroup$ Thanks for you answer. I realise that what I wrote in my question sounds a little dumb, but I was wondering whether the definition via equivalence classes of curves just has that particular quirk, or whether it holds for all definitions. I realise now that of course it must, since otherwise these definitions would be inconsistent. $\endgroup$ – user35305 Feb 20 '18 at 15:16

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