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I saw the definition of tangent vector on a manifold given as:

A tangent vector $v$ at a point $m$ of a smooth $n$-manifold $M$ is a linear derivation of the algebra of germs of functions at $m$.

My question is: How can I see (say in a concrete example) that every linear derivation on this algebra corresponds to a tangent vector?

I know it is a definition but it is clear that every tangent vector say to $S^2$ in $\mathbb R^3$ defines a linear derivation. What is less clear to me is why the other direction also holds.

self-learner
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    This is proved in every textbook which defines tangent vectors in this way. I suggest you look at a couple! – Mariano Suárez-Álvarez Nov 16 '14 at 06:24
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    Please can you suggest a book to me? I have Warner "Foundations of Differentiable manifolds and Lie groups" and he doesn't prove it there (or at least I can't find it!) – self-learner Nov 16 '14 at 06:36
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    Warner is very terse. Lee's Smooth Manifolds book is the opposite, if you're looking for more discussion. –  Nov 16 '14 at 07:02
  • If you agree that every tangent vector of $S^2 \subset \Bbb R^3$ defines a unique linear derivation, and you agree that the linear derivations at a point of $S^2$ form a dimension 2 vector space, then you're done by linear algebra: you have a linear injection between two vector spaces of the same dimension. –  Nov 16 '14 at 07:04
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    A book that explains it well is "Differential Geometry and Symmetric Spaces" by Sigurdur Helgason. Note that this works for $C^{\infty}$ manifolds and not really for other $C^{k}$ 's. – orangeskid Nov 16 '14 at 07:47
  • @MikeMiller You're right. I guess I was hoping to gain some better geometric insight by working out some map into the other direction. Thank you. – self-learner Nov 16 '14 at 09:11
  • I second orangekid's suggestion for that. –  Nov 16 '14 at 10:46
  • @MikeMiller I just realised: I don't agree that linear derivations at a point of $S^2$ form a dimension 2 vector space. That's what I want to derive... – self-learner Nov 16 '14 at 23:31
  • @self-learner As Mariano said, every differential geometry book does this. I looked at my copy of Warner and in my edition this is the content of theorems 1.16 and 1.17. –  Nov 16 '14 at 23:34
  • @MikeMiller But these two theorems only show it indirectly: by appealing to linear algebra and using finiteness of dimensions. I am hoping for a direct proof constructing an injective map from the set of linear derivations into the set of tangent vectors. Since the two are by definition in the book equal, I mean to see this only in a concrete example. Example where one can work with concrete tangent vectors instead of linear derivations. – self-learner Nov 16 '14 at 23:38

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