why are tangent vectors like this?

jostpuur

Why are the tangent vectors of smooth manifolds defined as mappings [itex]C^{\infty}(p)\to\mathbb{R}[/itex] that have the similar properties as derivations?

If a manifold is defined as a subset of some larger euclidean space, then the tangent spaces are simply affine subspaces of the larger space, but if the manifold is instead defined without the underlying larger euclidean space, then I don't understand what the tangent spaces even should be like.

Hurkyl

There are many practical, philosophical, and conceptual advantages to using intrinsic definitions in differential geometry -- definitions that do not involve an ambient space at all.

If you work through the details, you'll see that the tangent bundle as you know it is isomorphic to the definition via derivations.

(FYI, this definition is not the only way to define the tangent bundle)

jostpuur

The tangent space of a point [itex]p\in M[/itex] turns out to be a vector space spanned by the derivations [itex](\partial_i)_p[/itex]. But why like this? Why isn't it simply a vector space [itex]\mathbb{R}^n[/itex] spanned by the vectors [itex]e_i[/itex]?

If I didn't know better, I would have thought, that when moving from manifolds in ambient spaces to manifolds without ambient spaces, the obvious modification to the tangent spaces would have been to remove the information about the location of the affine subspace, and simply call a vector space [itex]\mathbb{R}^n[/itex] the tangent space.

Hurkyl

The problem is that Rⁿ is a completely unrelated structure; it doesn't 'know' anything about your manifold. You can't, for example, use this definition to define a "directional derivative" operator on scalar functions.

DeadWolfe

Well, in classical extrinsic geometry, for any chart x from R^n into your manifold M, the tangent manifold can be thought of as the space spanned by partial derivatives of x. If you prefer, you can think of this as the span of derivatives of all maps from R into M. The "intrinsic" definition is just an easy way of defining that when we don't know what differentiation is yet.

Namely, we can note that for any map from R into M we can compose with a map from M into R, and then differentiate that, and so any curve gives rise to a derivation. Clearly in the case that M already has an embedding in R^n this derivation clearly depends only on the derivative of your curve. So it's just a clever way of defining the same thing.

As for defining the tangent space to be R^n, *which* R^n. There are many isomorphic copies, but they are not canonically isomorphic. You'll learn the importance of this very quickly.