# Why are tangent vectors like this?

1. Oct 28, 2007

### jostpuur

Why are the tangent vectors of smooth manifolds defined as mappings $C^{\infty}(p)\to\mathbb{R}$ 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.

2. Oct 28, 2007

### Hurkyl

Staff Emeritus
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)

Last edited: Oct 28, 2007
3. Oct 28, 2007

### jostpuur

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

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 $\mathbb{R}^n$ the tangent space.

4. Oct 28, 2007

### Hurkyl

Staff Emeritus
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.

5. Oct 29, 2007