# Thread: Tangent space, derivation defifinition

1. ## Tangent space, derivation defifinition

I'm reading "tensor analysis on manifolds" by Bishop and Goldberg. I have taking a course in differential geometry i R^3. The course was held on Do Carmos book. Do carmo deffined the tangent at a point on a surface as all tangents to all curves on the surface going through that point (or something like that). This diffenition is pretty clear and easy to imagine.

In the other book it is defined as all darivations on all tions from the manifold at the point to R. Then they define the tangent of a curve in the manifold to be and operator. And then they define the tangent to the ith coordinate curve of gamma, and proof a theorem about that these coordinate tangents at a point m on the manifold is a basis for all tangents.

The proof are not that hard, but my problem is that i can't see that the two definitions are the same when i choose one of the "manifolds" we worked on in do carmo, fx. the unit sphere in R^3. I now the one is a operator and the other are vectors, but there most be a clear connection?

I've seen an example where they use R^2 as there manifolds, and then state that the directional derivative corospones to the operators. And then it is clear that given and directional derivative in the direction v lets call it Dv, then i have the operator and the if i define a function f(Dv)=v then i get all tangent vectors to all points in R^2.

But somehow i can't find the connection when my surface gets more complicated, like the unit sphere. I can see that the operator is connected to the tangent of curves through m, but it is not so clear to me, in my mind there most be a function (possible a bijection) on the "tangent operators" such that i get all tanget vectors, when my manifold is so simple like the unit sphere that a "normal" tangent vector/plane makes sence. My point is that if it makes sence to define the tangents like operaors, I most have just as much information as if I had defined it the "usual" way (of cause the new defintion makes more sence when i have obscure manifolds where tangentplanes are not clear, but in R^3 there most be a clear connection in my head).

I hope someone can understand what my problem is, anbd have the commitment to read the whole post (it got a bit long, sorry). And by the way, i'm sorry for the bad englsih.

2. Your confusion is understandable (I was experiencing almost the exact same problem recently). What I seem to gather from you is that, you see how a traditional tangent vector V can be thought of as a derivation (by the act of taking the directional derivative through V), but you fail to see how an abstract derivation can necessarily be thought of as a traditional tangent vector.

This is understandable... I don't find it at all obvious that an arbitrary derivation at a point p is always basically the taking of a directional derivative. That this is in fact the case is probably best embodied in the following theorem:

If a is in R^n, let Dx1[a], ... , Dxn[a] denote the derivations defined by taking the partial derivative of f(x1, ..., xn) with respect to the variables x1, ..., xn. Then ANY derivation at the point a can be written as a unique linear combination of these derivations. That is, Dx1[a], ... , Dxn[a] form a basis for tangent space of all derivations at a on the manifold R^n.

Again, I don't find this obvious. But once you've convinced yourself of it (I'm sure your book has the proof), the rest follows fairly easily. We can identify the basis for our "traditional" tangent vectors with corresponding derivations. For example, the vector V = (1, 0, 0, ..., 0) is identified with Dx1[a] (convince yourself that taking the directional derivative of of f with respect to the vector V is precisely the same as taking the partial derivative of f with respect to x1).

Of course, everything I've said so far only goes for the rather boring manifold R^n. But really, it goes for any smooth manifold, since we are ultimately identifying open subsets of the manifold with R^n anyway. When we deal with a function f : M --> R, we are probably just going to write it in terms of its coordinate representation anyway, i.e., to think of it as a map f : R^n --> R. And in that sense we are quite justified in carrying over our previous results about R^n into an arbitrary manifold M.

Finally, keep in mind I am no expert in Differential Geometry, so if I've contradicted anything you might read in your book, you should probably believe it over me. In any case, I hope it helped, let me know if it did (or didn't). Take care

3. Thanks for taking your time to answer my post. It helped a great deal. I think I know how to convince myself that it is a good definition.

I have also found a book called:

Modern Differential Geometry for Physicists 2nd ed. - C. Isham

He defines the tangent space with deriviations and the more geometrical intuative way, and then compare them, in fact show that the two spaces are isomorph when we are looking in R^n, so this book may help others too.

'thanks again, Anders Berthelsen