# Definition of tangent space on smooth manifolds

1. Nov 7, 2011

### Alesak

Hi,

I'm having trouble understanding why is tangent space at point p on a smooth manifold, not embedded in any ambient euclidean sapce, has to be defined as, for example, set of all directional derivatives at that point.

To my understanding, the goal of defining tangent space is to provide linear approximation of a manifold near certain point. Why not to just say that tangent space of n-manifold at point p is R^n? In the end, the set of all directional derivatives is isomorphic to it anyway. And after all, manifolds key characteristic is that it is localy similar to R^n, so why to not use it?

But there has to be some reason it is defined the way it is, since everybody is using it, so I just wonder what I'm missing here...

2. Nov 7, 2011

### lavinia

At a point the tangent space is just R^n but globally the tangent spaces may fit together to form a new space that is not the same as R^n.

If you just associate R^n with a point on the manifold you do not know how to identify the vectors with directions on the manifold. the key is interpreting the vectors as directions.

There are other ways to associate vectors with points on manifolds that can not be interpreted as tangential directions.

Last edited: Nov 7, 2011
3. Nov 7, 2011

### homeomorphic

It doesn't have to be defined that way. There are at least 3 definitions that I know of. The first is in terms of differentiation operators. The second is in terms of curves. And the third is actually in terms of R^n, but you need a little more than that.

Essentially, it IS R^n, since in finite dimensions, anything is R^n, but only up to isomorphism. You shouldn't forget about the isomorphisms.

Think about a vector field. You don't want to just think about tangent vectors at one point.

Isomorphic. But how do you get the isomorphism? That depends on your charts. You don't want it to depend on charts.

The similarity with R^n depends on your charts.

4. Nov 7, 2011

### Deveno

the way i've always thought about it is: the earth isn't flat. i mean locally, it is pretty flat, so much so that "geometry" (the euclidean plane kind) was originally thought to be a good abstraction of the earth.

but when we started making maps of the whole world, we ran into problems; you can't embed a sphere into a euclidean plane faithfully (isometrically), you get some distortion no matter how you approach it. in manifold terms: you can paste together a set of charts in a consistent way (think of taping different maps together), but what you get when you do that globally (ooh, i made a pun) is something that can have properties the individual tangent spaces don't.

a circle is only locally a line. if you cut it, and straighten it out, so that it fits nicely into R, you wind up breaking paths that were just one piece. as you get farther and farther away from the point the tangent space is defined at, you can wind up no longer having a proper normal or tangent vector. time to switch to a different local map, one without the distortion that creeps in (linear approximations aren't THAT good, especially on manifolds with a lot of twists and curves).

5. Nov 7, 2011

### lavinia

Using charts is not the only way to identify the tangent space. If the manifold is embedded in Euclidean space then the tangent space at a point is the hyperplane of all lines that touch the manifold at that point without cutting through it. This definition is geometrical.

Each tangent line gives you two directions along the manifold.

6. Nov 7, 2011

### Deveno

and then you get a whaddayacallit...tangent bundle of all those hyperplanes, and that doesn't look at ALL like euclidean space (unless your manifold is euclidean space, but maybe that's not such a useful example), although it's kinda cool to try to picture it in your head.

7. Nov 7, 2011

### homeomorphic

I was going to mention that, but then I realized he specifically asked about manifolds that are not embedded in Euclidean space, so we want an intrinsic definition.

The tangent bundle is a manifold, so it is, again, locally like Euclidean space. Well, at least with the appropriate topology on it, but without the topology, all bundles may as well be trivial, so there's nothing interesting from a bundle point of view. But you would still have the coordinate independence problem. You have a tangent vector sitting at some point in the manifold and you want to say it's in R^n, but which vector in R^n is it? Depends on your isomorphism. So, you still have to sort of mod out by coordinate changes, to make it a coordinate-independent concept.

8. Nov 7, 2011

### Sina

I misunderstood the question so my previous answer is down there. The real answer is mainly because of the usage. We define tangent vectors to curve as "infinitesimal operations" that describe translation along that curve. Well the amount of translation is infact given in terms of the derivative with respect to the curves parameter which will be d/dt or in coordinates some a^id/dx^i. So basically the reason is that probably the people building this sturucture had dynamical systems in mind and they wanted the vectors do describe translations on the curves. As I said the best operation for that is derivative along the curve.

You can define the tangent plane in terms of germs, derivations or by embedding it in some ambient space even usual tangent vectors etc. In the end though the strucure is that the tangent space at p is indeed isomorphich to Rn. However this isomorphism helps you only when you are working right at that point.

If you want to make more global statements about the structure of your manifold then you have to give the collection of all those tangent spaces some structure. For instance to define (0,n) tensor fields on your manifold (or the reverse I am not sure which is which) you really need to defined a collection of tangent spaces over open sets. If you do not give manifold structure to your tangent bundle, how will you know what is the results of change of coordinates to an arbitrary tensor field for instance? By giving the tangent bundle a manifold structure explicitly and by defining charts etc you can manipulate vector fields and tensor fields defined on the whole of manifold. Or else you will be restircted to making a change of basis at single points then patching them together etc etc. It is all basically done by giving it the manifold structure.

Last edited: Nov 7, 2011
9. Nov 7, 2011

### Alesak

Thanks for all answers. I'm going through the Introduction to Smooth Manifolds by Lee, and this whole modern differential geometry complex involving vector\covector spaces\bundles and combining them into tensors\differential forsms is making my head spin. But hey, on wikipedia they say Einstein had learned about them "with great difficulty":)

I think the tangent space is clear to me now: we want to have an arrow at a point, that exist in itself, without any reference to any coordinates. I like to imagine that whatever coordinate chart we choose, the same arrow just sits there glued, oblivious to whatever happens around it.

So while tangent bundle is a vector bundle, it carries this additional interpretation, and it is entirely up to us how we establish correspondence between a point in tangent space and a point in a same-dimensional vector space(i.e. what coordinates it will have).

Personaly I find the old definition from physicists most revealing, becouse it is straghtforward: we pick some coordinate chart around point p that maps p to 0. Now we automaticaly get basis $(x_{i})$ for tangent space, which consists of vectors tangent to lines, that map to canonical axes in R^n. So each tangent vector is uniquely represented as $v = a^{i}x_{i}$ (do I have the summation convention right?). If we pick another coordinate chart which induces tangent space basis $(\widehat{x_{i}})$ and construct change of coordinates matrix (while understanding $x_{i}$ as i-th basis vector, or i-th coordinate function\i-th basis vector of dual space as convenient)
$$A = \begin{pmatrix} \frac{\partial x_{1}}{\partial \widehat{x_{1}}} & \frac{\partial x_{2}}{\partial \widehat{x_{1}}} & ... & \frac{\partial x_{n}}{\partial \widehat{x_{1}}}\\ \frac{\partial x_{1}}{\partial \widehat{x_{2}}} & . & . & .\\ . & . & . & . \\ \frac{\partial x_{1}}{\partial \widehat{x_{n}}} & . & . & \frac{\partial x_{n}}{\partial \widehat{x_{n}}} \end{pmatrix}$$
we can comfortably write $\widehat{v} = Av$, which is in fact the same tangent vector. I suppose it can be remembered as a matrix having rows as old basis vectors expressed in new coordinates. In other words, $\widehat{v^{j}} = v^{i}\frac{\partial x_{i}}{\partial \widehat{x_{j}}}$.

I guess defining covectors and tensors won't be that different.

10. Nov 7, 2011

### Sina

in that case too it is very natural to denote the basis vectors as $\frac{\partial}{\partial x_i}$ because according to chain rule it just changes with the matrix you have given :) in the chapters on 1-parameter flows and lie derivatives, you will see that this definition is much much more physical then what you have given. For instance given a curve γ on your manifold and some function f, to see the rate of change of f along the curve you just apply the tangent vector of the curve to the function that is $\frac{d\gamma^i}{dt}\frac{\partial}{\partial x_i}f$. In your definition, there is no direct way of applying the vectors to functions etc.
You have to take the vector $v_ie^i$ and map it to something like $v_i\frac{\partial}{\partial x_i}$ to make it a derivation along the curve but that is exactly the tangent vector we are talking about. And infact this mapping can be shown to be an isomorphism between derivations and usual R^n.

Lee's book is a very good book, very pedagogical and easy to follow even though it handles some quite non-trivial topics in a completely rigorous way. No wonder lee has many teaching prizes :) Be sure to solve some questions as it has important results and covers the usage of techniques learned in the chapter.

Last edited: Nov 7, 2011
11. Nov 23, 2011

### mathwonk

no matter what definition you take for the tangent space V of a manifold at the point p, you must answer this question: given a smooth curve passing through p at time t=0, which element of V is the velocity vector at p for t=0?

I.e. you must give a surjective map from the set of all smooth curves passing through p to the vector space V, such that 2 curves have the same velocity vector in V at p, if and only if for every coordinate chart taking a neighborhood of p to a neighborhood of 0 in R^n, the corresponding curves through 0 have the same velocity vector in R^n, in the usual sense of calculus.

Any definition of the tangent space V, and of velocity vectors in V of curves, that has this property will do fine.

I.e. all the tangent space needs to do for you is tell when two curves should have the same velocity vector at p. E.g., if we let the tangent space be the space of differential operators on functions at p, then given a curve through p, we define for each function the tangent vector to be differentiation along the direction defined by the curve.

Then by the chain rule, indeed two curves have the same velocity vector at p if and only if they give the same directional derivative for all functions at p.

Last edited: Nov 23, 2011
12. Nov 24, 2011

### mathwonk

Each chart at p does indeed give a way to view the tangent space as R^n. But two different charts give two different ways to do this. Then if the vector v in R^n is associated to a given tangent vector at p by one chart, we must know which vector w in R^n is associated to the same tangent vector by another chart. This is done by using the derivative of the composition of the two charts, and letting w be the vector image of v under this derivative.

Thus a tangent vector at p is not one vector v in R^n, but is a collection of pairs {(f,v), one for each chart f at p}, where v is a vector in R^n and f is a chart. And it must be true that for any two pairs, (f,v) and (g,w) in the collection, that the derivative of gof^-1 takes v to w.

13. Nov 26, 2011

### mathwonk

in my clueless opinion i have completely answered this question. but i have the sinking feeling that my answer is totally a mystery. ?????

14. Nov 27, 2011

### lavinia

Every vector bundle has coordinate charts. In each bundle there are functions called coordinate transformations which say how to identify vectors above one chart with vectors on an overlapping chart.

In the case of the tangent bundle these transformations are the differentials of the change of coordinates. But with other vector bundles they are not.

When the coordinate transformations are the differentials of the change of coordinate functions it makes sense to define the derivative of a function with respect to a vector in the bundle. This is because the Chain rule will automatically work so the derivative of the function in one chart will transform into the derivative in the other.

With other bundles it is not possible to differentiate function precisely because the Chain Rule fails.