Yet more elementary questions about the tangent space

1. Jan 20, 2014

orion

I am trying to self-study some concepts in differential geometry to try to update my knowledge from the old-style index games to something more meaningful. I know that there are many threads that have in some way addressed this, but I am still not understanding it completely. I'm new to this, and I could be way off base so please bear with me as I am feeling my way in the dark. I'm looking for more conceptual help so rigor is not that important.

Consider a tangent space of a manifold M at point p. For simplicity, assume TpM is Euclidean and of dimension 2. I understand that {∂i} forms a basis of TpM for some coordinate chart around p.

Question 1: What does the dot product between two vectors in this space look like when the basis vectors are partial derivatives?

Question 2: How can I show that the ∂i form a basis?

Question 3: According to what I have read, the dxi form a basis of Tp*M, the dual of TpM. The Euclidean metric (for the two dimensional case) is expressed differentially as ds2=dx2+dy2.

This looks like length is formed from covectors in Tp*M and not vectors. Why is this?

Question 4: Consider the following definition:

dxi(∂j) = ∂jxi

I realize that this is a definition, but is there any background to this definition? Where did it come from? (I have a suspicion that the directional derivative has something to do with it, but I'm not there yet.)

2. Jan 20, 2014

jgens

There are no standard basis vector for tangent spaces in general so there usually is no natural choice of inner product. Once you have one fixed, however, you just need information about <∂i,∂j> to compute.

This is a local question so it suffices to assume M = Rn. Now clearly the ∂i are linearly independent. So only the spanning condition remains. For this assume D is an arbitrary derivation at p and let vi = D(xi). The claim now is that D = ∑vii. To see this note that every function f:RnR has an expansion of the form:

f(x) = f(p) + ∑∂if(p)(xi-pi) + ∑gi(x)(xi-pi)

where gi(p) = 0. Now just apply D to this expansion and simplify. The result is Df = ∑viif(p).

This is essentially because we can integrate covectors. The same is true in ordinary calculus. In that context, anytime you wanted to integrate a vector field, you first used the natural Riemannian metric (a consistent choice of inner product between all the tangent spaces) to turn the vector field into a covector field and integrated from there.

I cannot speak authoritatively about the historical reasons for the definition. I can say, however, that we want the natural pairing to between vectors and covectors to be nice and that this definition accomplishes just that.

3. Jan 21, 2014

orion

Thank you very much for your reply and I really do appreciate it though I must admit that a lot of what you wrote went over my head. What I'm trying to do is to relate these concepts from differential geometry to what I already know from elementary vector analysis. What usually aids in my understanding is to consider a particular, concrete case and then later generalize it to higher levels of abstraction.

I'd like to consider a sphere with a tangent plane at a particular point p. I assume the tangent plane is Euclidean with the standard Euclidean metric. (I am thinking of the Earth and a local street map lying in the tangent space.) Assume that I have placed some type of coordinates around p. Now, I want to take two vectors from that tangent space and form a dot product. I would like to understand how to relate this case to base vectors that are partial derivatives and what such a dot product would look like. It seems to me that the approach from differential geometry should reduce somehow to the familiar approach from vector analysis.

4. Jan 21, 2014

jgens

Here is the problem. The expression for the standard Euclidean metric depends heavily on our choice of basis. For Euclidean spaces we have a natural choice of basis so this poses no problem. For arbitrary tangent spaces we have no natural choices however. Each chart around p is going to give us a new set of basis vectors and there is no reason to prefer one of these over the other.

The dot product refers to a very specific inner product on Euclidean space. So the relevant concept here is an inner product. You form these guys in exactly the same ways you formed them in vector analysis.

5. Jan 22, 2014

orion

I guess what I am driving at is how can it ever be possible that ∂ij = δij assuming a basis of ∂k and a tangent space that is Euclidean. In other words, I want to see how base vectors ∂k can be related to the basis B = {(1,0), (0,1)} of a Euclidean plane or how we can get the basis B from the partial derivatives.

I want to take the abstract formalism of differential geometry and apply it to a specific, well-known case.

6. Jan 22, 2014

jgens

You can simply define your inner product that way. Just beware that with a difference choice of coordinates your inner product will no longer have this nice form.

7. Jan 22, 2014

Staff: Mentor

Even in Euclidean space, you don't need to use coordinate axes that are perpendicular to one another, and the variations of the coordinate values along the coordinate axes to not have to be equal to actual distances in space. Consider a set of affine coordinate axes that are straight lines and make a fixed angle with one another. Before you move on to arbitrary tangent spaces, first work everything out and understand it for Euclidean space. After you have finished with affine coordinate axes, consider orthogonal curvilinear coordinate axes. Then, finally consider arbitrary curvilinear coordinate systems. This should solidify your understanding.

8. Jan 22, 2014

WWGD

One of the issues here is that tangent spaces at different points of $\mathbb R^n$ with basis vectors $e_i$ and with the standard rectangular system, are naturally-isomorphic to each other; naturally isomorphic meaning that there is an iso. between them that does not depend on the choice of basis. Specifically, tangent spaces at any two points a,b are just translates of the tangent space at the origin, and this translation does not affect the essential vector space structure. This natural isomorphism makes the process of differentiating very simple; just translate the vector at (x+h) to that at x . This is not the case with most manifolds; while any two tangent spaces are isomorphic ( being of the same dimension) , they are not naturally-so. To make up for that, you make a choice of isomorphism between them by using a connection.

Re the issue of the basis, you may also want to consider the case where the manifold is embedded in some ambient $\mathbb R^n$ . Then you can use the curve definition of tangent vectors; jgens covered the case when the manifold is stand-alone, where a tangent spaces makes no sense, since the tangent space at p is not part of the manifold ( outside of the point p, of course ). Then you can show that every derivation is a directional derivative, so that the basis you mentioned spans every derivation.

For question 3, notice that your standard Riemann integral is the integral of the n-form $fdx_!dx_2..dx_n$. If you know some geometric algebra, i.e., the algebra/geometry associated to wedge products, I think will help make sense of the integration of forms.

9. Jan 22, 2014

jgens

Just a quick note: Defining tangent vectors via curves works even when the manifold is not embedded in some ambient Euclidean space. This is actually quite an important fact since often times we want to consider derivatives and vector fields associated to curves.

10. Jan 22, 2014

WWGD

But, how do you define the vector fields? Their range is not necessarily in the manifold itself.

11. Jan 22, 2014

jgens

A vector field is just a smooth section M→TM. Sure the range is not the manifold itself, but nothing about the definition requires that M be embedded in some Euclidean space either.

12. Jan 22, 2014

WWGD

Ah, I see, I misunderstood ( misunderestimated?) your post.

13. Jan 23, 2014

orion

Thanks everyone for the replies. I appreciate the discussion.

I'm not sure, but I think that I have found the answer in Introduction to Smooth Manifolds by John M. Lee, page 53 (as part of a proof). If we assume a rectangular coordinate system and allow the ∂i vectors to operate on the coordinates we can obtain the three standard basis vectors, ek because ∂ixj = 0 except when i=j when it is equal to 1.

14. Jan 23, 2014

jgens

As has been mentioned several times already. Yes there is an isomorphism of vector spaces taking ∂k to ek. From there one can define <∂i,∂j> = δij. Just beware: in another choice of local coordinates this inner product will look very different.

15. Jan 24, 2014

orion

Ah, yes, I'm sorry. My problem was actually seeing the isomorphism.

I still have a question involving inner products, but I need to think on it a bit.

Last edited: Jan 24, 2014
16. Feb 3, 2014

orion

I've given it more thought, and my question is that I don't want to just define <∂i,∂j> = δij and leave it at that, but I want to show that it is true in the same way that one can demonstrate that <ej,ek>= δjk for an suitable local coordinate system.

In other words, I want to show that it is true by the action of the differential operators.

17. Feb 3, 2014

jgens

The problem is without a Riemannian metric on your manifold there is no natural choice of inner product on the tangent spaces. So you have to define an inner product first, before you can show anything! If simply setting <∂i,∂j> = δij is unsatisfactory, then how do you propose defining it?

18. Feb 3, 2014

orion

I am assuming a Riemannian metric. I think that the problem was that I was mistaking ∂ij to be equal to ∂2/∂xij when it is ∂if∂jf. If we let f be the standard coordinate functions, we have ∂ixkjxk = δikδjk = δij.

19. Feb 3, 2014

jgens

In that case to each coordinate chart corresponds a symmetric matrix (gij) where gij = <∂i,∂j> and one need only apply elementary linear algebra to diagonalize this matrix.

An important note: The resulting local vector fields are not guaranteed to be the "partial derivatives" corresponding to any coordinate system. In most applications this introduces no real problems, since local frames with the desired properties are sufficient. But if you wanted these frames to be induced from a choice of coordinates, then you are out of luck.

20. Feb 3, 2014

orion

This is the point of confusion for me. I don't understand how one can use the metric to "define" <∂i,∂j> when the ∂k are operators. It seems like it must be possible to obtain the metric elements from letting the operators operate on something. That was the point of my last post.

21. Feb 3, 2014

jgens

The metric is a map <-,->p:TMp x TMpR. Since ∂i and ∂j are elements of TMp it makes complete sense to consider <∂i,∂j>.

This is probably not feasible. You can endow your manifold with two very different Riemannian metrics and these operators will remain unchanged. So it is probably too much to hope for this approach to work out.

22. Feb 3, 2014

orion

I need to backtrack a bit because I am still trying to come to grips with operators being a basis and having their inner product defined by the metric. Well, no, I'm ok with having the inner product defined by the metric. But what really is troubling me is whether I should I consider the ∂k as operating on something?

I will tell you how my mind wants to think of this, and I hope you will point out the error in my thinking. My mind wants to say that the ∂k operates on coordinate functions and out pops a basis like we are familiar with from elementary linear algebra. The confusion right now for me is whether I should consider the ∂k as operating on something (what?) or whether I should think of it mostly as a symbol like ej. The problem is that I can express the ej in a particular representation (eg. a column vector) if I must.

23. Feb 3, 2014

jgens

Yes. They operate on maps f:M→R. The Riemannian metric is an additional structure, on top of this operator property, so it would be folly to assume it is somehow determined by the way these ∂k act on functions.

The coordinate functions induce the basis. Not the action of ∂k on the coordinate functions.

Both. As mentioned before the ∂k act on functions and at times understanding this action is crucial. Other times, like the case at hand, the operator property is unimportant. The important fact is that these guys form a basis for the vector space TMp.

You can do the same with the ∂k. In general this will not help you much, but it can be done.

Last edited: Feb 3, 2014
24. Feb 3, 2014

orion

Thanks. I'm getting there, but I'm not quite there yet. Can you recommend any good books where this is covered at an elementary (well, as elementary as possible) level?

25. Feb 3, 2014

jgens

The book Introduction to Smooth Manifolds by Lee is probably the most easy-going exposition about this material I know. His Riemannian Manifolds book is also good (and considerably more on-point for this topic) but it assumes quite a bit more.