More Manifold Questions

  • Thread starter Tedjn
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  • #1
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Hi everyone,

On the Wikipedia page for Tangent space there is a definition of the tangent space at a point x using equivalence classes of curves. It mentions that the tangent space TxM is in bijective correspondence with Rn.

My first question is simply: is there an easy geometric way using only the notion of tangent vectors as equivalence classes of curves to see this?

I can understand why TxM is contained in Rn but cannot see how to prove the bijection without constructing a curve in the tangent space corresponding to each vector in Rn. I don't know how to do that.

Thanks in advance. I know sometimes I don't have the time to or forget to follow up on these question posts. Know that I appreciate all the help.
 

Answers and Replies

  • #2
Dick
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Use a coordinate chart including the point x to map curves in R^n into curves in the manifold.
 
  • #3
Landau
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I can understand why TxM is contained in Rn
You probably mean "injects in" instead of "is contained in". Anyway, could you share you understanding of this fact? Because if you understand this, then the other way around is not so hard.
 
  • #4
HallsofIvy
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One difficulty with R^n is that it allows "position vectors". I used to spend a lot of time wondering what a "position vector" looked like on the surface of a sphere. Was it curved to stay on the sphere? Did it go through the sphere? Finally I learned that the whole idea of "position vectors" disappears in general spaces. In fact, all vectors are derivatives- or, same thing, all vectors are tangent vectors, in the tangent space at each point on the surface, not in the surface itself. If you have a surface in R^n, you can take a curve through a given point, written in terms of some parameter, differentiate with respect to that parameter and get the tangent vector, to that curve, at that point. The tangent space at that point is the set of all such derivatives.

Now, the difficulty with that is that it involves a choice of parameter as well as a choice of "coordinate system" for R^n. For a general manifold, we would like to avoid such choices. We can do that by saying that two curves, p(s) and q(t), through the point x (p(s0)= x and q(t1)= x for some s0 and t1) are "equivalent" if and only if in any choice of coordinate system and any parameter they have the same derivative at x. What happens is that such an equivalence set consists of all curves that have the same derivative vector at x and so"defines" the tangent vector at that point.
 
  • #5
hunt_mat
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One simple idea from analysis is that if V an an n-dimensional vector space that V is isomorphic to R^n.
 
  • #6
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Thank you everyone. This has been very helpful.

I feel ridiculous for forgetting that of course we can map curves in Rn back to curves in the manifold. I think I have convinced myself then why the vector space structure of Rn induces the vector space structure of TxM.

Landau, I did mean injects. I imagine you were also leading to the idea of mapping curves in Rn back to curves in M.

HallofIvy, your comment reminds me that naturally this construction of TxM should not depend on our choice of parametrization and charts. Would it be correct in saying that there is a vector space isomorphism between any two constructions of our tangent space in this way through curves and choices of charts? Is this the way in which the tangent space is unique?


I loathe to make another topic for a series of questions of various levels I will certainly have, so I hope no one minds that I ask whatever questions related to manifolds I encounter here. In truth I am trying to understand several things in manifolds toward a goal, but of course there is a lot of background and related material I am trying to cover in a short time, so this forum can be quite helpful.

Right now I am reading the beginning of an introductory book on complex geometry.

Here are some related and probably very simple linear algebra questions that have me stumped. Given a real vector space V of even dimension n, I may be able to give it an almost complex structure J (an endomorphism of V such that J2 = -id). This can turn V into a complex vector space. First, is the dimension of this complex vector space n/2? I can't quite wrap my head around why or why not, even though I suspect it is true. Is it because there are n/2 basis vectors such that v1, Jv1, v2, Jv2, ..., vn/2, Jvn/2 forms a basis for the original real vector space V?

Another result was that J gives V a natural orientation. How does it do this? Does it order the basis vectors with respect to the basis vectors described above? (Or is this result specific to my book?)

I have even more questions regarding this basic topic, but this is probably a good start. As you can tell, there's a lot I don't know!
 

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