Tangent space and tangent plane

[kingwinner]

Homework Statement

So I'm a little confused about what a tangent space is. Is it the same as the equation of the tangent plane in lower dimensions?

My notes define the tangent space as follows.
Let M be a hypersurface of R^d.
Let x(s) be a differentiable curve in M such that x(0)=x₀ is in M.
Then x'(0) is a tangent vector.

The tangent space at x₀ of M is
T_xoM= { v in R^d: there exists a curve x(s): [-1,1]->M such that x(0)=x₀, x'(0)=v}

T_xoM is a vector space of the same dimension as M.
====================

Now I don't understand why T_xoM is a vector space. A vector space MUST contain the zero vector, which is the origin.

But even in lower dimensions, say M is the unit sphere in R³, the tangent plane at (0,0,1) clearly does not pass through the origin (0,0,0). How can this be a vector space?

Homework Equations

Tangent Space

The Attempt at a Solution

N/A

Any help/clarifications would be much appreciated!

 

[A. Bahat]

I think you are confusing the origin of the tangent space and that of the manifold itself. Generally speaking, a manifold doesn't have to be a subset of Euclidean space, so it usually doesn't even make sense to speak of the "origin" for the manifold. Even if the manifold is a subset of Euclidean space as you are using, the manifold may not contain the origin of R^d.

As for the tangent space itself, I'll use your example of a 2-sphere in R³ and the tangent space at (0,0,1), as is roughly depicted by this picture:
http://upload.wikimedia.org/wikipedia/commons/6/66/Image_Tangent-plane.svg

The zero vector for this particular tangent space would just be (0,0,1), that is, the p in T_pM.

 

[Dick]

$T_{x_0}M$ is a vector space. The curve x(t)=x0 is a curve with derivative the zero vector. In R^n you can write the tangent plane at x0 as $T_{x_0}M + x_0$. The tangent space and the tangent plane are a little different.

 

[kingwinner]

OK, now I think I understand it better. So in my sphere example, the tangent plane at (0,0,1) and the tangent space at (0,0,1) are just NOT the same.


T_xoM= { v in R^d: there exists a differentiable curve x(s): [-1,1]->M such that x(0)=x₀, x'(0)=v}

But now, it is still not clear to me why (0,0,...,0) must be an element of the tangent space. What kind of differentiable curve on M going through the point x₀ would have x'(0)=(0,0,...,0) as the tangent vector?

 

[Dick]

OK, now I think I understand it better. So in my sphere example, the tangent plane at (0,0,1) and the tangent space at (0,0,1) are just NOT the same.


T_xoM= { v in R^d: there exists a differentiable curve x(s): [-1,1]->M such that x(0)=x₀, x'(0)=v}

But now, it is still not clear to me why (0,0,...,0) must be an element of the tangent space. What kind of differentiable curve on M going through the point x₀ would have x'(0)=(0,0,...,0) as the tangent vector?

I told you in my last post. x(t)=x0 goes through x0 at t=0 and has derivative 0.

 

[kingwinner]

For 0 to be in T_x0M, there must exist a curve x(s): [-1,1]->M such that x(0)=x0, x'(0)=0.

But for any smooth curve, x'(0) can't be 0, right? So if the hypersurface is smooth at x₀, every curve going through x₀ will be smooth and x'(0) can never be 0?

 

[Dick]

For 0 to be in T_x0M, there must exist a curve x(s): [-1,1]->M such that x(0)=x0, x'(0)=0.

But for any smooth curve, x'(0) can't be 0, right? So if the hypersurface is smooth at x₀, every curve going through x₀ will be smooth and x'(0) can never be 0?

You are making up that bit about a smooth curve never having derivative 0. What's not smooth about x(s)=x0? All the derivatives exist and are 0.

 

[kingwinner]

Yes, I screwed up...sigh. x'(0) nonzero => smooth, but the converse is false.

One more question: In the definition, we have x(s): [-1,1]->M. Why is the [-1,1] part necessary?

Thanks.

 

[Dick]

Yes, I screwed up...sigh. x'(0) nonzero => smooth, but the converse is false.

One more question: In the definition, we have x(s): [-1,1]->M. Why is the [-1,1] part necessary?

Thanks.

The specific interval [-1,1] is unimportant. Any domain for s including an interval around the origin will do. The intent is just to say x(s) doesn't have to be defined for all s. Just near s=0.