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Tangent space and tangent plane

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    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 Rd.
    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
    TxoM= { v in Rd: there exists a curve x(s): [-1,1]->M such that x(0)=x0, x'(0)=v}

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

    Now I don't understand why TxoM 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?

    2. Relevant equations
    Tangent Space

    3. The attempt at a solution
    N/A

    Any help/clarifications would be much appreciated!
     
  2. jcsd
  3. Feb 2, 2012 #2
    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 Rd.

    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 TpM.
     
  4. Feb 2, 2012 #3

    Dick

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    [itex]T_{x_0}M[/itex] 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 [itex]T_{x_0}M + x_0[/itex]. The tangent space and the tangent plane are a little different.
     
  5. Feb 2, 2012 #4
    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.


    TxoM= { v in Rd: 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?
     
  6. Feb 2, 2012 #5

    Dick

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    I told you in my last post. x(t)=x0 goes through x0 at t=0 and has derivative 0.
     
  7. Feb 2, 2012 #6
    For 0 to be in Tx0M, 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?
     
  8. Feb 2, 2012 #7

    Dick

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    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.
     
  9. Feb 3, 2012 #8
    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.
     
  10. Feb 3, 2012 #9

    Dick

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    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.
     
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