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Math Help - Bijection between tangent spaces

  1. #1
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    Bijection between tangent spaces

    Hello,

    We have defined the tangent space at a point p of a manifold as the set of all derivatives.
    Now i want to show, that there exist a bijection between all derivatives and the "Tangentvectors" defined by (equivalent classes of) paths.
    More precisely:
    We call two paths equivalent, iff they define the same derivation at a point p. Now i want to show that there is a bijection between the set of all equivalent classes and the Tangentspace (defined as the set of derivatives).

    My Approach was the following.
    Every path c:I->M into the manifold M defines a derivation by
    D(f):=\frac{d}{dt}_{|t=0}(f\circ c)(t)

    so i try to show that \phi([c])=D with D(f)=\frac{d}{dt}_{|t=0}(f\circ c)(t) is a bijection.

    Of course \phi is well defined and injective, but i couldn't show that
    \phi is surjective!

    If D is some arbitrary derivative at the point p. How can i find a path c, s.t. \phi(c)=D?

    Regards
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  2. #2
    Super Member Rebesques's Avatar
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    If you fix a point p and express everything in a local chart, then the required path c(s), \ -\epsilon<s<\epsilon is just the unique solution of the initial value problem c'(0)=D, \ c(0)=p.
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  3. #3
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    Hello Rebesques,

    thank you for your help! But i don't see what you mean. Ok lets choose a chart (U, \phi) around the fixed point p.
    I think by your notation you mean c'(0)=\frac{d}{dt}_{|t=0}(f\circ c)(t), is it right?

    But How can i solve this differential equation D=c'(t)?

    But i know by a thm., that every derivation D is of the form
    D=\sum_{i=1}^d D(x_i) \frac{d}{dx_i}_{|p}

    whereas x_i are the coordinate functions of our selected chart \phi
    Perhaps this equation can help? But i don't know how.
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  4. #4
    Super Member Rebesques's Avatar
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    Not quite. We are looking for a curve -\epsilon,\epsilon)\rightarrow M" alt="c-\epsilon,\epsilon)\rightarrow M" /> with c(0)=p, \ c'(0)=(\frac{\partial}{\partial t}\vert_{0})c=D\in T_pM. There are many curves with this property; Let's construct one.

    Choose a chart (U,\phi), \ \phi=(x^1,\ldots,x^n), and w.l.o.g. assume \phi(p)=0, \ D(\phi)\vert_{p}=I, where D(\phi)\vert_{p}=(\frac{\partial}{\partial x^i}\vert_p(x^j)) and I is the identity matrix.

    Then, by letting D=a^i\frac{\partial}{\partial x^i} and identifying D=(a^1,\ldots,a^n)\in R^n, the curve c(t)=\phi^{-1}(tD), \ -\epsilon<t<\epsilon has the required properties.

    Ps. And c is unique in the sence it belongs to the equivalence class of curves passing through p and having tangent D\in T_pM.
    Last edited by Rebesques; December 17th 2010 at 02:50 AM. Reason: notation
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