Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isomorphic Tangent Spaces

  1. Feb 3, 2006 #1
    Hello,

    I am trying to understand how the geometric tangent space [tex]\mathbb{R}^n_a[/tex] given by

    [tex]
    \begin{displaymath}
    \mathbb{R}^n_a = \{(a,v) | v \in \mathb {R}^n\}
    \end{displaymath}
    [/tex]

    is isomorphic to the space of all derivations of [tex]C^{\infty}(\mathbb{R}^n)[/tex] at [tex]a[/tex], denoted by [tex]T_a(\mathbb{R}^n)[/tex].

    According to the book "Introduction to Smooth Manifolds" by John M. Lee, an isomorphism between these spaces is given by a map that sends each [tex]v_a[/tex] in [tex]\mathbb{R}^n_a[/tex] to the operator that represents the directional derivative evaluated at the point [tex]a[/tex] in the direction of [tex]v[/tex]. If [tex]\phi$[/tex] denotes the proposed isomorphism, we can write

    [tex]
    \begin{displaymath}
    \phi (v_a)(f) = \widetilde{v}_a(f) = v^i \dfrac{\partial f}{\partial x^i}(a)
    \end{displaymath}
    [/tex]

    for any [tex]f \in C^{\infty}(\mathbb{R}^n)[/tex].

    To show that the (clearly linear) map [tex]\phi[/tex] is an isomorphism we must show that it is a bijection. Now, to prove that [tex]\phi[/tex] is 1-1, from linear algebra we know that it is sufficient to show that the kernel of [tex]\phi[/tex] contains only the [tex]0[/tex] vector in [tex]\mathbb{R}^n_a[/tex], denoted by [tex]0_a[/tex]. This means that we must show that the only element of [tex]\mathbb{R}^n_a[/tex] that satisfies [tex]\phi (v_a)(f) = 0[/tex] is [tex]0_a[/tex].
    So, suppose
    [tex]
    \begin{displaymath}
    v^i \dfrac{\partial f}{\partial x^i}(a) = 0
    \end{displaymath}
    [/tex]
    If we can show that each component [tex]v^i = 0[/tex] injectivity will follow - and this is where I am stuck. Any ideas?

    Thanks
     
  2. jcsd
  3. Feb 3, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    why not choose to evaluate the directional derivatives on nice choices of f, eg the function that picks out components.
     
  4. Feb 3, 2006 #3
    Indeed; that is what the author does. He uses the standard coordinate functions for this purpose. But how can such a selection be justified? The relationship should hold not just for particular choices of f but for any f. Obviously, I'm being dense here, but I just can't see it...
     
  5. Feb 3, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Eh? You want to show that the resulting object (thing with [itex]\partial_x[/itex]) is not zero, it suffices to demonstrate so by finding a function, f, on which it doesn't vanish. Since v_i is not zero for some i, it follows that evaluated on the function that gives the i'th coordinate (the derivative is just v_i) is not zero.
     
    Last edited: Feb 3, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Isomorphic Tangent Spaces
  1. Tangent space (Replies: 17)

Loading...