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Math Help - Equivalent Definitions of Tangent Vectors

  1. #1
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    Equivalent Definitions of Tangent Vectors

    Hi, I'm working with different definitions of tangent vectors and trying to show they are equivalent. Here M is a smooth manifold of dimension n and C^{\infty} (p) is the set of smooth, real valued functions defined on an open neighbourhood of p \in M.

    Definition 1: A tangent vector at p \in M is the tangent vector to some curve c: (-\varepsilon ,\varepsilon ) \longrightarrow M, i.e. a map c'(0): C^{\infty}(p) \longrightarrow \mathbb{R} defined by c'(0)(f) := \frac{\mathrm{d} (f\circ c)}{\mathrm{d} t} \bigg\vert_{t=0}.

    Definition 2: A tangent vector at p \in M is a linear functional X: C^{\infty}(p) \longrightarrow \mathbb{R} satisfying the Leibnitz product rule: X(fg) = X(f)g(p) + f(p)X(g) for all f,g \in C^{\infty}(p).

    Definition 1 \Longrightarrow Definition 2 is obvious. To show Definition 2 \Longrightarrow Definition 1, let (U,\phi ) be a chart around p with coordinate functions x^1,\cdots , x^n. Define X^i = X(x^i) so that X = X^i\frac{\partial}{\partial x^i} \Big\vert_p.

    Now define a curve \gamma (t):= \phi^{-1} ( \phi(p) + t(X^1,\cdots ,X^n)) so that, for any f\in C^{\infty}(p), \gamma '(0) (f) = \frac{\mathrm{d} f (\gamma (t))}{\mathrm{d} t} \Big\vert_{t=0} = X^i \frac{\mathrm{d} f\circ \phi^{-1}}{\mathrm{d} x^i} \Big\vert_{\phi (p)} = X^i \frac{\partial}{\partial x^i}\Big\vert_p (f). Hence \gamma is the corresponding curve to X.

    I have two questions. First of all: is this correct? Secondly: if so, then why have we not used the Leibnitz product rule (and so could this not be true for linear functionals X: C^{\infty}(p) \longrightarrow \mathbb{R} not satisfying the product rule)?

    Any help would be appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Equivalent Definitions of Tangent Vectors

    Quote Originally Posted by markwolfson16900 View Post
    Hi, I'm working with different definitions of tangent vectors and trying to show they are equivalent. Here M is a smooth manifold of dimension n and C^{\infty} (p) is the set of smooth, real valued functions defined on an open neighbourhood of p \in M.

    Definition 1: A tangent vector at p \in M is the tangent vector to some curve c: (-\varepsilon ,\varepsilon ) \longrightarrow M, i.e. a map c'(0): C^{\infty}(p) \longrightarrow \mathbb{R} defined by c'(0)(f) := \frac{\mathrm{d} (f\circ c)}{\mathrm{d} t} \bigg\vert_{t=0}.

    Definition 2: A tangent vector at p \in M is a linear functional X: C^{\infty}(p) \longrightarrow \mathbb{R} satisfying the Leibnitz product rule: X(fg) = X(f)g(p) + f(p)X(g) for all f,g \in C^{\infty}(p).

    Definition 1 \Longrightarrow Definition 2 is obvious. To show Definition 2 \Longrightarrow Definition 1, let (U,\phi ) be a chart around p with coordinate functions x^1,\cdots , x^n. Define X^i = X(x^i) so that X = X^i\frac{\partial}{\partial x^i} \Big\vert_p.

    Now define a curve \gamma (t):= \phi^{-1} ( \phi(p) + t(X^1,\cdots ,X^n)) so that, for any f\in C^{\infty}(p), \gamma '(0) (f) = \frac{\mathrm{d} f (\gamma (t))}{\mathrm{d} t} \Big\vert_{t=0} = X^i \frac{\mathrm{d} f\circ \phi^{-1}}{\mathrm{d} x^i} \Big\vert_{\phi (p)} = X^i \frac{\partial}{\partial x^i}\Big\vert_p (f). Hence \gamma is the corresponding curve to X.

    I have two questions. First of all: is this correct? Secondly: if so, then why have we not used the Leibnitz product rule (and so could this not be true for linear functionals X: C^{\infty}(p) \longrightarrow \mathbb{R} not satisfying the product rule)?

    Any help would be appreciated.
    What does 'equivalent' mean? They obviously aren't the same concept, are you trying to show that the tangent spaces are canonically linearly isomorphic?
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    Re: Equivalent Definitions of Tangent Vectors

    Quote Originally Posted by Drexel28 View Post
    What does 'equivalent' mean? They obviously aren't the same concept, are you trying to show that the tangent spaces are canonically linearly isomorphic?
    Yes, sorry, by equivalent I mean that they define canonically isomorphic tangent spaces.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Equivalent Definitions of Tangent Vectors

    Quote Originally Posted by markwolfson16900 View Post
    Yes, sorry, by equivalent I mean that they define canonically isomorphic tangent spaces.
    Then, I don't understand why there is a both directions matter. Let T_pM stand for the usual tangent vector (not derivations) tangent space and D the derivation one. Define f:T_PM\to D by taking \gamma'(0)\to D_\gamma where D_\gamma(f)=(f\circ \gamma)'(0). It's clear that this is a monomorphism. Why don't you try proving it's a surjection.
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    Re: Equivalent Definitions of Tangent Vectors

    Quote Originally Posted by Drexel28 View Post
    Then, I don't understand why there is a both directions matter. Let T_pM stand for the usual tangent vector (not derivations) tangent space and D the derivation one. Define f:T_PM\to D by taking \gamma'(0)\to D_\gamma where D_\gamma(f)=(f\circ \gamma)'(0). It's clear that this is a monomorphism. Why don't you try proving it's a surjection.
    Ok, so to show it's surjective we take a derivation X \in D and try to find a curve \gamma such that X = D_{\gamma}. To do this we proceed as above:

    Let (U,\phi ) be a chart around p with coordinate functions x^1,\cdots , x^n. Define X^i = X(x^i) so that X = X^i\frac{\partial}{\partial x^i} \Big\vert_p. Now define \gamma (t):= \phi^{-1} ( \phi(p) + t(X^1,\cdots ,X^n)) so that, for any f\in C^{\infty}(p), \gamma '(0) (f) = \frac{\mathrm{d} f (\gamma (t))}{\mathrm{d} t} \Big\vert_{t=0} = X^i \frac{\mathrm{d} f\circ \phi^{-1}}{\mathrm{d} x^i} \Big\vert_{\phi (p)} = X^i \frac{\partial}{\partial x^i}\Big\vert_p (f). Hence X = D_{\gamma}.

    My question then is: where do we use the fact that elements of D satisfy the product rule? If we haven't used this fact, have we not defined an isomorphism between T_pM and the space of all linear functionals acting on C^{\infty}(p) (including ones which do not satisfy the product rule)?
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