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Thread: Derivations

  1. #1
    Senior Member slevvio's Avatar
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    Derivations

    Let $\displaystyle U \subseteq M$ be an open subset of a smooth manifold. Let $\displaystyle C^{\infty}(U)$ be the collection of smooth functions $\displaystyle f: C^{\infty}(U) \rightarrow \mathbb{R}$. Let $\displaystyle \operatorname{Der}_p (C^{\infty}(U)) $be the set of derivations $\displaystyle v: C^{\infty}(U) \rightarrow \mathbb{R}$, i.e. $\displaystyle v$ is linear and and$\displaystyle v(fg) = f(p)v(g) + g(p)v(f)$. We define $\displaystyle T_p M$ to be this space.

    Now I am trying to show that if $\displaystyle W \subseteq U$ is open, with $\displaystyle p \in W$, then the linear map $\displaystyle r: \operatorname{Der}_p(C^{\infty}(W)) \rightarrow \operatorname{Der}_p (C^{\infty}(U))$, given by $\displaystyle r(v)(f) = v (f|_W)$ for all $\displaystyle f \in C^{\infty}(U)$ is an isomorphism.

    However I am not sure how to show it is injective. If we take $\displaystyle v \in \operatorname{ker}r$, then $\displaystyle r(v) = 0$, i.e. $\displaystyle r(v)(f) = v(f|_W) = 0$ for all $\displaystyle f\in C^{\infty}(U)$. But why does this tell us that $\displaystyle v(f) = 0$ for all $\displaystyle f \in C^{\infty}(W)$? Is every function in $\displaystyle C^{\infty}(W)$ of the form $\displaystyle f|_W$ for some $\displaystyle f \in C^{\infty}(U)$ ?

    Any help with this would be appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Derivations

    Quote Originally Posted by slevvio View Post
    Let $\displaystyle U \subseteq M$ be an open subset of a smooth manifold. Let $\displaystyle C^{\infty}(U)$ be the collection of smooth functions $\displaystyle f: C^{\infty}(U) \rightarrow \mathbb{R}$. Let $\displaystyle \operatorname{Der}_p (C^{\infty}(U)) $be the set of derivations $\displaystyle v: C^{\infty}(U) \rightarrow \mathbb{R}$, i.e. $\displaystyle v$ is linear and and$\displaystyle v(fg) = f(p)v(g) + g(p)v(f)$. We define $\displaystyle T_p M$ to be this space.

    Now I am trying to show that if $\displaystyle W \subseteq U$ is open, with $\displaystyle p \in W$, then the linear map $\displaystyle r: \operatorname{Der}_p(C^{\infty}(W)) \rightarrow \operatorname{Der}_p (C^{\infty}(U))$, given by $\displaystyle r(v)(f) = v (f|_W)$ for all $\displaystyle f \in C^{\infty}(U)$ is an isomorphism.

    However I am not sure how to show it is injective. If we take $\displaystyle v \in \operatorname{ker}r$, then $\displaystyle r(v) = 0$, i.e. $\displaystyle r(v)(f) = v(f|_W) = 0$ for all $\displaystyle f\in C^{\infty}(U)$. But why does this tell us that $\displaystyle v(f) = 0$ for all $\displaystyle f \in C^{\infty}(W)$? Is every function in $\displaystyle C^{\infty}(W)$ of the form $\displaystyle f|_W$ for some $\displaystyle f \in C^{\infty}(U)$ ?

    Any help with this would be appreciated.
    Let me give you a hint, and see if you can figure it out from there. What is the other descrption of $\displaystyle T_pM$ and how does this relate to the last post I helped you with?
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  3. #3
    Senior Member slevvio's Avatar
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    Re: Derivations

    Can't this be proved purely algebraically, using the fact that the function f is smooth?
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