1. ## 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.

2. ## Re: Derivations

Originally Posted by slevvio
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?

3. ## Re: Derivations

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