> [!definition] > > Let $X$ be a smooth [[n-Manifold|n-manifold]], and let $\xi: X \to TX$ be a rough [[Vector Field|vector field]]. For any [[Function on Manifold|function on]] $X$, $f \in C^\infty(X)$, define $\xi f$ by > $ > \xi f(p) = \xi(p) \cdot f > $ > [!theorem] > > The following are equivalent: > 1. $\xi$ is smooth. > 2. For every [[Function on Manifold|function]] $f \in C^\infty(X)$, the function $\xi f$ is smooth. > 3. For every open subset $U \subset X$ and $f \in C^\infty(U)$, $\xi f \in C^\infty(U)$. > > *Proof*. Suppose that $\xi$ is smooth. Let $f \in C^\infty(X)$, $p \in X$, $(U, \varphi) \in X$ be a [[Atlas|chart]] at $p$, then $\xi f$ has the local representation > $ > \hat p \mapsto D(\varphi^{-1}f)(p) \cdot (\xi_{\widehat U, \widehat U \times E})_2 > $ > where both components are smooth, so $\xi f$ is smooth as well. > > Suppose that $(2)$ holds. Let $f \in C^\infty(U)$, and $p \in U$. Let $\psi$ be a smooth bump function that is equal to $1$ in a neighbourhood of $p$ and supported in $U$, then by filling in $0$ elsewhere, $\tilde f = \psi \cdot f \in C^\infty(X)$. Therefore $\xi f = \xi \tilde f$ is smooth on that neighbourhood, so $\xi f \in C^\infty(U)$. > > Lastly, let $p \in X$, $(U, \varphi) \in X$ be a chart at $p$. We can use the coordinate maps $\bracs{x^i}_1^n$ on $U$ to extract each coordinate of the vector field. By assumption each $\xi x^i$ is smooth, so $\xi$ itself is smooth as well. > [!theorem] > > Let $X$ be a smooth $n$-manifold and $D: C^\infty(X) \to C^\infty(X)$ be a [[Derivation on Manifold|derivation]]. Then there exists a vector field $\xi \in \vf(X)$ such that $Df = \xi f$ for all $f \in C^\infty(X)$. > > *Proof*. Fix $p \in X$, then there exists $v_p \in T_pX$ such that $(Df)(p) = v_p \cdot f$. Let $\xi$ be such that $p \mapsto v_p$, then $\xi$ is a rough vector field. By the above theorem, $\xi$ is actually a smooth vector field. > [!theorem] > > Let $X$ be a manifold and $(U, \varphi)$ be a chart. Let $\xi \in \vf(X)$ be a vector field. If $\xi f = 0$ for all $f \in C^\infty(U)$, then $\xi|_U = 0$. Therefore vector fields are uniquely determined by their actions as derivations. > > *Proof*. Suppose that $\xi(p) \ne 0$. By the [[Hahn-Banach Theorem]], there exists a continuous linear functional $\hat f: E \to \real$ such that $\hat f(\widehat{\xi(p)}) \ne 0$. Let $f = \hat f \circ \varphi$, then $f \in C^\infty(U)$ and $\xi(p) \cdot f \ne 0$. Therefore $\xi f \ne 0$.