> [!definition] > > Let $X$ be a [[n-Manifold|n-manifold]], $U \subset X$ be [[Open Set|open]] and $f \in C^\infty(U)$ be a [[Function on Manifold|smooth function]]. The pair $(f, U)$ is a **smooth function element** on $X$. > [!definition] > > Let $p \in X$, and $(f, U)$, $(g, V)$ be smooth function elements. Define $(f, U) \sim (g, V)$ if there exists a [[Neighbourhood|neighbourhood]] $W \in \cn^o(p)$ such that $f|_W = g|_W$. The equivalence class of $(f, U)$ is the **germ** of $f$ at $p$, denoted as $[f]$. The set $C^\infty_p(X)$ is the space of all germs at $p$, which is an [[Algebra over Ring|algebra over]] $\real$ with > - $(f, U) + (g, V) = (f + g, U \cap V)$ > - $(f, U) \times (g, V) = (fg, U \cap V)$ > [!theorem] > > A **derivation** of $C^\infty_p(X)$ is a [[Linear Functional|linear functional]] $v \in C^\infty_p(X)^*$ such that > $ > \angles{v, [fg]} = f(p) \cdot \angles{v, [g]} + g(p) \cdot \angles{v, [f]} > $ > then the space $\mathcal{D}_pX$ of derivations of $C^{\infty}_p(X)$ is naturally isomorphic to the [[Tangent Space|tangent space]] $T_pX$. > > *Proof*. Let $\Phi: \mathcal{D}_pX \to T_pX$ and $v \in \mathcal{D}_pX$, define $\Psi v$ such that for any $f \in C^\infty(X)$, > $ > \angles{\Psi v, f}_{T_pX} = \angles{v, [f]}_{\mathcal{D}_pX} > $ > For any $v \in T_pX$, define $\Psi^{-1}v$ such that for any $[f] \in C^\infty_p(X)$, > $ > \angles{\Psi^{-1}v, [f]}_{\mathcal{D}_pX} = \langle v, \tilde f \rangle_{T_pX} > $ > where $\tilde f$ is any extension of $f$ to $C^\infty(X)$. Let $(f, U) \in [f]$, then there exists a sufficiently small [[Coordinate Ball|coordinate ball]] $B \in \cn^o(p)$ such that $\ol{B} \subset U$. Therefore $f$ is smooth on $\ol{B}$ and [[Extension Lemma for Smooth Functions|admits a smooth extension]] $\tilde f \in C^\infty(X)$. Since any extension of any two elements in $[f]$ agree on a neighbourhood of $p$, the derivation is well-defined. > > Moreover, since the extension agrees with $f$ on $B$, $\tilde f \in [f]$, so > $ > \angles{\Psi^{-1} \Psi v, [f]}_{\mathcal{D}_pX} = \langle\Psi v, \tilde f\rangle_{T_pX} = \angles{v, [f]}_{\mathcal{D}_pX} > $ > we have indeed found an inverse for $\Psi$, and it is an isomorphism.