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Let $M$ be a manifold and $C_x^{\infty}(M,\mathbb R)$ be the algebra of germs of smooth functions on $M$ at $x$. A derivation of $C^{\infty}(M,\mathbb R)$ at a point $x$ is a linear map $D:C^{\infty}(M,\mathbb R)\rightarrow \mathbb R$ which satisfies $$D(f\cdot g)=Df\cdot g(x)+f(x)\cdot Dg$$

How can I show that a derivation of $C^{\infty}(M,\mathbb R)$ at a point $x$ factors to a derivation of $C_x ^{\infty}(M,\mathbb R)$?

Edit: I now understand the intended formulation is

Show that a derivation of $C^{\infty}(M,\mathbb R)$ at a point $x$ factors to a derivation of $C_x ^{\infty}(M,\mathbb R)$ at $x$.

Hence, the factoring is just through the projection taking a smooth function defined in an open neighborhood of $x$ to its germ.

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For any global derivation $D\colon C^{\infty}(M,\mathbb{R})\to C^{\infty}(M,\mathbb{R})$, we can form a derivation $D_x\colon C_x^{\infty}(M,\mathbb{R})\to \mathbb{R}$ by specifying how it should act on germs. So let $[f]_x$ be a germ, i.e. an equivalence class of partial functions $(f,U)$ on neighborhoods $U\ni x$ that agree on their overlaps. We will simply pick any partial function $(f,U)$ in the equivalence class and define $$ D_x([f]_x)=\left. D(f)\right|_x. $$ It is clear that this satisfies the properties of a derivation at $x$, since $D$ satisfies these properties pointwise.

Checking that it factors amounts to showing that this definition is well-defined; that is, it should be independent of which partial function $(f,U)$ we chose to represent $[f]_x$ by. But this is simply the statement that derivatives are local: $$ \text{If two functions agree on a neighborhood of $x$, then their derivatives agree at $x$.} $$

pre-kidney
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  • Yep, that's exactly what I added in the edit. I was confused at first because the word "at $x$" did not explicitly appear a second time, but there's no other possible meaning. –  Dec 31 '14 at 20:08
  • Yes your interpretation is correct; in this case the "at $x$" is implied. – pre-kidney Dec 31 '14 at 20:11
  • In the second paragraph (i.e., 'Checking that ... '), why do we assume that a derivation at a point $x \in U$ is a derivative? – rainman Sep 24 '20 at 08:31
  • How can we even make this definition, when $f \not\in C^\infty(M)$, but instead in a class of smooth functions of some smaller subset? – It'sNotALie. Oct 20 '22 at 00:14