For a manifold $M$ a vector field is a derivation of the algebra $C^{\infty}(M)$ of smooth functions on $M$. What happens if look instead as derivations on the continuous functions of a manifold. I guess we get fewer derivations . . . but I'm not sure how one might prove this.
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- $\begingroup$ This is also answered in one of the related questions: mathoverflow.net/questions/15448/… $\endgroup$M.G.– M.G.2022-05-08 22:04:58 +00:00Commented May 8, 2022 at 22:04
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2 More is true: if $X$ is a topological manifold, then in fact $\operatorname{Der}(C(X)) = 0$, where $C(X)$ denotes the $\mathbb{R}$-algebra of $\mathbb{R}$-valued continuous functions on $X$. In particular, this is so for smooth manifolds $M$.
Here is one proof: https://ncatlab.org/nlab/show/derivation#DerOfContFuncts
- 2$\begingroup$ I expected that the set of derivations would be smaller, but the fact that there are none is quite surprising! Thanks for the answe and the link! $\endgroup$Dave Shulman– Dave Shulman2022-05-08 22:16:31 +00:00Commented May 8, 2022 at 22:16
- $\begingroup$ @DaveShulman: You are welcome! Happy to help! $\endgroup$M.G.– M.G.2022-05-08 22:28:44 +00:00Commented May 8, 2022 at 22:28