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.


1 Answer 1


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! $\endgroup$ May 8 at 22:16
  • $\begingroup$ @DaveShulman: You are welcome! Happy to help! $\endgroup$
    – M.G.
    May 8 at 22:28

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