By an exotic vectorfield on $\mathbb{R}^n$, I mean a non-zero derivation on the algebra $C^0(\mathbb{R}^n)$.

Do such things exist?

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    $\begingroup$ See ncatlab.org/nlab/show/derivation#DerOfContFuncts $\endgroup$ – Mariano Suárez-Álvarez Nov 15 '16 at 3:36
  • $\begingroup$ Does it have to be in \mathbb{R}^n ? can it be quateriononic, or am I not understanding.. $\endgroup$ – Philip Oakley Nov 15 '16 at 15:53
  • $\begingroup$ @PhilipOakley, the important thing is that we're dealing with $C^0$ rather than $C^1$. $\endgroup$ – goblin GONE Nov 16 '16 at 3:35
  • $\begingroup$ @goblin, as an engineer/physicist, what effects do you expect of an "exotic" vector field? The definition is somewhat inward looking (as it needs to be), without the outward looking corollaries. $\endgroup$ – Philip Oakley Nov 25 '16 at 11:37
  • $\begingroup$ @PhilipOakley, I'm not an engineer/physicist :) When I was young, I slowly moved away from physics and toward mathematics because the physicists were just... not using symbols properly. My plan was to return to physics once I knew a lot more math, and clean up the mess they have made. But, I've kind of lost interest over the years. My current plan is to clean up the mess mathematicians are making, by refounding mathematics. A day may come when I start thinking about physics again. But it is not this day. $\endgroup$ – goblin GONE Nov 26 '16 at 0:29

No, they don't. Suppose $d:C^0(\mathbb{R}^n)\to C^0(\mathbb{R}^n)$ is a derivation. First, suppose $f\in C^0(\mathbb{R}^n)$ is nonnegative everywhere. Then the nonnegative square root $\sqrt{f}$ is continuous, and $$d(f)=d(\sqrt{f}^2)=2\sqrt{f}d(\sqrt{f}).$$ It follows that if $f(p)=0$, then $d(f)(p)=0$ (since $\sqrt{f}(p)=0$).

Now let $f$ be any continuous function and suppose $f(p)=0$. Then $d(f)=-d(|f|-f)+d(|f|)$. Since $|f|-f$ and $|f|$ are both nonnegative and vanish at $p$, it follows that $d(f)(p)=0$.

But now for any $f$ and any $p$, we can consider the function $g=f-f(p)$. Then $g$ vanishes at $p$, so $d(g)$ vanishes at $p$. But $d(g)=d(f)$ since they differ by a constant. Thus $d(f)$ vanishes at $p$. Since $f$ and $p$ were arbitrary, this means $d=0$.

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    $\begingroup$ You can actually use $f = (\text{sign}(f)\sqrt{|f|})(\sqrt{|f|})$ for any continuous $f$, I think. Then $d(f) = \text{sign}(f)\sqrt{|f|}d(\sqrt{|f|}) + d(\text{sign}(f)\sqrt{|f|})\sqrt{|f|}$. If $f(p) = 0$, that's $0$. $\endgroup$ – Balarka Sen Nov 15 '16 at 13:00

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