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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?

goblin GONE
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    See https://ncatlab.org/nlab/show/derivation#DerOfContFuncts – Mariano Suárez-Álvarez Nov 15 '16 at 03:36
  • Does it have to be in \mathbb{R}^n ? can it be quateriononic, or am I not understanding.. – Philip Oakley Nov 15 '16 at 15:53
  • @PhilipOakley, the important thing is that we're dealing with $C^0$ rather than $C^1$. – goblin GONE Nov 16 '16 at 03:35
  • @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. – Philip Oakley Nov 25 '16 at 11:37
  • @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. – goblin GONE Nov 26 '16 at 00:29
  • @goblin, I agree that often the physicists don't get the maths right, expecially for EM/QM/relativity which has a quaternion basis (Maxwell Vol2, Part 4, Ch9, art 618, p236 http://posner.library.cmu.edu/Posner/books/pages.cgi?call=537_M46T_1873_VOL._2&layout=vol0/part0/copy0&file=0263) hence my quaternion question. I also have a thing about Dimenional analysis and radians being an Angle, and hence a supplementary Dimension that must be tracked, because they squeezed a 3d space into a single Dimension of Length. I work in electro optics where such mistakes (forgetting Angle) are quite common. – Philip Oakley Nov 27 '16 at 16:29
  • @PhilipOakley, sounds interesting. You seem to be proposing that we should replace angles with an altogether different concept. If so, what is this alternative concept? If not, then what are you proposing? – goblin GONE Nov 28 '16 at 05:42
  • For "Angles", they represent the collapse of what was two dimensions into none, yet still within a 2d space. One would (normally) never collapse m/s to be unitless, but m(height)/m(width) appears fair game. For mathematicians it is to realise that values on the number line ($\pi$) are distinct from those on the number circle ($\pi rad$ ), and every time one says 'values in radians' one is implicitly noting that angles are 'special'. For the SI community, the issue is seen from the other side, that Length is a 3d space, not a single dimension, and Angle indicates a lost Dimension – Philip Oakley Nov 28 '16 at 09:59
  • @goblin see http://math.stackexchange.com/a/2034581/128985 for a longer answer about angles. – Philip Oakley Nov 28 '16 at 17:42

1 Answers1

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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 $d=0$.

Eric Wofsey
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    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 $0$. – Balarka Sen Nov 15 '16 at 13:00