It is well-known that one can describe tangent vectors to a smooth manifold $M$ as derivations on germs of functions, see this question. However, according to "Riemannsche Geometrie im Großen" by Gromoll, Klingenberg and Meyer, §1.1.3, A(i), there is an infinite-dimensional vector space of $\mathbb R$-linear maps $\partial\colon C^r(M)\to\mathbb R$ satisfying the usual product rule $$\partial(fg)=\partial f\cdot g(p)+f(p)\cdot\partial g$$ for function germs $f$, $g$ at $p$, where $p$ is a point on a finite-dimensional $C^r$-manifold $M$ with $1\le r<\infty$. Is there any accessible description of "strange" derivations that do not originate from tangent vectors in the classical sense, that is, either from velocities of $C^r$-curves, or from a coordinate description behaving correctly under coordinate changes?
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asked Oct 21 at 16:41
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1$\begingroup$ I do not see how it would be possible since it would yield a "description" of the algebraic dual of some infinite dimensional vector space. Then you get into issues related to the axiom of choice... I am actually unsure if the existence of exotic derivations is independent of the axiom of choice. Maybe logicians here can help with this... $\endgroup$– Moishe KohanCommented Oct 21 at 17:05
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1$\begingroup$ @MoisheKohan I would be surprised if this was a logics problem. The only proof that I know rewrites a $C^k$ function as a sum of products $g_i\cdot x^i$, where the $g_i$ are $C^{k-1}$ functions and the $x^i$ are coordinates. But then $g_i$ is no longer in the domain of the derivation, unless $k=\infty$. So for finite $k$, you cannot apply the product rule anymore. $\endgroup$– Sebastian GoetteCommented Oct 21 at 19:52
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1$\begingroup$ Take a look at my answer here where this is spelled out in the case $r=1$. When one says "extend the given linearly independent subset to a basis" one uses AC. $\endgroup$– Moishe KohanCommented Oct 21 at 22:29
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1$\begingroup$ @MoisheKohan my answer does not answer your question of "do there exist exotic derivations independent of choice," so I don't see why you call it "quite wrong". It answers the OP's question of whether or not there is an accessible way of describing an exotic derivation. I gave one example and the OP seemed to find it helpful. If you think using choice makes it not accessible then that would be a matter of opinion. $\endgroup$– user113407Commented Oct 21 at 22:53
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1$\begingroup$ @OrangeMushroom: The original question asked for a "description" of accesible derivation. Your answer only proves existence; not a single exotic derivation is described. What you described is a linearly independent subset of the dual space of the space of derivations. But, I see that even this is enough to OP, so be it. $\endgroup$– Moishe KohanCommented Oct 22 at 11:03
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2 Answers
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See this note describing the algebraic tangent space of a $C^k$ manifold. In particular, on $\mathbb{R}$, there exists a basis of the cotangent space $\mathfrak{m}_p\, / {\mathfrak{m}_p}^2$ containing the functions $|x-p|^{k+\alpha}$ for all $\alpha \in (0, 1)$.
We can therefore define a derivation at $p$ which maps $|x-p|^{k+\frac{1}{2}}$ to $1$ and every other basis element to $0$.
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2$\begingroup$ To my understanding, the paper establishes that the functions $|x - p|^{k+\alpha}$ are linearly independent. To extend this to a basis seems to require some choice principles. I'm not sure if this counts as accessible. (If the real question was why m/m^2 is infinite dimensional this works very well.) $\endgroup$– mmeCommented Oct 21 at 22:10
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1$\begingroup$ Even with AC, your answer at best describes "some" exotic derivations. In fact, instead of describing derivations, you describe a linearly independent subset of the space dual to derivations. If you think through your answer, you will see why. I spelled this out in the case $k=1$ in my answer here. $\endgroup$ Commented Oct 21 at 22:42
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3$\begingroup$ I know that it only describes "some", I was merely giving an example of some derivations. $\endgroup$– user113407Commented Oct 21 at 22:43
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2$\begingroup$ @MoisheKohan I don’t really understand your point. OP never specified any preference as to whether an answer should use choice or not? And your second comment does not make sense to me either. Describing “some” exotic derivations is exactly what OP asked to do. $\endgroup$ Commented Oct 21 at 23:05
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2$\begingroup$ @MoisheKohan And OP even said “this is exactly what i was looking for”… $\endgroup$ Commented Oct 21 at 23:08
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I think there are no "derivations that do not originate from tangent vectors in the classical sense".
Indeed, any derivation $D$ must satisfy the chain rule $$(Df)(p)=\sum_{i=1}^n(D_i f)(p)(Dx^i)(p)$$ at every point $p\in M$, where $f$ is a smooth enough function on $M$, $x^i$ is the $i$th coordinate function near $p$, and $(D_i f)(p)$ is the partial derivative of $f$ at $p$ wrt the $i$th coordinate. Letting now $v=(v^1,\dots,v^n)=((Dx^1)(p),\dots,(Dx^n)(p))$, we get $$(Df)(p)=\sum_{i=1}^n(D_i f)(p)v^i=(D_v f)(p),$$ the directional derivative of $f$ at $p$ in the direction of $v$.
answered Oct 21 at 17:41
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3$\begingroup$ The only proof of your first formula that I know uses functions similar to $g(x)=\begin{cases}(f(x)-f(0))/x&x\ne0\\f'(0)&x=0\end{cases}$ in dimension $1$. But $g$ has one derivative less than $f$, so one cannot apply $D$ to it. But maybe you do this differently? $\endgroup$ Commented Oct 21 at 19:49
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$\begingroup$ @SebastianGoette : This is a good point. Let me think about this. $\endgroup$ Commented Oct 21 at 20:47