What does an "exotic" derivation at a point $x_{0}\in \mathbb{R}^{n}$ look like?

Question

for $x_{0}\in \mathbb{R}^{n}$ I denote by $G^{k}(x_{0})\: (1\le k\le \infty )$ the $\mathbb{R}$-algebra of germs of real-valued $C^{k}$-functions at $x_{0}$. Recall that a derivation at $x_{0}$ is a linear form $D:G^{k}(x_{0})\to\mathbb{R}$ which satisfies the product rule $D(f\cdot g)=f(x_{0})\cdot Dg+g(x_{0})\cdot Df$. I call a differentiable function $c:\mathbb{R\to \mathbb{R}^{n}}$ with $c(0)=x_{0}$ a curve at $x_{0}$. Recall that each such curve defines a derivation $D_{c}$ at $x_{0}$ by $D_{c}(f)=\frac{d }{dt}(f\circ c)_{|t=0}$ (where$f$ is any representative of the respective germ), called the derivative in the direction of the tangent to $c$ at $x_{0}$.The well-known (and not hard to prove) fact that in the case $k=\infty $ every derivation at $x_{0}$ is actually a directional derivative is sometimes used to define the tangent space at a point of a $C^{\infty }$-manifold. But already as a student I was baffled to learn that this does not work for $k\lt \infty $: It has been proved that the vector space of all derivations is infinite-dimensional in this case. Just recently I have studied a rather short proof of this but have not been able to deduce a definite example for such an "exotic" derivation which is not a directional derivative. Can anybody present such an example here?

And I would like to know: are there derivations $\neq 0$ on $G^{0}(x_{0})$ ?

Answer 1

I will consider the case of derivations on $C^1({\mathbb R})$ at $0$. (I will also work with functions instead of germs, just the the notational ease.) The general case is similar. Let ${\mathfrak m}$ denote the maximal ideal in the ring $C^1({\mathbb R})$ consisting of functions vanishing at $0$. Then the space of derivations $Der_0$ on $C^1({\mathbb R})$ is isomorphic to the (algebraic) dual of the vector space $V={\mathfrak m}/{\mathfrak m}^2$. (In what follows, no continuity will be assumed for linear functionals.) The vector space $V$ is infinite-dimensional, an infinite set of linearly independent vectors in $V$ is given by $$ |x|^\alpha, 1<\alpha<2. $$ Thus, $V^*$ is also infinite-dimensional.

In order to get a (more) concrete example of an "exotic" derivation on $C^1({\mathbb R})$ at $0$, fix some $\alpha\in (1,2)$ and consider the subspace $W\subset {\mathfrak m}$ consisting of functions $f$ satisfying:

• $f'(0)=0$.

• $f(x)=|x|^\alpha f_1(x)$, for a function $f_1\in C^0({\mathbb R})$.

(If you prefer, instead of $W$ you can take the subspace of $C^{1,\alpha}({\mathbb R})$ consisting of functions vanishing at $0$ together with their first derivative.)

Now, define the linear functional $\varphi\in W^*$ by $\varphi(f)=f_1(0)$. In other words, $$ \varphi(f)=\lim_{x\to 0} \frac{f(x)}{|x|^\alpha}. $$

Lastly, and this is where I need some form of the Axiom of Choice, represent $C^1({\mathbb R})$ as a direct sum $W\oplus U$, where $U$ contains the functions $1$ and $x$, and extend $\varphi$ to a linear functional on $C^1({\mathbb R})$ by letting it be identically zero on $U$. (Below I will check that $\Phi$ vanishes on the ideal ${\mathfrak m}^2$ and, hence, defines a derivation on $C^1({\mathbb R})$ at $0$.)

Every such extension $\Phi$ is an "exotic" derivation on $C^1({\mathbb R})$ at $0$. Indeed, the "standard" derivations are scalar multiples of the linear functional $$ f\mapsto f'(0). $$ Clearly, $\Phi$ is not of this form since it vanishes on $f(x)=x$.

Let's check that $\Phi$ satisfies the Leibnitz Rule. It suffices to show that for any two functions $f, g\in \mathfrak m$, $\Phi(fg)=0$. We have $$ f(x)=xf_1(x), \ g(x)=x g_1(x), $$ where $f_1, g_1$ are continuous. Then $h=fg=x^2 f_1(x) g_1(x)$ belongs to $W$ and, hence, we have $$ \Phi(fg)=\varphi(fg)= \lim_{x\to 0} \frac{x^2 f_1(x) g_1(x)}{|x|^\alpha}= \lim_{x\to 0} |x|^{2-\alpha} f_1(x) g_1(x)=0, $$ since $2-\alpha>0$.

I do not know if there is a construction just in ZF, avoiding any form of the Axiom of Choice.

As for your last question: $G^0(x_0)$ has only zero derivations. Indeed, it suffices to consider the case $x_0=0$. Now, for every continuous function $f$ vanishing at $0$, the function $f^{1/3}$ (the cube root of $f$) is also continuous and vanishes at zero. Hence, $f\in {\mathfrak m}^2$, where ${\mathfrak m}\subset C^0({\mathbb R}^n)$ is the maximal ideal consisting of functions vanishing at $0$. Hence, ${\mathfrak m}= {\mathfrak m}^2$, which implies that the algebraic dual to ${\mathfrak m}/{\mathfrak m}^2$ is also zero. Hence, there are no nonzero derivations.

Addendum. Here is a proof of a linear isomorphism
$$ ({\mathfrak m}/{\mathfrak m}^2)^* \to Der_0$$ where $Der_0$ is the space of derivations on $C^k({\mathbb R}^n)$ at $0$ and ${\mathfrak m}$ is the maximal ideal in $C^k({\mathbb R}^n)$ consisting of functions vanishing at $0$.

Let $\varphi: {\mathfrak m}\to {\mathbb R}$ be a linear functional vanishing on ${\mathfrak m}^2$. Extend $\varphi$ to a linear functional $\varphi$ on $C^k({\mathbb R}^n)$ by requiring $\varphi(const)=0$ for all constant functions. Now, define $$ D_\varphi: C^k({\mathbb R}^n)\to {\mathbb R}$$ by $$ D_\varphi(f)= \varphi(f) = \varphi(f- f(0)). $$ It is clear that $D_\varphi$ is linear and that the map $\mu: \varphi\mapsto D_\varphi$ is linear as well. Let's verify that $D=D_\varphi$ satisfies the product rule at $0$. Take two functions $f, g\in C^k({\mathbb R}^n)$ and define $$ \tilde f= f - f(0), \tilde g= g- g(0). $$ Then $D(f)=D(\tilde f), D(g)=D(\tilde g)$ and $$ D(fg)= D((\tilde f + f(0))(\tilde g + g(0))= \varphi( \tilde f \tilde g) + f(0) D (\tilde g) + g(0) D(\tilde f) + 0= $$ $$ 0+ f(0) D(g) + g(0) D(f) $$ since $\varphi(\tilde f \tilde g)=0$ (because $\tilde f \tilde g\in {\mathfrak m}^2$). The inverse to $\mu$ is given $\nu$, where $\nu(D)$ is the restriction of the derivation $D$ to ${\mathfrak m}$. Since every derivation vanishes on ${\mathfrak m}^2$, this restriction descends to a linear functional on ${\mathfrak m}/{\mathfrak m}^2$. qed