Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a derivation of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$.
If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$.
Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and
$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$

Question: I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?

Examples: (with sketches of the proofs)
1. Algebra $C^\infty(M)$ of smooth functions on a closed manifold $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using Picard-Lindelof theorem.
2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.
3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.
4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.
5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.
6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.

share|cite|improve this question
Do you know if it's true for the algebra of polynomial functions on a real affine variety? (A compact real affine variety?) – Qiaochu Yuan Sep 19 '10 at 1:17
At least for one of them the answer is no. I have added an explanation in examples 5 and 6. – Fiktor Sep 19 '10 at 20:00
For 1 don't you need that the derivation be local? – Mariano Suárez-Alvarez Sep 19 '10 at 23:26
For 1 it is always local. Moreover, it can be proven, that locally every derivation $\xi$ of $C^\infty(M)$ has the form $\sum_i\alpha_i(x)(\partial/\partial x_i)$ with smooth functions $\alpha_i$. – Fiktor Sep 20 '10 at 7:17
As was pointed to me on my definition of derivative (over t) is bad. In the sense, stated above, the derivative rarely exists. – Fiktor Sep 28 '10 at 16:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.