Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

1 Answer 1

up vote 0 down vote accepted

@ fiktor: due to the low amount of reputation points I have at the moment, I can't comment yet. I just wanted to point out that, since I suspect this is a research-level question, there's a website called which might be an (even) better platform to launch this question from.

share|cite|improve this answer
Yes, I've realized this, when there were no answers here and posted it there:… . –  Fiktor Sep 28 '10 at 16:01

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.