In which commutative algebras does any derivation possess a flow?

Question

Definitions
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
$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$

The question
1. I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?
2. Is there any simple condition for an algebra and it's derivation, from which it follows, that this derivation possesses a flow?

Examples
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.
1'. Algebra $C^\infty(M)$ of smooth functions on a non-compact manifold without boundary --- no (see example 2), but a derivation possesses a flow if it preserves some function $H\in A$, such that for any $c\in\mathbb{R}$ subspace $\{x\mid H(x) < c \}$ of the topological space $M$ is compact.
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.
7.(from Greg Muller's answer) Algebra $A$ is finite-dimensional --- yes, every derivation possesses the flow.

This question was already posted here on math.stackexchange.com, but it has received no answers even with a bounty. Any help is appreciated, both in the theme of question and in improving its wording.

Answer 1

If you really mean to get away with treating the algebra as an algebraic object -- no topology is given on this vector space -- then you don't even have the smooth compact manifold example. The given definition of $a=\gamma'(\tau)$ is rarely satisfied. There are too many linear maps from $A$ to $\mathbb R$. A function $\gamma:\mathbb R\to A$ cannot have a derivative in your sense at $\tau$ unless there is some $\epsilon$ such that the vectors $\gamma(t)$ for $|t-\tau|<\epsilon$ span a finite-dimensional vector subspace of $A$.

On the other hand, it is true that a smooth compact manifold (with its (sheaf of) smooth functions) can be reconstructed from the purely algebraic object which is the algebra of global functions, and also that the only derivations of that algebra are the ones arising from tangent vector fields. I don't how to describe (without introducing a topology on the algebra, or something like that) the relation between the derivations and the corresponding homomorphisms from the additive group of real numbers into the automorphism group of the algebra.

Answer 2

My guess is this is the kind of algebra you don't care about (since they aren't subrings of real-valued functions), but algebras of the form $\mathbb{R}[x]/x^n$ will have your property. When looking for a flow corresponding to a derivation $\xi$, consider the formal power series $$ \sum_i \frac{t^i\xi^i}{i!} $$ The problem, of course, is showing that this power series converges in the algebra of appropriate endomorphisms. However, a lazy trick for guaranteeing its convergence is to hope that $\xi^n=0$ for some $n$. This happens for every derivation in the ring $\mathbb{R}[x]/x^n$, or more generally when the nilradical is maximal.

Edit: As is pointed out in the comments, this is a correct claim but the wrong reason. The real reason flows exponentiate in these rings, as well as all finite dimensional ones, is that a derivation is given by a matrix (after some choice of basis), and all matrices can be exponentiated (that is, the above series converges).

Answer 3

If you want to work in the algebraic category you should think about an algebraic notion of flow. If $M$ is a smooth manifold then a flow is an $\mathbb{R}$-action $\mathbb{R} \times M \to M$; dualizing this map gives a coaction $C^{\infty}(M) \to C^{\infty}(\mathbb{R}) \otimes C^{\infty}(M)$ where by $\otimes$ I mean a suitably completed tensor product. (Note that the group structure on $\mathbb{R}$ gives $C^{\infty}(\mathbb{R})$ a Hopf algebra structure, at least with respect to a suitably completed tensor product as above, so we can talk about comodules over it.)

So the most algebraic notion of flow would be a "polynomial flow," namely a coaction $A \to k[t] \otimes_k A$, where $A$ is a $k$-algebra and we are thinking of $k[t]$ in its incarnation as the ring of functions on the additive group scheme $\mathbb{G}_a$ over $k$. In fact in this language a derivation is precisely a coaction $A \to k[t]/t^2 \otimes_k A$ and the problem of integrating this to a flow is a lifting problem.

A sufficient condition for a polynomial flow to exist is that the original derivation $D$ be nilpotent, but you knew that already. A slightly more interesting sufficient condition is that it be locally nilpotent in the sense that for every $a \in A$ there is some $n$ such that $D^n a = 0$. For example this is true of the derivation $\frac{\partial}{\partial x}$ on $k[x]$. I think this condition is also necessary when $A$ is an integral domain.

There are variations on this theme, e.g. we can talk about formal flows $A \to k[[t]] \otimes_k A$ but these always exist in characteristic zero (formally exponentiate) so this is in some sense uninteresting. Replacing $k[[t]]$ with other variations of $k[t]$ give other notions of flow.

For example, to talk about flows on smooth manifolds in this algebraic language we should really talk about smooth algebras. Smooth algebras admit a smooth tensor product (this is the suitably completed tensor product I wanted) with respect to which $C^{\infty}(M) \otimes C^{\infty}(N)$ really is just $C^{\infty}(M \times N)$, and in particular $C^{\infty}(\mathbb{R})$ really is a Hopf algebra in the category of smooth algebras.

Now a flow is a coaction $A \to C^{\infty}(\mathbb{R}) \otimes A$, and these exist for smooth compact manifolds but also for other smooth algebras, e.g. any embedding of a finite-dimensional real commutative algebra $A$ into a real matrix algebra gives it a smooth algebra structure and all derivations on such things exponentiate to smooth flows.