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?

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.

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    $\begingroup$ There is some typo in "1'. Algebra $C^\infty(M)$ of smooth functions on a compact manifold without boundary". Presumably you mean "non-compact"? In any case, nice question overall. $\endgroup$ – Theo Johnson-Freyd Sep 23 '10 at 16:32
  • $\begingroup$ @Theo Johnson-Freyd Yes, "non-compact". Corrected. Thank you. $\endgroup$ – Fiktor Sep 23 '10 at 17:27
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    $\begingroup$ I may have misunderstood your question; but if I recall correctly there are commutative algebras A which admit no non-zero derivation from A to A -- in fact every semisimple Banach algebra has this property, this is Marc Thomas' improvement on the Singer-Wermer theorem. (Note that I'm not assuming the derivation is continuous.) I suspect that if A has no non-zero derivations, then every derivation will be a flow -- but this is somehow not the kind of example you are looking for... $\endgroup$ – Yemon Choi Sep 23 '10 at 19:09
  • $\begingroup$ Formal power series rings? Rings of analytic functions? $\endgroup$ – Tom Goodwillie Sep 24 '10 at 2:03
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    $\begingroup$ But surely the usual flow associated to a smooth tangent vector field on a smooth closed manifold does not satisfy the condition written down in the question for an <i>arbitrary</i> $\mathbb R$-linear map $h:A\to\mathbb R$. $\endgroup$ – Tom Goodwillie Sep 24 '10 at 2:18

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.

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  • $\begingroup$ Yes, you are right, I've mistaken with that. I will think, how to correct it. $\endgroup$ – Fiktor Sep 24 '10 at 17:51

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).

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    $\begingroup$ "$\xi^n=0$" is wrong. Consider $\xi=xQ(x)\frac{\partial}{\partial x}$. It is well defined. In order to show this, we should ensure that $[\xi P]$ does not depend on the polynomial $P$ from equivalence class $[P]\in\mathbb{R}/x^n$. We can check this for $[P]=0$. In this case P is multiple of $x^n$ then $\xi P$ is also multiple of $x^n$, so $[\xi P]=0$. If, for example, $Q(x)=1$, then $\xi^n(x)=x\neq 0$. However from your approach I see, that all derivations in finite-dimensional algebras possess a flow, because for matrices $\xi$ the power serie you wrote converges. $\endgroup$ – Fiktor Sep 23 '10 at 14:33
  • $\begingroup$ Note that $\frac\partial{\partial x}$ is not a derivation of $\mathbb R[x]/x^n$. It would be the derivation that sends $x\mapsto1$ if it exists, and this determines the derivation, but then $0=x^n\mapsto nx^{n-1}\neq0$ by the Leibniz rule. On the other hand, as fiktor points out, $x\frac\partial{\partial x}$, which is the derivation extending $x\mapsto x$, does exists: it acts on polynomials by $x^k\mapsto kx^k$. (Under $x=e^z$, $x\frac\partial{\partial x}=\frac\partial{\partial z}$.) All the derivations of $\mathbb R[x]/x^n$ are of the form $p(x)\frac\partial{\partial x}$ for $p(0)=0$. $\endgroup$ – Theo Johnson-Freyd Sep 23 '10 at 16:41
  • $\begingroup$ Also, I object to the idea that $\mathbb R[x]/x^n$ "contain[s] no geometric information". It is precisely the ring of $\mathbb R$-valued functions on an order-$n$ formal neighborhood of a point in a line. In particular, the "Taylor series" map $\mathcal C^\infty(\mathbb R) \to \mathbb R[x]/x^n$ is the restriction of a smooth function to a very small formal neighborhood. The restriction remembers precisely the first $n$ derivatives of the function: this is definitely "geometric" information. $\endgroup$ – Theo Johnson-Freyd Sep 23 '10 at 16:44
  • $\begingroup$ Fine, `no geometric information' was needlessly severe. I meant that it wasn't a ring of real-valued functions on some space, which seemed to be the motivation behind to examples given; particularly the choice to use R instead of C. $\endgroup$ – Greg Muller Sep 23 '10 at 18:24
  • $\begingroup$ @Greg Muller I care about this kind of algebras too. I can explain, why. I'm reading a course about algebraic approach to Hamiltonian mechanics. All definitions are given there for general commutative (and supercommutative) algebras. The classical Hamiltonian mechanics appears here as the case $A=C^\infty(M)$ and I'm interested in other useful and/or interesting cases. The problem in the question is the most important problem with many of them. Although finite-dimensional algebras are not so important for the goal of the course, they are good for simple examples and exercises. $\endgroup$ – Fiktor Sep 23 '10 at 19:00

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.

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