When does a derivation of a ring admit the exponential?
Exponentials of (real/complex) vector fields are classically defined via the vector field integration. Take a k-algebra k[x] \subset R\subset k[[x]], where k\supseteq \Q is a local domain. Suppose a derivation \xi is x-adically nilpotent. Define the exp-operator via the Taylor expansion, e^\xi:=\sum \frac{\xi^j}{j!}. It is a formal automorphism, e^\xi\in Aut_k(k[[x]]). When does e^\xi act on R? When does the formal power series e^\xi x\in k[[x]] belong to R? We address this question for the following rings. i. The algebraic power series, R=k\bl x\br, differentially finite (holonomic) power series, D(k[x]), and their higher versions, Picard-Vessiot extensions D^\bullet(k[x]), Picard-Vessiot closure D^\infty(k[x]), and differentially-algebraic power series D^{alg}(k[x]). ii. Power series over normed fields. In particular, power series with coefficients of controlled growth, e.g. analytic/Denjoy-Carleman/Gevrey classes. iii. Germs of smooth functions C^\infty(\R^n,o)/J, for arbitrary ideal J\subset C^\infty(\R^n,o). In case i. the operator e^\xi is transcendental, and the power series e^\xi x is ``usually" far from being algebraic. We give various criteria on e^\xi x to belong to k\bl x\br, D(k[x]), D(R), or D^{alg}(k[x]). In case ii. the answer is positive (i.e. e^\xi acts on R ) under rather weak assumptions on R. In case iii. the answer is ``totally negative". For any \xi\neq0 the operator e^\xi (defined as before) does not act on the quotients of the ring of germs of smooth functions, C^\infty(\R^n,o)/J.