Is there a notion of exponentiation that subsumes the well known versions, and in particular the versions on

  • tangent spaces (e.g., of Lie groups and Riemannian manifolds), in which the exponential map sends a vector to a point on a curve naturally defined in terms of the vector;
  • unital Banach algebras?

(NB. I am not conversant with category theory beyond the words "morphism" and "functor". But a categorically flavored answer that takes my limited knowledge base into account would be preferable. An internet search led me to the notion of a "Cartesian closed category", which doesn't seem to be the sort of thing I have in mind.)

  • $\begingroup$ Exponentiation can mean 1. the function $x \rightarrow \exp(x)$ or 2. the function $(a,b) \rightarrow a^b$. Cartesian closed categories are about the latter. $\endgroup$
    – Dan Piponi
    Jan 23, 2010 at 15:25
  • $\begingroup$ That's why it didn't seem like it to me. $\endgroup$ Jan 23, 2010 at 15:54

1 Answer 1


I think the exponential function for unital Banach algebras is a special case of the exponential functions for Lie groups modeled on topological vector spaces: the set of invertible elements in an unital Banach algebra naturally is a Banach Lie group, and the exponential function of this Lie group is the "classical" exponential function for the Banach algebra.

  • $\begingroup$ Nice, but if there is a group structure on generic (nice, finite-dimensional) Riemannian manifolds I don't know about it. $\endgroup$ Jan 26, 2010 at 21:50
  • $\begingroup$ @Hanno Becker @Steve Huntsman just a note: the group of invertible elements in a Banach algebra has natural structure of manifold modeled on the underlying Banach space, and I know Serge Lang's Foundamentals of Differential Geometry as the reference for Banach Manifolds. There the exponential map is associated to any spray on a Banach manifold. $\endgroup$
    – agt
    Apr 16, 2011 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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