Does there exist an infinite dimensional Lie group $G$, with Lie algebra $\mathfrak g$, such that the exponential map $exp:\mathfrak g \to G$ is not defined?
If so, can one provide an example of such a group.
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Sign up to join this communityDoes there exist an infinite dimensional Lie group $G$, with Lie algebra $\mathfrak g$, such that the exponential map $exp:\mathfrak g \to G$ is not defined?
If so, can one provide an example of such a group.
Sure. Let $G$ be the group of diffeomorphisms of the real line. Then $\frak{g}$ is the Lie algebra of vector fields on the real line. However, many vector fields on the real line cannot be exponentiated to a $1$-parameter subgroup of $G$. For example, $$ X = x^2\frac{\partial\ \ }{\partial x} $$ is not tangent to any $1$-parameter subgroup of $G$.
About Robert's example $Diff(\mathbb R)$ one can argue that it is not a Lie group, since it does not admit charts: For the compact $C^\infty$-topology it is not open in $C^\infty(\mathbb R)$. In the Whitney $C^\infty$-topology it is not locally contractible.
Aside: A setting where $Diff(\mathbb R)$ is a Lie group in the category of manifolds based on smooth curves etc (where the finite dimensional ones are exactly the known ones, as are the Banach ones) is: Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178.(pdf of I) (pdf of II)
Definition of regular Lie groups:
We consider a smooth Lie group $G$ with Lie algebra $\mathfrak g=T_eG$ modelled on convenient vector spaces. The notion of a regular Lie group is originally due to Omori and collaborators (see [Omori Maeda Yoshioka 1982], [Omori Maeda Yoshioka 1983]) for Frechet Lie groups, was weakened and made more transparent by [Milnor 1984] and carried over to convenient Lie groups in (here), see also 38.4 of (here). A Lie group $G$ is called regular if the following holds:
$\bullet$ For each smooth curve $X\in C^{\infty}(\mathbb R,\mathfrak g)$ there exists a curve $g\in C^{\infty}(\mathbb R,G)$ whose right logarithmic derivative is $X$, i.e., $$ g(0) = e, \qquad \partial_t g(t) = T_e(\mu^{g(t)})X(t) = X(t).g(t),\quad\text{where } \mu(a,b)=\mu_a(b)=\mu^b(a) = a.b. $$ The curve $g$ is uniquely determined by its initial value $g(0)$, if it exists.
$\bullet$ Put $\operatorname{evol}^r_G(X)=g(1)$ where $g$ is the unique solution required above. Then $\operatorname{evol}^r_G: C^{\infty}(\mathbb R,\mathfrak g)\to G$ is required to be $C^{\infty}$ also.
Note that for $X$ constant in time, $\operatorname{evol}^r_G(X)=\exp(X)$. So each regular Lie group admits an exponential mapping.
The family of regular Lie groups is remarkably stable under constructions like extensions and quotients.
I do not know a Lie group modeled on convenient vector spaces which is not regular.
A quasi-counter-example is due to Wuestner: Consider the space of trigonometric rational functions on $S^1$ which are everywhere positive and have no pole, with multiplication. This is not regular since $\exp(X)=e^X$ is real analytic and not trigonometric rational any more. But the modelling space is not convenient, since it is not Mackey sequentially complete for any suitable topology.