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.

  • 1
    $\begingroup$ For Banach Lie groups (see e.g. encyclopediaofmath.org/index.php/Lie_group,_Banach ) the finite-dimensional argument that defines the exponential map goes through without difficulty to the infinite dimensional case. Most interesting examples of infinite-dimensional Lie groups, though, are not Banach. $\endgroup$ – Terry Tao Feb 12 '13 at 21:48

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 $G$.

  • 1
    $\begingroup$ @Robert: can you please explain how the group of diffeomorphisms of the real lie is a Lie group (Which topology do you use? What is topological the vector space on which it is modeled? What is a chart near the identity element?), and can you also please explain why the Lie algebra is what you claim it is. $\endgroup$ – André Henriques Feb 12 '13 at 14:45
  • 2
    $\begingroup$ @André: It's a Lie group because it's the kind of group that Sophus Lie himself treated. Moreover, Cartan, Engel, and many others called such groups "groupes de Lie infinis" and explicitly listed its "Lie algebra of infinitesimal transformations" as the Lie algebra of vector fields on the line. Of course I know that modern treatments have introduced more restrictive definitions, requiring topologies and other properties, etc., but I was not referring to any of these, and the OP did not specify that he wanted a "regular Lie group" or any other such modern notion, so I didn't try to guess. $\endgroup$ – Robert Bryant Feb 12 '13 at 15:45

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.

  • 1
    $\begingroup$ @Peter: I suppose you can argue that the modern notion of Lie group should be whittled down to fit the Procrustean bed of modern mathematics, but I think that something is lost when you throw out the examples that motivated the theory in the first place. Certainly, if you forbid yourself to read Lie and Cartan, you will lose a lot. $\endgroup$ – Robert Bryant Feb 12 '13 at 17:59
  • $\begingroup$ @Robert. In fact I share your sentiment and I upvoted it. In my "aside" I described a an attempt to include the full diffeomorphism groups. I just wanted to point out that in the class of Lie groups with charts, up to now nobody found an example without exponential map. $\endgroup$ – Peter Michor Feb 17 '13 at 16:42

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.