For finite-dimensional Lie algebras, see this for a nice example, the exponential map is smooth and in particular, it is locally-Lipschitz onto its image. However, things are different when moving to the infinite-dimensional setting, as discussed for example in the answer to this post.

Let $G=\operatorname{Diff}_c(M)$ the space of compactly supported diffeomorphisms on $M$, say $M$ is Riemannian manifold diffeomorphic to $\mathbb{R}^k$, and let $\mathfrak{g}=C_c^{\infty}(M,M)$ be the $C^{\infty}$-vector fields on $M$. Again making use of comments in the answer to this post the exponential map taking a vector field $V$ in $\mathfrak{g}$ to its flow $\Phi^V:x\to x_1^x$ where $t\to x_t$ is well-defined by $$ \partial x_t^x = V(x_t^x), \, x_0^x=x . $$

  • Is this map continuous?
  • Moreover, when is it locally-Lipschitz, in the sense that, for every $\emptyset \subset K \subset M$ compact there exists some $L_K>0$ such that for every $U,V \in C^{\infty}_c(M,M)$ $$ \sup_{x \in K} d_M\left( \Phi^V(x),\Phi^U(x) \right)\leq L_K \sup_{x \in X} d_M\left( V(x),U(x) \right) $$ where $d_M$ is the metric induced by the Riemannian metric on $M$?

1 Answer 1


There is no need to be so specific in the question, you can indeed answer this in general for $M$ a (paracompact, finite-dimensional) manifold. It is well known, that in this setup $\mathrm{Diff}_c(M)$ is an infinite-dimensional Lie group with Lie algebra $\mathfrak{X}_c(M)$ (this was your space of $C^\infty_c(M,M)$. Before we continue, it needs to be mentioned that there are at least two settings in which the Lie group statement makes sense and is true:

  1. The convenient setting of global analysis (a la Kriegl and Michor, this is Theorem 43.1. in 1)
  2. The Bastiani setting of calculus (again Michor, Theorem 11.11 in 2)

In both settings the exponential is of course the flow map zou mention (this is said explicitely in the convenient calculus source 1, but NOT in the Bastiani source 2.

How does this relate to your question? Well $\exp$ is convenient smooth according to 1, but we can not use that! Convenient smooth mappings are in general NOT continuous with respect to the original topology. Note that this phenomenon happens only outside of Frechet spaces and some other nice classes of spaces. However, the model space of compactly supported vector fields is not one of the nice spaces, so we can not deduce it from this result. In the Bastiani setting I know several sources wich mention Bastiani smoothness of the exponential map (this is a consequence of regularity of the Lie group, here regularity means that the flow map for curves in the Lie algebra exists and is smooth, in your picture this means that you take the flow map of time dependent vector fields) but only feature proofs in more restricted settings (i.e. $M$ compact). To my knowledge a proof of the regularity of $\mathrm{Diff}(M)$ in the Bastiani setting for paracompact $M$ appeared for the first time in print in my thesis 3 (as a special case of the orbifold diffeomorphism group discussed there). Regularity of a Lie group (in the Bastiani setting) guarantees smoothness (in Bastiani sense) of the exponential map. Now as Bastiani smooth implies continuity, this answers your first question affirmatively.

First of all the second question seems to be off, or at least contain some non-trivial identification since the vector fields map into the tangent bundle and are still comared using the metric induced by the Riemannian metric on $M$. Anyway, a closer analysis of the problem should show that whatever you want is probably a finite-dimensional problem anyway. Therefore it might be misleading to think about your second property as Lipschitzness of the exponential.
To see this, note that the source of the differentiability of the exponential map is the so called exponential law (for function spaces, has nothing to do with Lie group exponentials). Here this translates roughly to the insight that you do not need to solve the defining equation on an infinite-dimensional manifold, but can place yourself in a situation where you are solving a differential equation on the underlying manifold $M$ (see e.g. 4 for more information on this topic). This is usually highly technical but since your question fixes the compact set, it should be doable (I would have to think more closely about how to do this though).


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