Reconciling some result about the exponential map, the Chow-Rashevskii theorem, and $\mathrm{Diff}_0(M)$

Question

Let $M$ be a $C^{\infty}$ manifold $C^{\infty}$-diffeomorphic to $\mathbb{R}^d$. I've recently come across some results which I'm trying to reconcile. Let $\mathfrak{X}(M)$ denote the set of Lipschitz vector fields on $M$ and $\exp:\mathfrak{X}(M)\rightarrow \mathrm{Homeo}_0(M)$ be the map taking a vector field to its integral curve (let's make similar notation for the $C^r$analogues when $r>0$).

• The Chow-Rashevskii Theorem says that the flow associated to a vector field satisfying the Hormander condition can attain any point $y \in \mathbb{R}^d$ from any other point $x \in \mathbb{R}^d$. In other words, there exists some $X \in \mathfrak{X}(M)$ such that $\exp(X)(x)=y$. Does this imply density in the topology of pointwise convergence (point-open)?

• Let $\{X_i\}_{i=1}^{\infty}$ be a collection of Lipschitz vector fields for which $\{X_j\}_{j=1}^i$ generate $d_i$ dimensional Lie sub-algebras $\mathfrak{g}_i$ of $\mathfrak{X}(M)$ (where $d_i<d_{i+1}$). This is a relaxation of the Hormander condition. Moreover, I ask that $\{X_i\}_{i=1}^{\infty}$ generates a dense linear subspace of $\mathfrak{X}(M)$. Moreover, $\exp|_{\mathfrak{g}_i}(\mathfrak{g}_i)$ defines a $d_i$-dimensional Lie subgroups of $Homeo_d(M)$ and therefore $ \cup_{i=1}^{\infty} \exp|_{\mathfrak{g}_i}(\mathfrak{g}_i) \subset Homeo_d(M), $ is "infinite dimensional". But when is its closure, in the compact-open topology, the entire space?

• The result referenced in this answer states that the group generated by $\exp(\mathfrak{X}(M)) $ in $\mathrm{Diff}(M)$, we denote it by $\langle \exp(\mathfrak{X}(M))\rangle$, is $\mathrm{Diff}_0(M)$ the identity component therein. How can this be reconciled against this paper this result which shows that $\exp(\mathfrak{X}(M))$ is meager in the $C^1$ topology on $\mathrm{Diff}^1(M)$? I.e.: $$ \langle \exp(\mathfrak{X}(M))\rangle - \exp(\mathfrak{X}(M)), $$ is topologically non-trivial.

Is there any topology stronger than pointwise convergence for which $\exp(\mathfrak{X}(M))$ is dense in $\langle \exp(\mathfrak{X}(M))\rangle$?

Answer 1

This was a bit long for a comment, thus I post it as an answer.

First of all, you have to be very careful with what you actually mean by the exponential here. The flow map to time 1 does NOT exist in your setting, not even on an arbitrarily small zero-neighborhood. The reason for this is that you assume your manifold $M$ to be diffeomorphic to euclidean space and thus it is non-compact. For every non-compact manifold, it is well known that one can construct smooth vector fields whose integral curves explode before reaching time 1. Thus the mapping is ill-defined on all Lipschitz vector fields. The usual remedy for non-compact manifolds is to pass to the space $\mathfrak{X}_c (M)$ of compactly supported vector fields (of your favourite regularity). There the flow can be defined (obviously if $M$ is compact nothing goes wrong).

Concerning your last point: I think one should be careful here, at least when it comes to the notation. Your $\mathfrak{X}(M)$ is the set of all Lipschitz vector fields and by the exponential you mean the flow map, whereas in the references you gave, the same symbol means the set of all smooth vector fields (and the exponential also the flow map at time one). This difference might already be an essential piece of the puzzle. On the other hand, it is well known that in the smooth setting the image of the exponential is not an open neighborhood of the identity diffeomorphism (the strongest result is due to Grabowski who showed that one can approach the identity with continuous curves who intersect the image of the exponential only in the identity). This is an essential point in infinite-dimensional Lie theory as it shows that contrary to the finite-dimensional setting, there are infinite-dimensional Lie groups such that the Lie group exponential is not a local diffeomorphism.

Coming back to your question: This shows that the image of the exponential is topologically speaking a quite complicated subset of all diffeomorphisms. However, it generates the whole component of the identity (as pointed out by your references). This means that there are diffeomorphisms in this component which can not be reached by the exponential but only approximated arbitrarily well by (arbitrarily high) products of exponentials. Note: As pointed out by the OP in general the statement referenced gives one only that the finite products are dense [more information on this can be found in Banyagas book, referenced in one of the links of the post above]

Since the group is not locally exponential this is not even locally a contradiction near the unit . This is a new infinite-dimensional phenomenon as in finite-dimensions you could just take the logarithm if you are near the identity.

Answer 2

To answer one of your questions:

In this example you have a topological group $G$, a meager subset $S\subset G$ such that $S$ generates the entire $G$ (as an abstract group, not just as a topological group). Nothing wrong with this. For an easier example of this phenomenon consider the additive group $G={\mathbb R}$ and a Cantor subset $S\subset G$ of positive measure. By the Steinhaus theorem, $S+S$ has nonempty interior, hence, $S$ generates $G$.