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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 $\langle \exp(\mathfrak{X}(M))\rangle$?

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    $\begingroup$ I don't know about the core of your question, but for a single vector field to satisfy the Hömander condition, you must have $d=1$. The Chow-Rashhevskii theorem, in your terms, says that given $X_1,\ldots,X_k$ satisfying the Hörmander condition (if they are not smooth I am not sure what it means), the sub-semigroup of $\mathrm{Homeo}_0(M)$ generated by $\exp(\langle X_1,\ldots,X_k\rangle)$ acts transitively on $M$. And you'll need another argument to say that this sub-semigroup is dense in the pointwise convergence topology, so I'm not sure your first point holds. $\endgroup$ – Pierre PC Apr 22 at 20:07
  • $\begingroup$ The fact that for any $x,y\in M$, there exists $X\in\mathfrak X(M)$ such that $\exp(X)(x)=y$ is elementary though, more than CR theorem itself. But I am not sure how to go from here to the density. $\endgroup$ – Pierre PC Apr 22 at 20:18
  • $\begingroup$ @PierrePC I added some thoughts but no clear conclusion, just some reasons I have to question the possibility of getting density. $\endgroup$ – James_T Apr 23 at 11:14
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    $\begingroup$ The answer to the question "how can this be reconciled.." is that if you have a meagre subset $S$ of a topological group $G$, the subset $M^2$ consisting of products of elements of $M$ need not be meagre. This is what happens in this case when you compose elements of $exp({\mathfrak X}(M))$. $\endgroup$ – Moishe Kohan Apr 23 at 22:42
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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.

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  • $\begingroup$ I do have a question about your last point. The fact that the entire identity component is "generated by the image of $\exp$" means that all (multiple-fold) possible compositions of $\exp$ diffeomorphisms generate the component? Or can we only take the smallest group containing all those combinations (I mean are there some diffeomorphisms in the identity component which can never be obtained by composing diffeomorphisms in the image of exp)? $\endgroup$ – James_T Apr 23 at 18:13
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    $\begingroup$ AH crap, I was thinking algebraic generating set here. So in general for topological groups the definition is that the finite products and inverses are in general only dense. To find out what is really happening, one should reference the proof in Banyagas book. It might be that you get something stronger, but it has been some years since I last looked into the book. So indeed my third statement was probably too strong $\endgroup$ – Alexander Schmeding Apr 23 at 19:02
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    $\begingroup$ The theorem of Thurston has nothing to do with being dense. It literally says that time-1 maps of flows generate the identity component of the diffeomorphism group of a compact manifold, as an abstract group (for noncompact manifolds, his theorem applies to compactly supported diffeomorphisms). The proof does not make use of approximation arguments (at least not directly). $\endgroup$ – Andy Putman Apr 23 at 22:40
  • $\begingroup$ @AndyPutman: thanks for clarifying that! $\endgroup$ – Alexander Schmeding Apr 24 at 7:02
  • $\begingroup$ @AlexanderSchmeding In your answer you state that the exponential map in the $C^k$ setting isn't subjective. However, is it known to be surjective/continuous in the $C^0$ setting (with compact supported)? $\endgroup$ – James_T May 7 at 13:07
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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 $G$.

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