Consider the following claim:

Let $p:M \to N$ be a (surjective) submersion of finite-dimensional smooth manifolds. Let $J$ denote one of $[0,1],\ [0,1),\ (0,1]$. Then $p_*:M^J \to N^J$ is a submersion of Frechet manifolds, where $X^J$ denotes the usual manifold of smooth paths in $X$.

The intuition is that given a lift of a path $\gamma$ through $p$, one can find a neighbourhood $U$ of the image of $\gamma$ and a neighbourhood $V$ of the image of the lift such that for every path in $U$ one can smoothly choose a lift to $V$.

Here I suppose we need a submersion of Frechet manifolds to be a map that admits local sections through every point in the domain, if that is the 'correct' notion of submersion in that setting (certainly not the 'surjective on tangent spaces' version).

I think the proof would use the characterisation of submersions as maps which look locally (on both the domain and codomain) like projections $U \times \mathbb{R}^n \to U$, and the existence of good open covers with smooth contractions.

I think I'm able to prove that there are continuous sections through every point in the domain, thinking of everything as a topological space, and using the compact-open topology on the mapping spaces. But I don't know off the top of my head that the compact-open topology on the space of smooth paths is the same as the topology inherited from the Frechet manifold structure. (My guess is that it is.)

My question: is the claim true?

As Andrew Stacey points out in the comments, the mapping space is not a manifold for non-compact intervals. However, I think I really only need maps which have all derivatives uniformly bounded (but a different bound for each derivative!). Since the topology on the mapping space for compact intervals uses uniform convergence, I'm betting that this set has the structure of a Frechet manifold.

Question 2: am I right?

Question 1': if so, is the claim true for this (putative) map of Frechet manifolds?

  • 4
    $\begingroup$ Two technical comments: 1. that $M^J$ is a Frechet manifold holds only for $[0,1]$ (of your candidates, in general you need some form of compactness). 2. If by the "compact-open topology" you mean uniform convergence of arbitrary (but finite) derivatives on compact sets then, yes, that is the same as the manifold topology. $\endgroup$ Commented Jan 18, 2012 at 9:18
  • $\begingroup$ Do we know $X^J$ is any sort of manifold for (half-)open intervals? $\endgroup$
    – David Roberts
    Commented Jan 18, 2012 at 10:50
  • 3
    $\begingroup$ It isn't. You must have (sequential) compactness. The non-compact end can wander all over the manifold and that destroys the possibility of any sort of nice local structure. In the compact-open topology then locally, I cannot know anything about the open end because I can only know what's happening on a compact set. So two nearby paths need not be physically near at all. You can change the topology to get a manifold, but then you get uncountably many components (according to what the open end is doing). $\endgroup$ Commented Jan 18, 2012 at 11:04
  • $\begingroup$ This might be too trivial, but if $M$ is either a disjoint union of open sets of $N$, or a principal bundle for a Lie group, then $p_{*}$ is a submersion. $\endgroup$ Commented Jan 18, 2012 at 14:15
  • $\begingroup$ Actually, I think I'm only interested maps from the open interval which are uniformly smooth, in the sense that all derivatives are bounded. Is there a Frechet manifold structure on the set of these? $\endgroup$
    – David Roberts
    Commented Jan 18, 2012 at 22:03

3 Answers 3


This is answered affirmatively in Yet More Smooth Mapping Spaces and Their Smoothly Local Properties. Specifically:

Theorem 1.1

Let $M$ be a finite dimensional smooth manifold. Let $S$ be a Frölicher space with the property that there is a non-zero smooth function $C^\infty(S,\mathbb{R}) → \mathbb{R}$ with support in $C^\infty(S,(-1,1))$. Then $C^\infty(S,M)$ is a smooth manifold which is locally modelled on its kinematic tangent spaces.

Suppose, in addition, that $N$ is another finite dimensional smooth manifold and $f \colon M \to N$ a regular smooth map. Then $C^\infty(S,f) \colon C^\infty(S,M) \to C^\infty(S,N)$ is a regular smooth map.

  • $\begingroup$ Thanks, Andrew. It's been a journey and half to get this answer :-) $\endgroup$
    – David Roberts
    Commented Jun 7, 2013 at 1:24

I think I can answer my own question:

First, the issue of the (half-)open intervals versus closed intervals. There is a Frechet space of sections of a vector bundle over a manifold $X$ (not necessarily compact!), where we require that the sections have bounded derivatives of all order. Then the usual family of seminorms (sums of sups over $X$ of norms of derivatives up to order $n$) is well defined. Then we can consider the set $M^{J,b}$ of paths (using $J$ as above) with bounded derivatives (which is all paths in the case of a compact interval), and give local charts which are of the form $(\mathbb{R}^m)^{J,b} \simeq \Gamma_b(J,\gamma^\ast TM)$, sections (with bounded derivative) of the trivialisable vector bundle $\gamma^\ast TM \to J$. The usual arguments should (I haven't check all the details, but there don't seem like any obstacles) give the result that $M^{J,b}$ is a Frechet manifold, which reduces to the usual manifold of paths in the case that $J$ is a closed interval.

So in what follows we will only consider paths and sections both with bounded derivatives, but will drop the 'with bounded derivatives' for brevity.

Let $f:M \to N$ be a surjective submersion. There exists an open cover $U=\coprod_I U_i$ of $M$ such that $U_i = \mathbb{R}^m \times V_i$ for an open cover $V = \coprod_I V_i$ of $N$ and the induced map $U \to V$ is projection on the second factor. Thus the induced map on tangent bundles $TU \to TV$ is split.

Let $\gamma:J \to N$ be a path and $\gamma':J\to M$ be a lift through $f$. Then the induced map $f_\ast$ restricts to a map on charts $$ \Gamma_b(J,\gamma'^\ast TM) \to \Gamma_b(J,\gamma^\ast TN) $$ The local splitting of the map of tangent bundles $TM \to TN$ gives rise to a local splitting of the map (over $J$!) of tangent bundles.

Now consider a pair of vector bundles $E \to J$, $F\to J$ (the following argument works over any manifold, and lets us generalise the result from $J$ to an arbitrary finite dim, paracompact manifold), a map of vector bundles $E\to F$, and an open (numerable) cover $c:U\to J$ such that $c^\ast E \to c^\ast F$ is a split map of vector bundles over $U$. Then the following chain of maps gives us a section of the map $\Gamma_b(J,E) \to \Gamma_b(J,F)$ of Frechet spaces: $$ \Gamma_b(J,F) \stackrel{c^\ast}{\to} \Gamma_b(U,c^\ast F) \stackrel{\sigma}{\to} \Gamma_b(U,c^\ast E) \stackrel{\sum\phi\cdot}{\to} \Gamma_b(J,E) $$ where $\sigma$ is a splitting of $c^\ast E \to c^\ast F$, $\phi$ is a partition of unity subordinate to $U$ and the last map takes a family of sections $s_i$ and sends it to $\sum_I \phi_i\cdot s_i$.

Thus there are charts of the path spaces such that the map in question, $f_\ast:M^{J,b} \to N^{J,b}$ looks like a split projection when restricted to those charts, hence is a submersion if we take the definition given in Hamilton's "The inverse function theorem of Nash and Moser".

In particular, for the case when $J$ is a compact interval we have a submersion of the usual path manifolds.

  • $\begingroup$ Minor comment on those derivatives: if derivatives of all orders are bounded then derivatives of all orders exist. So your path space is just the path space with domain a closed interval. $\endgroup$ Commented Jan 23, 2012 at 13:36
  • $\begingroup$ hmm, ok. But ignoring the issue of open paths, and just considering the usual path space, does this look alright? $\endgroup$
    – David Roberts
    Commented Jan 23, 2012 at 22:32
  • $\begingroup$ It looks a little over-complicated, to be honest! Your splitting map needs to vary nicely with $\gamma$, so I don't like having to choose an open cover of $J$. But I don't see why you need that. The splitting of $\gamma^* T M \to \gamma^* T N$ tells you that it splits. You can even get a nicely varying splitting map by splitting $T M \to T N$ globally - use a metric on $M$ and take orthogonal complements. (ctd) $\endgroup$ Commented Jan 24, 2012 at 11:15
  • $\begingroup$ (ctd) However, I disagree that the map $f_*$ restricts to the given map on charts. The map $f_*$ is given by $\gamma \mapsto f \circ \gamma$ whereas the chart map is given by $\alpha \mapsto T f \circ \alpha$. These need not agree - one depends on $f$ in a neighbourhood of $\gamma$ whilst the other only depends on the jet of $f$ at $\gamma$ - and will only agree if the chart map is chosen extremely carefully. I believe that the chart map can be chosen that carefully, but I need a little time to work out the details. So I believe the result, but not for this reason. $\endgroup$ Commented Jan 24, 2012 at 11:17
  • $\begingroup$ Incidentally, on the open/closed issue on paths. You do need the paths themselves to be extendible to the closed interval. Otherwise no matter what bounds you put on the derivatives, it will not be a manifold. And once it extends continuously to the closed interval then my remark about derivatives of all orders existing will apply, and by being extendible to a closed interval everything happens in a compact subset of the target manifold so varying metrics produces equivalent conditions. $\endgroup$ Commented Jan 24, 2012 at 11:19

A different version of this result, inspired by Andrew's paper, is Lemma 2.4 in

  • Habib Amiri, Alexander Schmeding A differentiable monoid of smooth maps on Lie groupoids, Journal of Lie Theory 29 (2019), No. 4, 1167–1192, arXiv:1706.04816.

It uses more classical methods of global analysis and differential geometry, rather than generalised smooth spaces and smoothly-local methods.

  • $\begingroup$ Lets mention that due to the sources of inspiration, Lemma 2.4 is also known as the Stacey-Roberts Lemma. $\endgroup$ Commented Aug 7, 2020 at 5:20
  • $\begingroup$ @AlexanderSchmeding modesty forbids... $\endgroup$
    – David Roberts
    Commented Aug 7, 2020 at 6:11

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