Induced map on path manifolds: is it a submersion? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:39:35Z http://mathoverflow.net/feeds/question/85960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85960/induced-map-on-path-manifolds-is-it-a-submersion Induced map on path manifolds: is it a submersion? David Roberts 2012-01-18T05:01:32Z 2013-06-06T12:43:06Z <p>Consider the following claim: </p> <blockquote> <p>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 $X$.</p> </blockquote> <p>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 $V$.</p> <p>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).</p> <p>I think the proof would use the characterisation of submersions as maps which look locally (on both the <em>domain</em> and codomain) like projections $U \times \mathbb{R}^n \to U$, and the existence of good open covers with smooth contractions.</p> <p>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.)</p> <p>My question: is the claim true?</p> <hr> <p>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.</p> <p>Question 2: am I right?</p> <p>Question 1': if so, is the claim true for this (putative) map of Frechet manifolds?</p> http://mathoverflow.net/questions/85960/induced-map-on-path-manifolds-is-it-a-submersion/86416#86416 Answer by David Roberts for Induced map on path manifolds: is it a submersion? David Roberts 2012-01-23T01:18:00Z 2012-01-23T01:18:00Z <p>I think I can answer my own question:</p> <p>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.</p> <p>So in what follows we will only consider paths and sections both with bounded derivatives, but will drop the 'with bounded derivatives' for brevity.</p> <p>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.</p> <p>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.</p> <p>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 $\sum_I \phi_i\cdot s_i$.</p> <p>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".</p> <p>In particular, for the case when $J$ is a compact interval we have a submersion of the usual path manifolds.</p> http://mathoverflow.net/questions/85960/induced-map-on-path-manifolds-is-it-a-submersion/132937#132937 Answer by Andrew Stacey for Induced map on path manifolds: is it a submersion? Andrew Stacey 2013-06-06T12:43:06Z 2013-06-06T12:43:06Z <p>This is answered affirmatively in <a href="http://arxiv.org/abs/1301.5493" rel="nofollow">Yet More Smooth Mapping Spaces and Their Smoothly Local Properties</a>. Specifically:</p> <h3>Theorem 1.1</h3> <p>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.</p> <p>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.</p>