Ind-Frechet manifolds? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:44:19Z http://mathoverflow.net/feeds/question/79256 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79256/ind-frechet-manifolds Ind-Frechet manifolds? David Roberts 2011-10-27T11:31:05Z 2011-10-28T06:56:08Z <p><strong>Short version:</strong> has anyone done geometry on something that is the formal filtered colimit of Frechet manifolds?</p> <p><strong>Longer version:</strong> A colleague and I came up with a concept today that seems like we require generalising smooth Frechet manifolds to something that is a filtered colimit of such manifolds: Ind-Frechet manifolds if you will.</p> <p>A fragment of this idea is that we need to consider a space of smooth paths $I \to M$ in a smooth manifold (poss. compact) with a finite number discontinuities (poss. zero). As a topological space one can simply take the (filtered) colimit over the spaces of paths with discontinuities at <em>specified number</em> of points in $I$ (this should be a Frechet manifold). The result is a space over the infinite simplex (i.e. $colim_n \Delta^n$). But there is no guarantee that this colimit exists in the category of Frechet manifolds and smooth maps. </p> <p>We really need this to be something like a smooth space, so we can discuss connections and the such like. I'm aware that we could use something like diffeological spaces or Chen spaces or the like, but since people talk about ind-schemes, I wondered if there was something similar for manifolds.</p> http://mathoverflow.net/questions/79256/ind-frechet-manifolds/79263#79263 Answer by Andrew Stacey for Ind-Frechet manifolds? Andrew Stacey 2011-10-27T13:37:44Z 2011-10-28T06:56:08Z <p><strong>Short version:</strong> Yes, me.</p> <p><strong>Slightly Longer Version:</strong> </p> <p><em>Added in Edit</em>: What I write below is concerned with the space of <em>continuous piecewise-smooth</em> paths or loops. That is, continuous maps which are smooth except on a finite subset of the interval (or circle). Dropping the continuity requirement is only a cosmetic change: all the regularity statements shift down by one degree. Indeed, differentiation is an isomorphism from <em>continuous and piecewise-smooth</em> to just <em>piecewise-smooth</em> (providing you've chosen your end-point conditions compatibly). In particular, the analysis in the linked paper will readily adapt to this case.</p> <hr> <p>There <em>is</em> a manifold of piecewise-smooth loops or paths, but it isn't pretty. It's just about as bad as you can get and still be called a manifold.</p> <p>The first thing to decide is exactly what you mean by "piecewise-continuous smooth". The basic question is as to what you want a smooth map $[0,1] \to \mathbb{R}$ to be. Should it be smooth on $(0,1)$ and continuous on $[0,1]$ or should the derivatives exist at the end points?</p> <p>If the first ("open"), then you're in for a nasty shock. The space of piecewise-smooth paths in $\mathbb{R}^n$ when considered as a locally convex topological vector space turns out to be a topological subspace of the space of continuous paths in $\mathbb{R}^n$. That's right: <em>topological</em> subspace. So all that information that you thought you had about derivatives is thrown away: you're using the $\|-\|_\infty$ norm. This means that any manifold structure that you are looking at is as a subspace of the Banach manifold of continuous paths (or loops). So since completeness is quite important, you should really work with continuous paths.</p> <p>So then we look at the second ("closed"). In this case, the situation looks a bit better. What happens here is that adding additional breakpoints plays nicely: the space with breakpoints at ${t_1,...,t_n}$ is a closed subspace of the space with breakpoints at ${t_1,...,t_n,t_{n+1}}$. The difficulty here is that the family that you are taking the colimit over is uncountable (all finite subsets of the interval or circle) which means that all the "nice" theorems about inductive limits of LCTVS's don't (necessarily) hold.</p> <p>Nonetheless, the colimit is a locally convex topological vector space and the space of piecewise-smooth loops or paths in an arbitrary manifold is a smooth manifold modelled on it. But that's the best you can say. I've not been able to show yet that it is paracompact, nor even that it is smoothly regular (like Tychanoff but with smooth functions).</p> <p>However, there's a worse problem with this manifold. Let's take piecewise-smooth loops. Then the obvious property that you'd like is for the circle to act nicely on this space. Any given circle element acts by translation and this is a diffeomorphism. However, the assignment $\lambda \to \tau_\lambda$ is not continuous. The image of the circle is discrete. In fact, the image of the diffeomorphism group of the circle acting by reparametrisation is totally disconnected. So that's pretty bad as it means that all the standard reparametrisation homotopies don't work <em>as is</em>. They only work by homotoping the whole space to the space of smooth loops and working there - in which case you should probably work with smooth loops in the first place.</p> <p>If you only allow breaks at <em>rational</em> points then life is a bit better in terms of the manifold's structure (it admits smooth partitions of unity, for example) but the circle action is still bad.</p> <p>Fixing the circle action requires adding in lots more loops, right up to the space of differentiable loops with derivative of bounded variation. So that's almost as bad as the continuous loops that we had earlier.</p> <p>Conclusion: either way, it's a Bad Space.</p> <p><strong>Long Version:</strong> Read <a href="http://www.math.ntnu.no/~stacey/Research/Preprints/piecewise.html" rel="nofollow">The Smooth Structure of the Space of Piecewise-Smooth Loops</a>.</p>