# David Roberts which smooth paths do I use

This page is where we will work out what manifold of smooth paths in a smooth manifold has smooth operations of concatenation, inclusion of constant paths, evaluation and possibly exponentiation for $\left[0,1\right]$.

Andrew Stacey: Right, so my first question is this: how much do you want your space of paths to be a manifold? You say above “what manifold of smooth paths” but I want to know at the outset how much you want actual manifolds.

It’s a question of approach. Either we can start by saying “We want manifolds!” and stick with it throughout, which will probably mean that we have to be creative with some of the other pieces of the structure, or we can say “Well, smooth space is enough” in which case we may well get a nice way of phrasing all of the rest of the structure, but the basic objects won’t be so nice. I think both could be interesting, but I’d like to choose one at the outset and see how far it can be pushed rather than change halfway through.

My - mild - preference is for an actual smooth manifold. I suspect that at some point the extra structure will be useful and it’d be nice to have it all the way through. Also, you’re starting with a smooth manifold at the base so there are smooth manifolds of paths around and even though we know what a “smooth space” is, it’s still nice to have nice objects when we can.

DR: I would also like an actual manifold - going to a smooth space is interesting, but people will respond better to objects that are more familiar.

AS: Great! Okay, so the next thing is for me to understand all the structure that you want to build. I guess the topological case is the first port of call.

DR: For which, see here.

AS: I’ve started on that, thanks.

This may or may not be relevant, but I thought I’d mention the parallel between what you’re doing and the basic constructions of string topology. There, they have the figure 8 and set it in between maps of the circle and maps of two circles. So there are maps:

$Map\left(\mathrm{OO},M\right)←Map\left(8,M\right)\to Map\left(O,M\right)$\operatorname{Map}(OO,M) \leftarrow \operatorname{Map}(8,M) \rightarrow \operatorname{Map}(O,M)

The question is: what is the smooth structure of 8? Is it two circles that coincide at a point, is it one circle that’s been pinched, or something even more complicated? (Insert brilliant pictures illustrating the choices here!)

The string topologists wish to “invert” one of those maps. They can do so if it has finite codimension (okay, okay, plus some other conditions). That determines how they think of the 8: it’s two circles that touch because they want to invert the first map.

So some of the actual pieces may well depend on what you want to do as well as what it should look like.

### Options

• Paths with sitting instants

• Any other…

DR: A quick question on smooth paths: what is the difference between paths with sitting instants and paths with all derivatives vanishing at endpoints? Clearly the first lot form a subspace of the other. Is it a submanifold? (actually I suspect they are dense) Are the latter well behaved, such as being a closed submanifold of all smooth paths?

AS: It gets a little confusing because there are a variety of topologies that could be chosen. It is possible to topologise paths-with-sitting-instants using the colimit topology over the sitting instant. That will be a different topology to that induced by the space of paths that are flat at the end points. My instinct is that paths-flat-at-end-points will be a smooth manifold (and a submanifold of all smooth paths) but that paths-with-sitting-instants will not. That’s only instinct, however. The quickest route to the former would be to show that the conditions in my paper “Constructing Smooth Manifolds of Loop Spaces” hold (the loop/path distinction is irrelevant). The quickest route to the latter would be to show that the associated linear space (paths in ${ℝ}^{n}$) is not a LCTVS - that should be easy, the colimit topology on a union is rarely locally convex and usually needs convexifying (that’s the key to the proof that one way of talking about piecewise smooth loops leads to something a long way from what would be expected: namely that the colimit LCTVS topology on piecewise smooth loops is the subspace topology from continuous loops!).

### Operations

• Concatenation: ${X}^{I}{×}_{X}{X}^{I}\to {X}^{I}$
• Evaluation: ${\mathrm{ev}}_{0},{\mathrm{ev}}_{1}:{X}^{I}\to X$
• Constant paths: $X\to {X}^{I}$

### Other requirements

• Exponentiation? $I\to {X}^{I}$ versus $*\to {X}^{{I}^{2}}$
• We also need the evaluation maps ${\mathrm{ev}}_{0},{\mathrm{ev}}_{1}:{X}^{I}\to X$ to be submersions, or at least the pullback of ${\mathrm{ev}}_{0}$ along ${\mathrm{ev}}_{1}$ to exist.

### The Naive Structure

AS: Right, I’m going to have a go at defining the structure and see how far I get. I’ll base it on definition 5.21 in your thesis as bigroupoid doesn’t have any diagrams.

• Starting point: a manifold $M$.

• So we take ${B}_{0}=M$

• The category $\underline{{B}_{1}}$:

• $Obj\left(\underline{{B}_{1}}\right)={C}^{\infty }\left(I,M\right)$ ($I=\left[0,1\right]$)
• If ${\gamma }_{0},{\gamma }_{1}\in Obj\left(\underline{{B}_{1}}\right)$ are such that ${\gamma }_{0}{\mid }_{\left\{0,1\right\}}={\gamma }_{1}{\mid }_{\left\{0,1\right\}}$ then $Mor\left(\underline{{B}_{1}}\right)\left({\gamma }_{0},{\gamma }_{1}\right)$ is not empty and is ${\pi }_{0}\left({C}^{\infty }\left({I}^{2},M;\partial \right)\right)$ where $\partial$ is the condition that $\alpha {\mid }_{\partial {I}^{2}}={\gamma }_{0}*{\iota }_{{\gamma }_{0}\left(1\right)}*\overline{{\gamma }_{1}}*{\iota }_{{\gamma }_{1}\left(0\right)}$.

DR: I would just like to point out that we need a global description of $Mor\left(\underline{{B}_{1}}\right)$ - something like the quotient of the pullback of $M×M\to {C}^{\infty }\left(I,M\right)×{C}^{\infty }\left(I,M\right)$ and ${C}^{\infty }\left({I}^{2},M\right)\to {C}^{\infty }\left(I,M\right)×{C}^{\infty }\left(I,M\right)$. I’m not sure even that the quotient manifold is the right thing, this is why I constructed the topology on the space of 2-tracks from scratch in my thesis - is was assured to be locally homeomorphic to the loop space.

Actually speaking of loop spaces, would it be possible to define 2-tracks via representatives ${D}^{2}\to M$? But then I suppose that we lose any chances of having naive exponentiation for paths in the path space.

• The functor $\left(S,T\right):\underline{{B}_{1}}\to disc\left({B}_{0}\right)$ is the endpoint functor, $\gamma ↦\left(\gamma \left(0\right),\gamma \left(1\right)\right)$.
• The functor $I:disc\left({B}_{0}\right)\to \underline{{B}_{1}}$ is the “constant path” functor: $I\left(x\right)\left(t\right)=x$.
• The “conjugation” functor: $\underline{{B}_{1}}\to \underline{{B}_{1}}$ is path reversal: $\overline{\gamma }\left(t\right)=\gamma \left(1-t\right)$ with the obvious effect on homotopies.

At this point, we need the composition functor. The various natural transformations all depend on the composition functor at some point and will probably be obvious once that’s sorted out. With topological spaces, one simple concatenates loops and the associator is the reparametrisation functor. Given that reparametrisation is required anyway, I don’t see why one needs to muck about with the space of loops at all. There are two things that spring to mind to try:

1. Define concatenation as follows: if $\alpha$ and $\beta$ are such that $\alpha \left(1\right)=\beta \left(0\right)$ then

$\left(\alpha {*}_{\infty }\beta \right)\left(t\right)=\left(\alpha *\beta \right)\left(\varphi \left(t\right)\right)$(\alpha *_\infty \beta)(t) = (\alpha * \beta)(\phi(t))

where $*$ denotes the usual concatenation of continuous paths and $\varphi :\left[0,1\right]\to \left[0,1\right]$ is a smooth bijection which is flat at $1/2$.

When defining the natural transformations, one is allowed to use the original paths and thus the original parametrisation. This should mean that the rest of the structure is definable without any more finessing.

2. “Blow up” the manifold so that ${B}_{0}$ isn’t $M$ but is the infinite jet space over $M$. Then we only try to concatenate paths where the topological concatenation is smooth. Then the topological constructions should go over without any change.

Of the two, I would start with (1) because it’s the one that most is currently known about. I like (2), but the background theory isn’t so well established.

So, starting with (1), we look at the space $PM={C}^{\infty }\left(I,M\right)$. We have the following facts:

1. The evaluations ${e}_{0},{e}_{1}:PM\to M$ are locally trivial fibre bundles.
2. The inclusion of constant paths $\iota M\to PM$ is an embedding (with a tubular neighbourhood)
3. Exponentiation works just fine: ${C}^{\infty }\left(N×I,M\right)\cong {C}^{\infty }\left(N,PM\right)$

This answers almost all your questions above. Concatenation will be a smooth map, do you want any further properties? As I’ve defined it above, although it might be an embedding (not sure), it won’t have a tubular neighbourhood but that might not be important.

DR: I like (1) a lot! I don’t think concatenation needs any further properties, and indeed, my main follow-up construction (hurried along by Chris S-P’s interest) doesn’t need to concatenation map at all. I do want the fibre over a point of $S:{\underline{{\Pi }_{2}\left(M\right)}}_{1}\to M$ to be a Lie groupoid, but this should follow from general ideas. Once we have $\mathrm{PM}$ established, we (or you :), I have little clue how to do this with manifolds) need to define the charts on the space of 2-tracks so that we have a covering space.

DR: I was more wondering about exponentiation ${C}^{\infty }\left(N×I,PM\right)\cong {C}^{\infty }\left(N,{C}^{\infty }\left({I}^{2},M\right)\right)$, even if only for simple N, like $N=I$, or $N$ a finite 0-dim manifold.

I think that all the maps in at the 2-level will just go through: every 2-track is equivalent to one that is flat on its boundary, and these can be concatenated every which way they like.

DR: Good - we then need to know that this operation is smooth.

#### The Manifold of 2-Tracks

Andrew: The line just above separates out the new stuff from the old bits that it was replacing. I’ve not deleted any of the stuff below because not all of it is being replaced, but at the moment I can’t be bothered to go through and work out exactly what should stay and what should go.

We define the set $U{\prime }_{\sigma }$ by:

$U{\prime }_{\sigma }≔\left\{\tau :\left({\partial }_{0}\sigma ,{\partial }_{0}\tau \right)\in PV,\left({\partial }_{1}\sigma ,{\partial }_{1}\tau \right)\in PV\right)\right\}$U'_\sigma \coloneqq \big\{ \tau : (\partial_0 \sigma, \partial_0 \tau) \in P V, (\partial_1 \sigma, \partial_1 \tau) \in P V)\big\}

Then the claim is that $U{\prime }_{\sigma }$ is of the form ${U}_{{\partial }_{0}\sigma ,{\partial }_{1}\sigma }×{\pi }_{2}\left(M\right)$, where the first part are the domains of charts in $PM$.

AS: The ${\pi }_{2}$ bit might not be quite correct, depending on how connected $M$ is. The point is that it is discrete.

DR: Actually the ${\pi }_{2}$ should be replaced by a set that is (non-canonically) isomorphic to it, but for now this version is ok. I’m not sure what ${U}_{{\partial }_{0}\sigma ,{\partial }_{1}\sigma }$ is, so I’m going to flesh out the topological version in case I’ve not given a clear enough picture.

AS: I’ll flesh out what ${U}_{{\partial }_{0}\sigma ,{\partial }_{1}\sigma }$ when I next get a moment. The key point for you to notice now is that they are related to charts on $PM$ and so (once they’ve been made clearer) all the required properties of the structure maps ought to follow fairly easily.

The idea is that if we start with a pair of paths $\left(\alpha ,\beta \right)$ such that $\left({\partial }_{0}\sigma ,\alpha \right)$ and $\left({\partial }_{1}\sigma ,\beta \right)$ then we have canonical 2-tracks ${\tau }_{\alpha }$ and ${\tau }_{\beta }$ such that ${\partial }_{0}{\tau }_{\alpha }=\alpha$, ${\partial }_{1}{\tau }_{\alpha }={\partial }_{0}\sigma$, ${\partial }_{0}{\tau }_{\beta }={\partial }_{1}\sigma$, and ${\partial }_{1}{\tau }_{\beta }=\beta$. These come from specific charts on $PM$ defined as in smooth loop space with the ${\partial }_{0}\sigma$ and ${\partial }_{1}\sigma$ as centre points. So we get a 2-track from $\alpha$ to $\beta$ given by composition ${\tau }_{\beta }\sigma {\tau }_{\alpha }$.

AS: At this point, we can either modify $U{\prime }_{\alpha }$ so that it contains only one component, or proceed with splitting it up along components according to ${\pi }_{2}\left(M,{\partial }_{0}\sigma \right)$. Which is preferable from the final structure?

DR: Personally I like taking one component at a time, but I think that one can pass pack and forth easily enough - this should be a covering space after all.

AS: That’s what I was getting at: with the multiple component view it’s blindingly obvious that it’s a covering space; however, most people like to deal with contractible neighbourhoods (though there’s nothing in the definition of a manifold that requires that!)

### Comparison to continuous paths

Clearly a desirable requirement for using smooth paths is that every continuous homotopy class of continuous paths (and later, 2-tracks) in a manifold has a smooth representative, and that concatenation of smooth paths is related to continuous paths. This will help ensure that the smooth fundamental bigroupoid is weakly equivalent, as a topological bigroupoid, to the topological fundamental bigroupoid.

DR: I’m a bit hazy on the topology of the smooth version. Do we have a continuous map ${\Pi }_{2}\left(X{\right)}^{\mathrm{top}}\to {\Pi }_{2}\left(X{\right)}^{\mathrm{diff}}$ at each level (forgetting the smooth structure on ${\Pi }_{2}\left(X{\right)}^{\mathrm{diff}}$?