## What is the infinite-dimensional-manifold structure on the space of smooth paths mod thin homotopy?

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This question is motivated by the recent paper An invitation to higher gauge theory by Baez and Huerta, and the 2007 paper Parallel Transport and Functors by Schreiber and Waldorf.

Let $M$ be a smooth, finite-dimensional manifold. A lazy path in $M$ is a smooth function $\gamma: [0,1]\to M$ such that all derivatives of $\gamma$ vanish at $0,1$. A homotopy of lazy paths $\gamma_0,\gamma_1: [0,1] \to M$ is a smooth function $\Gamma: [0,1]^2 \to M$ such that all derivatives of $\Gamma(s,t)$ vanish near $t=0,1$, and such that $\Gamma(s,t) = \gamma_s(t)$ for $s = 0,1$. A homotopy of lazy paths is lazy if additionally we have that for each $t$, all the $s$ derivatives of $\Gamma(s,t)$ vanish near $s = 0,1$. A homotopy (of possibly non-lazy paths) is thin if $\text{rank}\, d\Gamma < 2$ everywhere. Note that (non-lazy) thin homotopies include all reparameterizations, and so any (possibly non-lazy) piecewise-smooth path is thinly homotopic to a lazy path. Note also that lazy paths concatenate smoothly, and the concatenation of lazy paths is associative up to thin homotopy. Note also that if $\gamma^{-1}(t) = \gamma(1-t)$, then the concatenation $\gamma^{-1}\gamma$ is thinly homotopic to a constant path. Note also that lazy thin homotopies concatenate, and so define an equivalence relation, and if two paths are thinly homotopic, then they are lazily-thinly homotopic. So define $\mathcal P^1(M)$ to be the set of all lazy thin homotopy equivalence classes of lazy paths in $M$. It is a groupoid with base $M$.

The idea is to consider $\mathcal P^1(M) \rightrightarrows M$ as not just a groupoid but an infinite-dimensional Lie groupoid. I think I understand the smooth structure on $\mathcal P^1(M)$: a curve in $\mathcal P^1(M)$ should be precisely a (non-thin) homotopy of lazy thin paths, up to thin homotopy. It's not entirely clear to me that this defines a smooth structure. But it probably works in some formalism.

But if I really want to think of $\mathcal P^1(M) \rightrightarrows M$ as a Lie groupoid, then I should treat $\mathcal P^1(M)$ not just as a smooth space, but actually as an (infinite-dimensional) manifold, and there are various things to check about the maps (the source and target maps should be surjective submersions, etc.). And it's not clear to me how to write down a smooth manifold structure on $\mathcal P^1(M)$.

Here's what I'd like. Given a point in $\mathcal P^1(M)$, I'd like a neighborhood of it and a "diffeomorphism" between that neighborhood and some (Fréchet, maybe?) vector space, and I'd like it to be clear that the gluings are smooth. I can make a start: it's clear that the space of lazy paths in a finite-dimensional vector space is a vector space, and that thin homotopies respect addition, so that $\mathcal P^1(\mathbb R^n)$ is a vector space. It's not clear to me how to put a topology on it, and it's not clear that I can approximate $\mathcal P^1(M)$ by chopping $M$ into trivializable pieces, take $\mathcal P^1$ of each piece, and try to glue back together — thin homotopies can take a path in one trivializable patch and make it wrap around $M$ in a complicated way, providing it wraps back, for example.

Hence the question:

What is the manifold structure on $\mathcal P^1(M)$?

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You know, you could probably also send an e-mail to Urs. He's quite friendly. – Harry Gindi Mar 11 2010 at 6:07

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@fpqc: Yes, of course. But he's sometimes here, and Konrad and John Baez and so on are as well --- I mean, I could also walk down the hall and talk to Konrad. But then I wouldn't get, I don't know, Andrew Stacey's answer, and other folks wouldn't get to listen in. – Theo Johnson-Freyd Mar 11 2010 at 6:19

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+2 for linking to my paper, -1 for not linking to any nLab pages! – Andrew Stacey Mar 11 2010 at 9:03

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+1 for your reasons for asking this here, rather than email. If everyone starts doing this (especially asking the questions so well!), the world will be a better place. – Scott Morrison♦ Mar 11 2010 at 16:55
I'm commenting here just to ensure that Theo gets notified that I've added a link to my answer. – Andrew Stacey Mar 14 2010 at 15:27
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## 2 Answers

Question: What is the manifold structure on $P^1(M)$?

Answer: There isn't one.

Update: The biggest failing is actually that the obvious model space is not a vector space. The space of paths mod thin homotopy in $\mathbb{R}^n$ does not inherit a well-defined addition from the space of paths in $\mathbb{R}^n$. Full details at the nLab page http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.

(Update added here, rather than at the end, as it's the most direct answer to the specific question; the rest should be viewed as extra for those interested in more than just whether or not this space is a smooth manifold.)

It is, as you say, a smooth space. This is formal: whatever category of generalised smooth spaces you like, take the quotient of $P(M)$ by thin homotopies. All the proposed categories of generalised smooth spaces admit quotients, so the quotient exists and is a smooth space. Depending on your choice of category, the description of this smooth space may vary. For example, its Frolicher structure and its diffeological structure are very different.

But it is not "locally linear" in any sense. The basic problem is that, as you say, within an equivalence class you have paths wrapping all the way around the manifold. This destroys any hope of local linearity.

As for the proposed local model, you hit the nail on the head when you say:

It's not clear to me how to put a topology on it,

Absolutely! Topologising these spaces can lead to quite strange behaviour. You want a LCTVS structure, else you haven't a hope of even starting, and that can distort the topology from what you expect. For example, if you take piecewise-smooth paths (with no quotient) then the LCTVS topology on that is the $C^0$-topology! Indeed, simply taking so-called "lazy paths" could be fraught with difficulties (I notice that you define "lazy" slightly differently to how I've seen it done before with sitting instances). Is that space a manifold? (I know the answer to this one, but if you don't then you should start with that one as it is a much easier question and will hone your skills a little.)

If you really want a manifold, the solution is to go one step further. Rather than quotienting out by thin homotopies, make your "thing" into a 2-structure and put the thin homotopies in at the 2-level. Keep all paths at the 1-level. Then each level has a manifold structure and by mapping into a 1-structure you effectively quotient out by the 2-structure but never actually have to consider the quotient itself.

To coin a phrase:

Quotients are horrible, it's a shame so many people think otherwise.

Lastly, that's not to say that there is no way of making $P^1(M)$ into a manifold. There may well be. But if there is, it'll be so convoluted and contrived that it won't look anything like the quotient of $P(M)$. A cautionary tale here is the case of all paths in a manifold, $C^\infty(\mathbb{R},M)$. That can be made into a manifold, but it has uncountably many components, for example, so looks absolutely horrid.

Okay, not quite lastly. There's lots of details here that have been glossed over. If you are really interested in working out the smooth space structure of this particular space then I (and I suspect Urs and Konrad) would be very interested in seeing it done and helping out. But MO isn't the place for that. Hop on over to the nLab, create a spin-off of http://ncatlab.org/nlab/show/path+groupoid, and start working.

### Further Reading

1. Constructing smooth manifolds of loop spaces. canonical page. The point of this is to figure out exactly when the "standard method" (alluded to by Tim) works. The distinction between "loop" and "path" is irrelevant.

2. The Smooth Structure of the Space of Piecewise-Smooth Loops. canonical page. Why you should be very, very nervous whenever anyone says "consider piecewise-smooth maps"; and take as a cautionary tale as to the inadvisability of going beyond smooth maps in general.

3. Work of David Roberts on the nLab. This is where I got the 2-idea that I mentioned above.

4. Other relevant nLab pages: http://ncatlab.org/nlab/show/generalized+smooth+space, http://ncatlab.org/nlab/show/smooth+loop+space and further.

5. Of course, the magnificent book by Kriegl and Michor. (I'm going to create a separate MO account for that book; its role will be to post an answer on relevant questions simply saying "Read Me".)

In response to Konrad's comment below, I've started an nlab page to work out the smooth structure of this space. The initial content considers the linear structure of the space of paths in some Euclidean space modded out by thin homotopy. The page is http://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid.

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Great answer, Andrew! But can we be more specific? Given any assumption on a possible manifold structure on $P_1(M)$ (e.g. that the projections to $M$ are submersions), can one prove that there is no such manifold structure? – Konrad Waldorf Mar 11 2010 at 16:58
That would be worth settling once and for all, I've wondered off-and-on about this.  However, I think that that's an nlab project rather than an MO one, and I'll only start it if I know that others (you and Theo?) will join in! – Andrew Stacey Mar 11 2010 at 21:23
Expanding Andrew's comment above about 2-structure, one may be able to describe a Lie groupoid which presents a smooth stack of paths up to thin homotopy. I echo the above sentiment, this would be a good nlab project (not necessarily linked with mine and Andrew's) – David Roberts Mar 12 2010 at 0:42

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If you want to prove something is a smooth manifold, a good way to begin is to decide what its tangent spaces ought to be. So let $\gamma_s$ be a (smooth) homotopy of lazy paths, say for $s$ in $(-\epsilon,\epsilon)$. Its derivative at $s=0$ is a vector field $\xi$ along $\gamma:=\gamma_0$. This is a section of $\gamma^\ast TM$, not necessarily a "lazy" one. The vector field is to count for nothing if $\gamma_s$ is a lazy thin homotopy. So we should take the quotient of $C^\infty(\gamma^* TM)$ by the subspace $L$ of those $\xi$ which have vanishing (higher) derivatives at $0$ and $1$ and such that $\dot{\gamma}(t)$ and $\xi(t)$ are linearly dependent in $T_{\gamma(t)}M$ for all $t$.

A standard method to produce smooth charts (on path spaces in particular) is to exponentiate tangent vectors. This requires an auxiliary choice, say of a metric $g$ on $M$, so the manifold structure won't be absolutely canonical; but it may well be canonical up to diffeomorphism (strategy: define a smooth structure on the family of manifolds parametrized by the contractible manifold $Met(M)$, and show it's a smooth fibre bundle).

Well, $g$ induces an $L^2$-metric on $C^\infty(\gamma^{\ast}TM)$, so we could take the orthogonal complement $L^{\perp}$ (isn't that the vector fields pointwise-orthogonal to $\dot{\gamma}$, vanishing where $\dot{\gamma}$ does?) and view that as our tangent space. That makes it a little clearer that it's a Frechet space (consider the $C^k$ norms on $L^\perp$...). Let $L^{\perp}_\epsilon$ be the vector fields in $L^\perp$ which, pointwise, have length $<\epsilon$. Assume $\epsilon$ is smaller than the injectivity radius of $g$ along $\gamma$. Then one has $Exp_g \colon L^\perp\to \mathcal{P}^1 M$ (since it defines a diffeo from $(T_{\gamma(0)}M)_{<\epsilon}$ onto its image, $\exp_g$ preserves laziness). This map is injective, and it's a reasonable candidate for a coordinate chart. Declare such charts to be our atlas, defining, as a by-product, a topology - the coarsest that makes the charts continuous.

Now you have several things to check. (I haven't - maybe it doesn't work...) One of those is that the topology is Hausdorff, so you might even want to make this into a metric space, perhaps via a Riemannian metric.

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