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As far as I can tell, ever since Milnor's Microbundles and differentiable structures (1961) paper, whenever people talk about $Diff(\mathbb R^n)$ or $PL(\mathbb R^n)$ or $Homeo(\mathbb R^n)$, they always bring in an auxiliary simplicial space, given various names like the associated "c.s.s. complex" (Milnor), or the associated singular complex $Sing_\cdot(Diff(\mathbb R^n))$. It seems to me like there are as many names for essentially the same thing as there are authors in this subject.

My question is, why do people bother with this complex? i.e. why not just use the original space $Aut^{CAT}(\mathbb R^n)$?

In Milnor's paper he does not address this question. It seems to be a trend in the literature to avoid this discussion.

So to be more precise, I suppose I want to ask the question,

  • Q: Is there any point in the smoothing literature where the associated simplicial complex to $Aut^{CAT}(\mathbb R^n)$ is absolutely necessary to a proof, and if so, why?

If there is an affirmative answer to that question, I would like to know what it would take to remove the usage of the associated simplicial complex. What information are we missing about $Aut^{CAT}(\mathbb R^n)$ that we can avoid when we use the associated simplicial complex?

The only reference I've found that makes any attempt to address something like this question is Lurie's notes on smoothing theory (lecture 6, bottom of page 1). But this isn't quite what I'm looking for.

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Ryan, is simplicial complex or simplicial set what you mean? –  Fernando Muro Oct 24 '13 at 12:14
Ryan, I don't think there's an obvious topoloogy on $\text{PL}(\Bbb R^n)$ such that $B\text{PL}(\Bbb R^n)$ has the correct homotopy type. –  John Klein Oct 24 '13 at 12:26
Yes, John's comment succinctly states the main point in my long-winded answer. –  Tom Goodwillie Oct 24 '13 at 12:48
Thanks John. I haven't seen why what you say is true but hopefully it will help me as I unwind the literature. –  Ryan Budney Oct 24 '13 at 12:55
@John: I have seen that statement before, and there seems to be no reason to expect otherwise. Nevertheless, I think it would still be very instructive to see a convincing, clear example or explanation of the veracity of that statement. Do you happen to know one? Thanks. –  Ricardo Andrade Oct 24 '13 at 19:52

2 Answers 2

In the topological category the usual compact-open topology does the job. (EDIT: Or rather I suppose you might have to modify it a little so that $h\mapsto h^{-1}$ is continuous.) At times you might want to replace a function space by the simplicial set called its total singular complex. This is an instance of the more general procedure of replacing a space by its total singular complex when convenient, a trend that began around that same time. But there may not be any great technical advantages in doing so when working with spaces of homeomorphisms.

In the smooth category you can give the space of diffeomorphisms the Whitney topology and then look at the total singular complex, but in some ways it's a better idea to look at the smaller simplicial set consisting of the "smooth" (rather than merely continuous) maps of simplices into the space, meaning those which correspond to smooth maps $\Delta\times M\to M$. Note that when dealing with smooth bundles with fiber $M$ over a base $B$ you really are interested more in "smooth" maps from open subsets $U\subset B$ into the group than in continuous maps, because it is the former that correspond to automorphisms of the trivial bundle $U\times M$. It is true that, just as for maps between finite-dimensional manifolds, smooth homotopy classes of smooth maps from $U$ to the group are in bijection with homotopy classes of continuous maps from $U$ to the group. It is not true that the compact-open topology on smooth automorphisms is just as good as the Whitney topology for these purposes: you don't want to be able to kill an exotic diffeomorphism of $D^n$ that fixes a neighborhood of the boundary sphere by the Alexander trick (which scales it through diffeomorphisms until at the last moment it becomes the identity).

In the $PL$ category it gets worse: I don't know a useful topology on the group of $PL$ automorphisms of $M$. Topologizing it as a subspace of the space of homeomorphisms can't be right, and there isn't something like a Whitney topology to fix things up.

EDIT: This seems about right as a reason why $PL$ homeomorphisms cannot be treated as a subspace of homeomorphisms. Suppose $h$ is a $PL$ automorphism of $\mathbb R^n$ fixing the origin. In a neighborhood of the origin it is radial (linear on each ray). By a sort of reverse Alexander trick you would be able to make it radial everywhere if you were allowed to use isotopies through $PL$ homeomorphisms rather than actual $PL$ isotopies. In this way you would get a deformation retraction to the subgroup consisting of radial $PL$ homeomorphisms. Then you could further deform to the smaller group that consists of those radial homeomorphisms that are made by coning $PL$ homeomorphisms of a ($PL$) $(n-1)$-sphere. That would mean that the canonical map from $Aut^{PL}(D^n)\simeq Aut^{PL}(S^{n-1})$ to $Aut^{PL}(D^n-S^{n-1})$, or to germs at the origin, is an equivalence. But the fiber of this map to germs is the pseudoisotopy space $P^{PL}(S^{n-1})$, and this is not contractible. Contradiction. Why is $P^{PL}(S^{n-1})$ not contractible? It fibers over the pseudoisotopy embedding space $PE^{PL}(D^{n-1},S^{n-1})$, with contractible fiber $P^{PL}(D^{n-1})$. The space $PE^{PL}(D^{n-1},S^{n-1})$ is not contractible because (1) the smooth analogue $PE^{Diff}(D^{n-1},S^{n-1})$ is contractible by Hatcher's light bulb trick, (2) the fiber of $PE^{Diff}(D^{n-1},S^{n-1})\to PE^{PL}(D^{n-1},S^{n-1})$ is equivalent by smoothing theory to $\Omega fiber (O_n/O_{n-1}\to PL_n/PL_{n-1})$, and (3) this last is not contractible. Why (3)? Because if it were contractible then the fiber of the map $P^{Diff}(D^{n-1})\to P^{PL}(D^{n-1})$, which is $\Omega^n fiber (O_n/O_{n-1}\to PL_n/PL_{n-1})$ by smoothing theory, would be contractible. But $P^{PL}(D^{n-1})$ is contractible while $P^{Diff}(D^{n-1})$ is not by Waldhausen (rationally it becomes $\Omega^2 K(\mathbb Z)$ as $n\to \infty$).

One other point:

What about the structure group for microbundles? You want to compare on the one hand the group of automorphisms of $\mathbb R^n$ fixing the origin (the structure group for open disk bundles with section) and on the other hand the group of invertible germs of maps $\mathbb R^n\to \mathbb R^n$ at the origin (the structure group for microbundles). As I understand the Kister-Mazur Theorem, bundles are "the same" as microbundles because the comparison map of groups is an equivalence. The kernel of the comparison map (i.e. the group of all automorphisms of $\mathbb R^n$ such that some neighborhood of the origin is pointwise fixed) is rather obviously contractible, while on the other hand the map is surjective (every invertible germ is the germ of a global automorphism). I think that in getting the details of that right it is convenient not to have to actually topologize the group of germs, but rather to speak of germs of maps from $\Delta^k\times \mathbb R^n$ to itself along $\Delta^k\times 0$.

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Thanks Tom, this might be starting to help. Is there a good example of how the compact-open topology on the PL automorphisms of $\mathbb R^n$ isn't appropriate, that $B(Aut^{PL}(\mathbb R^n))$ would not classify PL structures on topological manifolds? Orthogonally, is it true that there is no known useful topology on spaces of germs, appropriate for setting up smoothing theory? –  Ryan Budney Oct 24 '13 at 12:53
@Ryan: I have a vague comment concerning topologies on spaces of germs. The natural topology on a space of germs is the colimit or final topology. Obviously, a point in the filtered colimit $\operatorname{colim}_i X_i$ lifts to one of the spaces $X_i$. But this does not hold parametrically: a map into the filtered colimit does not usually factor through any of the spaces $X_i$. (to be continued) –  Ricardo Andrade Oct 24 '13 at 19:05
(continued) This parametric lifting property is often required. It holds if one uses, for example, simplicial sets instead of topological spaces: since filtered colimits of sets commute with finite limits, a map from a finite simplicial set to a filtered colimit $\operatorname{colim}_i X_i$ necessarily lifts to one of the simplicial sets $X_i$. Further, I believe this is one of the reasons why quasi-topological spaces are used in proofs of the h-principle: quasi-topological spaces also admit this parametric lifting property for filtered colimits, at least when the source is compact Hausdorff. –  Ricardo Andrade Oct 24 '13 at 19:09
I remember Terry Wall telling me very long ago that if you give the set of PL homeomorphisms of the n-sphere the subspace topology in Top(n), the inclusion is a homotopy equivalence. I don't remember him telling me a proof. –  Peter May Oct 24 '13 at 23:29
Ryan, I will edit my answer to sketch a proof that $Aut^{PL}$ as a subspace of $Aut^{Top}$ has the wrong homotopy type. –  Tom Goodwillie Oct 25 '13 at 0:27

I present here a reference for Peter May's comment to Tom Goodwillie's answer in this thread. It also corroborates the comment by John Klein below the question stating that there is no obvious topology on the set of PL homeomorphisms which recovers the correct homotopy type.

Corollary 2 in the article

Ross Geoghegan and William Haver, On the space of piecewise linear homeomorphisms of a manifold, Proceedings of the A.M.S., volume 55, number 1, 1976, pages 145-151

contains the following result:

Let $M$ be a compact PL manifold (without boundary) of dimension different from 4. Let $PLH(M)$ denote the space of PL homeomorphisms of $M$ with the compact-open topology. Further, let $H^\ast(M)$ denote the space of homeomorphisms of $M$ which are isotopic to PL homeomorphisms, equipped with the compact-open topology. In other words, $H^\ast(M)$ is the union of all the path components of the space of homeomorphisms of $M$ which contain a PL homeomorphism. Then the inclusion of $PLH(M)$ into $H^\ast(M)$ is a weak equivalence.

Unfortunately, I am not in a position to say anything about the proof of the above result. Perhaps someone else can provide more information.

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Thanks Ricardo, I'll try to take a look at this over the weekend. –  Ryan Budney Oct 25 '13 at 18:02
Curious. The cited result is for compact manifolds. I wonder if this is also true for noncompact manifolds, such as $\mathbb R^n$. If so, and if my sketched argument in my answer is correct, then that would seem to say that $PL$ automorphisms of an open ball is equivalent to homeomorphisms of a closed ball (which can't be right). –  Tom Goodwillie Oct 26 '13 at 13:58

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