14

Are infinite dimensional simplicial complexes manifolds locally modeled on $\mathbb R^\infty=\operatorname{colim}\mathbb R^n$? If they are homotopy equivalent, are they homeomorphic?

Of course not. Part of such a complex might not be infinite dimensional: glue an interval to $\mathbb R^\infty$ by one endpoint and it stops being homogeneous. Or take two copies of $\mathbb R^\infty$ and glue them at a point and that point is special. Also, for $\alpha$ an uncountable cardinal, $\mathbb R^\alpha$, topologized as the colimit of the finite dimensional subspaces, is not locally isomorphic to $\mathbb R^\infty$ because every open subset contains an uncountable discrete subset, namely an appropriate choice of basis. But are those three obstacles the only ones?

If a simplicial complex is countable, every link* is contractible, and every simplex is the face of another, does it follow that it is locally homeomorphic to $\mathbb R^\infty$?

Is it PL isomorphic? What if we generalize to CW complexes? This leads to a second class of question. If those hypotheses are sufficient for local homeomorphism with $\mathbb R^\infty$, then apply them to the open cone on a contractible space satisfying the hypotheses. The space is now homeomorphic to $\mathbb R^\infty$. Was it already homeomorphic before applying the cone? If two such $\mathbb R^\infty$ manifolds are homotopy equivalent, are they, in fact, homeomorphic?

Is this topic in the literature? What about the special case of the product of a finite simplicial complex with $\mathbb R^\infty$? What about the product with a finite CW complex?

* The homotopy link of a point $x$ in a space $X$ is the (pro?) homotopy type of the homotopy inverse limit of the deleted neighborhoods $U-\{x\}$. This is a homeomorphism invariant of a space. Asking for the local homology $H_*(X,X-\{x\})$ to vanish is not quite sufficient for the link to be contractible because of the possibility that it is not simply connected. In a simplicial complex, the link is a specific simplicial complex, namely the union of all simplicies that are opposite $x$ in some simplex; and its PL isomorphism type is a PL invariant of the space.

Here is a potential counterexample to the PL statement. Take a closed manifold with the homology of a sphere but with nontrivial fundamental group, such as the Poincaré homology 3-sphere. The Cannon-Edwards theorem says that the double suspension is homeomorphic but not PL isomorphic to a sphere. Equivalently, the double cone is homeomorphic but not PL isomorphic to a disk. It is not PL isomorphic because a link of the link of the cone point is the homology sphere, not a real sphere. Now cross the double cone with $\mathbb R^\infty$. The result is homeomorphic to $\mathbb R^\infty$, but is it PL isomorphic? Is the cone point special? Is there some way of extracting the original fundamental group from the space, or has it been pushed off "to infinity"?

Motivation:

Is there an abelian group structure on $\mathbb C\mathbb P^\infty$? There are a couple of other models for the classifying space of $S^1$ that do have abelian group structures, such as the bar construction $BS^1$ and $FS^2$ the connected component of the free abelian group on $S^2$; we’d like to transfer them over to our favorite model. I think that we managed to cobble together a proof that $BS^1$ is PL isomorphic to $\mathbb C\mathbb P^\infty$, but what about $FS^2$? For many groups there is a model for the classifying space as the union of manifolds, but the filtration of the bar construction is only by manifolds if the group is a sphere, as fails for, eg, $\mathbb Z/3$ and $SO(3)$. But even though each $B_n\mathbb Z/3$ fails to be homogeneous, the full $B\mathbb Z/3$ is a group, thus homogeneous. So, in some sense it is a manifold. How many types of manifolds are there among infinite dimensional simplicial complexes? Perhaps they are all locally $\mathbb R^\infty$? And what about $BSO(3)$? It is neither a group nor filtered by manifolds, so it has no reason to be homogeneous. But I think it is.

Ben Wieland
  • 8,667
  • 2
    In my answer to your earlier question I sketch a proof that $EG$ is homeomorphic (and hence $BG$ is locally homeomorphic) to $S^\infty$ for any nontrivial finite $G$. The strategy there is to show that every finite subcomplex is contained in some subcomplex homeomorphic to a PL ball, and in such a way that these balls are nested in a nice way. – Tom Goodwillie Apr 06 '16 at 12:21
  • One could also ask such a question about (realizations of) simplical sets. These are not precisely PL, right? but are a good deal more rigid than CW complexes. – Tom Goodwillie Apr 06 '16 at 12:23
  • I was going to say don't we need to assume something about skeleta being finite. But of course that condition makes no sense until you fix a cell structure; and while $S^\infty$ can be taken to have $n$-skeleton $S^n$ it is in fact homeomorphic to $\mathbb R^\infty$, which at first glance might appear to need infinitely cells in each dimension. – Tom Goodwillie Apr 06 '16 at 12:35
  • Is it clear that the link of a simplex in a PL space is well defined up to PL homeomorphism? It's certainly a standard result in the locally finite case, but in general? – Tom Goodwillie Apr 06 '16 at 12:44
  • You need not assume that every simplex is a proper face of another. This says that the link of a simplex is not empty, which follows from the link of a point being contractible. – Tom Goodwillie Apr 06 '16 at 13:43
  • If a triangulated space $X$ is homogeneous in the sense that its germ at a point is isomorphic to its germ at any other point, and if it is also infinite-dimensional, then the link of every point is contractible, because infinite dimension implies that there are points having arbitrarily highly connected links: the link of a point that lies in the (relative) interior of an $n$-simplex is an $n$-fold suspension and therefore $(n-2)$-connected. – Tom Goodwillie Apr 06 '16 at 17:04
  • @TomGoodwillie Lots of good points. . . As for the link, I'm not sure I really understand the definition of PL isomorphism. If it is just a common refinement, then the link of a refinement is the refinement of the link, so that's OK. I am nervous about the definition of refinement in the not locally finite case, but that is an orthogonal issue. – Ben Wieland Apr 06 '16 at 20:27
  • I'm thinking now that it might go like this: Let $X$ be a PL space (where by "PL" maybe I mean something more general, like the realization of a simplicial set).

    Conjecture: If $X$ is such that each point has contractible link, then $X$ is locally isomorphic to $\mathbb R^\infty$. If in addition $X$ itself is contractible, then $X$ is isomorphic to $\mathbb R^\infty$.

    The second statement applied to the link of a point would seem to give the first statement.

    A strategy for proving the second statement might go like this:

     – Tom Goodwillie Apr 07 '16 at 00:40

  • Prove that every compact PL subspace of $X$ is contained in a contractible compact PL subspace.

  • Improve this to show that every compact PL subspace is contained in a PL ball of some dimension.

  • Thus $X$ is the union of an increasing sequence of balls. Now show that the embedding of one ball in the next can be taken to be a particularly nice embedding.

    • – Tom Goodwillie Apr 07 '16 at 00:41
  • @TomGoodwillie What about my potential counterexample, the infinitely iterated cone on a homology sphere? Is the first cone contained in a PL disk? If anything, you've made me doubt the PL version. – Ben Wieland Apr 07 '16 at 03:51
  • Key Lemma: Let $(K,L)$ be a compact PL pair and let $f:K\to X$ be a PL map whose restriction to $L$ is an embedding. If links in $X$ are contractible, then $f$ is PL homotopic, by a homotopy fixed on $L$, to an embedding $K\to X$. (This looks right to me; contractible links should allow general-position arguments.) – Tom Goodwillie Apr 07 '16 at 13:30
  • On the other hand, I'm not sure what I mean by a PL space. I wish I understood a generalization of the notion of PL structure in which a structure is given by being the realization of a simplicial set, and when the simplicial set is subdivided then the structure is still the same. Things like $BG$ and $EG$ are of this kind. It's not clear to me what one would now mean by the link of a point (as a PL space of some kind). – Tom Goodwillie Apr 08 '16 at 01:04
  • This includes things like $K/L$ where $(K,L)$ is a compact PL pair in the usual sense. In the case of $BG$ or $EG$ with $G$ abelian, the thing is homogeneous in the sense that there is an automorphism group acting transitively (because it's a group). Acting in a "generalized PL" way, I suppose. – Tom Goodwillie Apr 08 '16 at 01:07
  • @TomGoodwillie I think that the link of a 0-simplex in a simplicial set has a standard definition as a simplicial set with a non-injective map to the barycentric subdivision. Probably its subdivision has an injective map to the second subdivision, justifying the name link. . . I don't think simplicial sets have more generality. The subdivision of a simplicial set doesn't have the property that its non-degenerate simplices inject, but the collapse of their boundaries is quite controlled. – Ben Wieland Apr 08 '16 at 02:17
  • I cannot see a sense in which $B\mathbb Z/2$ is a PL object. I can invent a fairly convincing extended PL category in which any simplicial set gives an example; it has the usual PL category as a full subcategory; but as soon as there is a nondegenerate simplex having a degenerate face you are outside that subcategory. $BG$ for abelian $G$ is a group object in this category, and all such examples for countable $G$ seem to be locally PL isomorphic in this sense, but they are not locally like $\mathbb R^\infty$. – Tom Goodwillie Apr 13 '16 at 18:58
  • @TomGoodwillie A (finite dimensional) PL object is a space with a distinguished class of functions, "locally" isomorphic to a simplex with the PL functions. A simplex with a face collapsed has the class of functions that pull back to PL functions on the simplex, ie, the subset of the functions on the simplex that are constant on the face. Is this a PL object? I think the question is equivalent to: consider a square mapping to a triangle by collapsing a face; are the pullbacks of the PL functions on the triangle the same as the PL functions on the square that are constant on the edge? – Ben Wieland Apr 13 '16 at 20:22
  • @TomGoodwillie I'm not sure what you mean by your 13:30, but one interpretation is false. Contractible links for just points do not allow PL general position arguments. The double cone on an acyclic space has contractible links, but the cone on a nontrivial circle in the classifying space cannot be PL moved away from the line of cone points ($K=D^2\cup D^1$, $L=\partial$). But maybe you only want to apply it to the case of $X$ an $\mathbb R^\infty$-manifold. – Ben Wieland Apr 13 '16 at 20:39
  • Good point. Concerning your previous comment: I would not call the quotient of a square by one side by a PL space for the reason you give. For similar reasons I would not call BG a PL space, and in particular I don't know what it means to ask if it is an infinite-dimensional PL manifold. But I was wondering whether there is something more general than PL (I don't have a good name for it) designed to include such examples, and in which the question about whether BG is a PL manifold would become a sensible question. – Tom Goodwillie Apr 13 '16 at 22:14
  • @TomGoodwillie I didn't give a reason, but asked a question. Are you answering that question no? Are you claiming that there is a PL function on the square, constant on the edge that is not the pullback of a PL function on the triangle? – Ben Wieland Apr 14 '16 at 01:46
  • If I understand the question right, you are thinking of some particular PL map $f:S\to T$ from square onto triangle such that the preimage of one vertex is an edge and the preimage of each other point is a point, and you are asking whether a function on $T$ must be PL if its composition with $f$ is PL. But there is no such PL map $f$. – Tom Goodwillie Apr 14 '16 at 02:25
  • @TomGoodwillie OK, right, there is no such PL map. But we can extend the morphisms to piecewise smooth (or piecewise polynomial?) morphisms. Then the morphism exists and I think that every simplicial set is isomorphic to a simplicial complex. Of course this is a topic in the literature because it is necessary to compare smooth with PL. But usually people prove that it is "the same" as PL and then back off of it. Why? Are they just picking sides, or Is there a drawback to allowing these collapses? – Ben Wieland Apr 14 '16 at 10:55
  • Harking back to your objection to my proposed general-position lemma: But you can move the line of cone points off itself (not by an ambient isotopy but by a homotopy) so that it no longer touches the cone on that circle. – Tom Goodwillie Apr 15 '16 at 13:11
  • @TomGoodwillie I don't think that ambient isotopies are a sensible concept before proving homogeneity. I interpreted "general position" to mean a small homotopy. A small homotopy is not possible in this example. A large homotopy is possible here, but I am uncertain about in general. I believe that it is possible to move the line off of the disk, but not the disk off of the line. That asymmetry makes me suspicious that there might be a counterexample, but I do not have one. – Ben Wieland Apr 15 '16 at 15:20
  • Re your second to last comment: The composition of smooth maps is not piecewise smooth. You can compose with PL on one side and with smooth on the other. – Tom Goodwillie Apr 15 '16 at 18:48
  • Wait, what? The double cone on an acyclic space does not have contractible links. The double cone on a disk does not have contractible links. – Tom Goodwillie Apr 15 '16 at 18:52
  • @TomGoodwillie If a simplicial complex has contractible links of all simplices, then general position should be OK. In particular, if it is the product of $\Bbb R^\infty$ with the cone on a homology sphere. . . I thought we were talking just about links of points. The double cone on a homology sphere has $n$-connected links, which ought to be enough to illustrate the problem. If the acyclic space has contractible links, then its double cone has contractible links. This is probably true for the bar construction on an acyclic group. – Ben Wieland Apr 15 '16 at 20:39
  • @TomGoodwillie Yes, not forming a category is a good reason to avoid PDIFF. So I can't ask that they be isomorphic in the natural category. But we can still define a PDIFF object as a space together with a distinguished class of functions, and ask if a simplicial set is homeomorphic to a simplicial complex via a map that identifies the functions. But which functions? Traditionally one only considers very special functions to $\mathbb R^n$, but I think we should use functions to $\mathbb R$. – Ben Wieland Apr 15 '16 at 21:14