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Apr
26
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Apr
15
comment Are infinite simplicial complexes all manifolds?
@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 $\mathbb R$.
Apr
15
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
15
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
14
comment Does $\mathbb C\mathbb P^\infty$ have a group structure?
Simplification: surgery theory (really just the h-cobordism theorem) is overkill. In our case, we know that the complement of the embedding is a disk and that is adequate to show that it is the standard embedding, at least in the real and complex cases where the normal bundle is small dimensional and thus easy to control.
Apr
14
comment Does $\mathbb C\mathbb P^\infty$ have a group structure?
Correction: where in my answer I talk a collapsing map between simplicial complexes, such a thing does not exist as a piecewise linear (PL) map. The target is not a simplicial complex, but only a simplicial set. Instead of working in the PL category I should work in the piecewise smooth category. I think that everything I said goes through there. And a piecewise smooth manifold has an essentially unique PL structure, so the conclusion is not much changed.
Apr
14
comment Are infinite simplicial complexes all manifolds?
@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?
Apr
14
comment Are infinite simplicial complexes all manifolds?
@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?
Apr
13
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
13
comment Are infinite simplicial complexes all manifolds?
@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?
Apr
10
comment When may “summand of” be dropped from the definition of perfect dg module?
Perhaps your question is: can one lift an idempotent from the homotopy category to a finitely generated model? . . . If the category has one object with endomorphisms the ring of functions on an affine elliptic curve, would you say that idempotents split?
Apr
9
comment Homotopy groups of Lie groups
Let me correct and consolidate my comments. There are two easy constructions of simplicial complex topological groups homotopy equivalent to $CP^\infty$. (1) The free abelian group on $S^2$ is nice because it is $P(C(x))$ (with the right topology), a projective space, though a different one. (2) The bar construction $BS^1$ can be easy to construct and we proved with difficulty that it is homeomorphic to $CP^\infty$. I conjecture that everything is homeomorphic.
Apr
8
comment Why do some literatures prefer right module to left module when dealing with DG modules?
As with the Yoneda embedding of the category into (contravariant) presheaves, there is an embedding of $A$ into right $A$-modules. Alternately, there is an embedding of $A^{op}$ into left $A$-modules: you can choose where to put the contravariance, but it has to end up somewhere. . . So that's one reason, but there may be others.
Apr
8
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
8
comment When may “summand of” be dropped from the definition of perfect dg module?
What do you mean by "should"? The choice of objects is a choice and which objects are semi-free depends on that choice. In generalizing from an algebra to a category, we just replace the single free module with the many free modules. A module is equivalent to a semi-free object if it is in the subgroup of $K$ generated by the representable modules. Of course, there is a universal choice: if $A'=D_{parf}(A)^{op}$ then then $D_{parf}(A')=D_{parf}(A)$ but now every module is semi-free, indeed representable.
Apr
7
answered When may “summand of” be dropped from the definition of perfect dg module?
Apr
7
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
6
comment Are infinite simplicial complexes all manifolds?
@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.
Apr
6
comment Does $\mathbb C\mathbb P^\infty$ have a group structure?
The uniqueness of $\mathbb R\mathbb P^n$ in $\mathbb R\mathbb P^{n+1}$ is a hard theorem (needs the h-cobordism theorem) but it is overkill. For both your answer and my answer we can prove that the complement is a $\mathbb R^{n+1}$ and then it becomes fairly easy. (Same for $\mathbb C\mathbb P^n$. $\mathbb H\mathbb P^n$ probably does require h-cobordism, but not surgery.)
Apr
6
comment Does $\mathbb C\mathbb P^\infty$ have a group structure?
The strategy suggested in my question was to prove (1) that everything is homogeneous, (2) that everything is locally modeled on $\mathbb R^\infty$, and (3) that for such manifolds homotopy equivalence implies homeomorphism. But since we pretty much managed to answer the question by explicit construction, I posted it as a separate question.