Timeline for Manifolds up to homeomorphisms VS Manifolds up to homotopy equivalence
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30 events
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| S Dec 7, 2021 at 1:01 | history | bounty ended | CommunityBot | ||
| S Dec 7, 2021 at 1:01 | history | notice removed | CommunityBot | ||
| S Nov 28, 2021 at 23:09 | history | bounty started | GSM | ||
| S Nov 28, 2021 at 23:09 | history | notice added | GSM | Canonical answer required | |
| S Aug 19, 2021 at 0:07 | history | bounty ended | CommunityBot | ||
| S Aug 19, 2021 at 0:07 | history | notice removed | CommunityBot | ||
| Aug 18, 2021 at 21:07 | comment | added | Ben Wieland | Mazur stabilized by multiplying by open intervals, leaving only the stable tangent bundle. Stabilizing by multiplying by the closed interval is the subject of pseudoisotopy theory, which is related to $K$-theory. Crossing with a manifold of multiplies $K$-theory by the Euler characteristic, so stabilizing by multiplying by $S^1$ or other odd sphere might be interesting with a tractable simplification. I suspect stabilizing with arbitrary manifolds doesn't destroy more | |
| Aug 18, 2021 at 20:57 | comment | added | Ben Wieland | Experience says that the problem is more coherent if you don't decategorify so early. Don't consider spaces up to homotopy equivalence, but consider specific homotopy eq. The structure set $S(X)$ is the set of manifolds with a homotopy eq to $X$ up to homeo $M\to M'$ that commute up to homotopy with the equivalences to $X$. The homotopy eqs of $X$ act on the structure set your object is the quotient. In particular, the Borel conjecture is that every homotopy equivalence between aspherical manifolds is homotopic to a homeomorphism, which is not obviously implied by your statement. | |
| Aug 18, 2021 at 20:47 | comment | added | Ben Wieland | @ConnorMalin a map between finite complexes is a simple homotopy equivalence if and only if it becomes homotopic to a homeomorphism after crossing with the Hilbert cube. But crossing with a finite number of intervals destroys much less information. In particular, Pontrjagin classes are an obstruction to homemorphism preserved by crossing with the interval. | |
| S Aug 10, 2021 at 23:01 | history | bounty started | Paris | ||
| S Aug 10, 2021 at 23:01 | history | notice added | Paris | Canonical answer required | |
| Aug 9, 2021 at 17:48 | comment | added | Ryan Budney | @MartinBrandenburg: correct. I suspect that's not the entire group completion. Presumably the total Stiefel-Whitney class of the tangent bundle would give some independent elements for the "manifolds up to homeomorphism" version of the question. I would imagine a chain of similar invariants coming from surgury theory. | |
| Aug 9, 2021 at 1:07 | comment | added | Martin Brandenburg | @Ryan Just to be sure, you mean symmetric polynomials in the sense that the coefficients form a symmetric sequence, and we have just one variable? With rational polynomials you mean rational functions? And is there already something which shows that the K-group is more than that? | |
| Aug 8, 2021 at 10:53 | history | edited | Greg Friedman | CC BY-SA 4.0 |
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| Aug 6, 2021 at 21:03 | history | edited | Paris | CC BY-SA 4.0 |
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| Aug 6, 2021 at 13:30 | comment | added | Paris | @mme I think you are right, thanks | |
| Aug 6, 2021 at 13:16 | comment | added | mme | I don't agree. It says that 2X = 2Y, so that X - Y is 2-torsion. But X - Y could be zero in your Grothendieck group. In my explicit example, I believe it is (these should be homeomorphic after crossing with a 3-sphere). | |
| Aug 6, 2021 at 12:54 | history | edited | Paris | CC BY-SA 4.0 |
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| Aug 6, 2021 at 12:48 | comment | added | Paris | @mme thanks, so that means that $K(Man)$ does have torsion elements! | |
| Aug 6, 2021 at 12:38 | comment | added | mme | Lens spaces give counterexamples to your question. I think X = L(7,1) and L(7,2) work. | |
| Aug 6, 2021 at 12:32 | history | edited | Paris | CC BY-SA 4.0 |
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| Aug 5, 2021 at 20:15 | comment | added | user171227 | @ConnorMalin I think $S^1$ and $[0,1]$ behave quite differently: for instance, any homotopy equivalence becomes a simple homotopy equivalence after crossing with $S^1$, see Theorem 23.2 in A Course in Simple Homotopy Theory by M. Cohen. | |
| Aug 5, 2021 at 16:10 | comment | added | Connor Malin | @user171227 Perhaps this says more about manifolds up to simple homotopy equivalence? If I recall correctly, simple homotopy equivalent is the same as stably homeomorphic with respect to $[0,1]$. | |
| Aug 5, 2021 at 14:51 | comment | added | user171227 | The previous comment was about smooth manifolds and diffeomorphism, where the map to $BO$ classifies the stable tangent bundle. For topological manifolds and homeomorphism, replace $BO$ with $BTOP$. | |
| Aug 5, 2021 at 14:31 | comment | added | user171227 | In between the two abelian semigroups in the question, one could consider $\mathcal{M}an^{th}$ defined as manifolds up to homotopy equivalence over $BO$. Maybe Proposition 3.3 of this paper can be used to show that $K(\mathcal{M}an) \to K(\mathcal{M}an^{th})$ is an isomorphism? I'm not sure how to proceed from there though, but at least it would be "purely homotopy theory". | |
| Aug 5, 2021 at 8:20 | history | edited | Greg Friedman | CC BY-SA 4.0 |
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| Aug 5, 2021 at 7:35 | comment | added | Ryan Budney | There is some partial understanding. For example, the Poincare Polynomial is a homomorphism to symmetric polynomials, so when you pass to the group completion you get a homomorphism to the rational symmetric polynomials. That is a fairly coarse invariant. But it gives you some constraints on the kernel. | |
| Aug 5, 2021 at 0:45 | comment | added | Connor Malin | The first step to understanding 2 is using surgery theory to classify manifolds in the homotopy type of $M$. The next step I presume would be understanding cancellation phenomena of manifolds and how it differs when one moves from homeomorphism to homotopy equivalence. | |
| Aug 4, 2021 at 21:52 | comment | added | Carl-Fredrik Nyberg Brodda | Side note: an abelian semigroup with a unit is usually called a commutative monoid (semigroups do not require an identity element. I don’t know why “abelian” did not catch on for semigroups/monoids.) | |
| Aug 4, 2021 at 21:43 | history | asked | Paris | CC BY-SA 4.0 |