Timeline for Sharpness of the $1/6$-constant in the Cancellation Theorem
Current License: CC BY-SA 4.0
17 events
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| Sep 6, 2018 at 12:51 | comment | added | H1ghfiv3 | I've just checked the very same book. It seems to be that, if $P$ satisfies the cancellation conditions for any $\lambda \in (1/6,1/5)$ (i.e, replacing $1/6$ by $\lambda$), then it is aspherical, since for any such $\lambda$, no relator $r \in R$ can be written as the product of less than $6$ pieces (this is the essential step in the proof). Moreover, if i understand correctly, every group admits a presentation $P_{\lambda}$ satisfying (2) with $\lambda$ replacing $1/6$ for any $\lambda > 1/5$. | |
| Sep 6, 2018 at 12:39 | comment | added | user35370 | I seem to recall Geometry of defining relations in groups having references to this(maybe even Lyndon and Schupp), but I don't have my copy with me. "1/6" is the "start" of notions of flat curvature (tilling plane with hexagons) | |
| Sep 6, 2018 at 12:15 | comment | added | HJRW | @BerniWaterman -- that's right! This is a standard application of the Euler characteristic of the sphere, which is closely related to how the 1/6 theorem is proved. | |
| Sep 6, 2018 at 11:31 | comment | added | H1ghfiv3 | Following your argument, the $1/6$-Cancellation Theorem therefore also implies that there cannot exist a platonic solid such that each face is bounded by $k \geq 6$ edges :D. | |
| Sep 6, 2018 at 10:53 | comment | added | HJRW | @FedorPetrov: that's why I only posted a comment (though it is the canonical example). But it should now be clear that the problem is about finding certain cellular decompositions of the sphere. I don't think it's terribly difficult (though I admit I haven't thought about it). | |
| Sep 6, 2018 at 9:58 | comment | added | Fedor Petrov | @HJRW but there exist numbers between $1/6$ and $1/5$? | |
| Sep 6, 2018 at 9:18 | comment | added | H1ghfiv3 | I didn't see that the universal cover $\tilde{P}$ ($D$ in my notation, but nevermind, yours is more appropriate) of $P$ is a infinite tree of copies of $D \cong S^2$. This is of course enough to see that $P$ is aspherical. | |
| Sep 6, 2018 at 9:00 | comment | added | HJRW | ... Homotopy Lifting Lemma, $\tilde{\pi}$ would also be null-homotopic, and so, applying $\rho$ to the homotopy, we get that the identity map of $S^2$ is null-homotopic, which is a contradiction. | |
| Sep 6, 2018 at 8:57 | comment | added | HJRW | If $D\cong S^2$ is the dodecahedron (clearly non-aspherical, I hope), then the presentation complex $P=D/\sim$ is obtained by identifying the vertices of $D$, so we have a quotient map $\pi:D\to P$. As I think you say, this defines the "obvious" non-trivial element of $\pi_2(P)$. There are various ways to see that it's non-trivial. For instance, the universal cover $\widetilde{P}$ is an infinite tree of copies of $D$, and $\pi$ lifts to an inclusion $\tilde{\pi}:D\hookrightarrow\widetilde{P}$. There is also a retraction $\rho:\widetilde{P}\to D$. If $\pi$ were null-homotopic, by the... | |
| Sep 6, 2018 at 7:52 | comment | added | H1ghfiv3 | This seems to be a very nice example. I just can't see why it is "clearly not aspherical". It the quotient map $\pi: S^2 \to D$ the "obvious" non-trivial element of the second homotopy ? | |
| Sep 6, 2018 at 7:35 | comment | added | HJRW | Sure. Take a dodecahedron. Identify all the vertices. The result is a presentation complex which is clearly not aspherical, and where each 2-cell (ie relator) overlaps its neighbours in precisely 1/5 of its length. | |
| Sep 6, 2018 at 5:56 | history | edited | H1ghfiv3 | CC BY-SA 4.0 |
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| Sep 6, 2018 at 5:46 | comment | added | H1ghfiv3 | Can you elaborate ? | |
| Sep 6, 2018 at 5:43 | history | edited | H1ghfiv3 | CC BY-SA 4.0 |
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| Sep 5, 2018 at 22:05 | comment | added | HJRW | The dodecahedron. | |
| Sep 5, 2018 at 18:52 | history | edited | H1ghfiv3 | CC BY-SA 4.0 |
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| Sep 5, 2018 at 18:36 | history | asked | H1ghfiv3 | CC BY-SA 4.0 |