Timeline for Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 22, 2016 at 22:50 | history | edited | user80296 | CC BY-SA 3.0 |
fixed typo in denominator, updated remark in light of a comment to another question.
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| Jan 20, 2016 at 11:16 | history | edited | Chris Schommer-Pries | CC BY-SA 3.0 |
fix error (8 to 7)
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| Jan 20, 2016 at 0:50 | comment | added | David Roberts♦ | @ChrisSchommer-Pries can someone edit the above answer to make sure the 7 vs 8 issue is fixed, and importantly, clarified that it has been fixed? | |
| Sep 16, 2015 at 8:35 | comment | added | Chris Schommer-Pries | It would be nice if there was a construction of this manifold that didn't require knowing a priori that $bP_8 = \mathbb{Z}/28$. I was hoping to use this manifold to prove that $240 \sigma = 0$, but since the computation that $bP_8 = \mathbb{Z}/28$ presupposes $\pi_7^{st} = \mathbb{Z}/240$, this is circular reasoning. | |
| Sep 15, 2015 at 19:57 | comment | added | Tom Mrowka | Indeed, I scanned the previous discussion too quickly and conflated the two appearances of 8, the $E_8$ and the number of $S^4\times S^4$s. Should indeed be 7. | |
| Sep 15, 2015 at 17:33 | comment | added | user80296 | Yes, 7 sounds right now. | |
| Sep 15, 2015 at 15:21 | vote | accept | Chris Schommer-Pries | ||
| Sep 15, 2015 at 15:20 | comment | added | Chris Schommer-Pries | That is a great way to see that $p_1$ is trivial! Thanks @TomMrowka btw it was your beautiful answer on the linked MO question that got me thinking along these lines. I will accept this answer, but I still think it is 7 copies of $S^4 \times S^4$ that are needed. If I am doing things right the first half of the construction (the 28 Milnor plumbings + cap) has $\chi = 8 *28 + 2 = 226$ while the connect sum of 8 $S^4 \times S^4$s has $\chi = 8 *2 +2 = 18$. The connect sum of these two parts then has $\chi(M) = 226 + 18 - 2 = 242$. If you use 7 $S^4 \times S^4$s, you get 240 instead. | |
| Sep 15, 2015 at 14:42 | comment | added | Tom Mrowka | To see that $p_1(TM)$ is zero note that it is enough to check this on the obvious generators for the 4-dimensional homology (the $S^4$'s). The tangent bundle of $M$ on these spheres is the sum of two copies of $TS^4$ and hence stably trivial so $p_1$ is zero. Nice question by the way. It would be nice if there was a nicer answer. Along this line there is Hirzebruch's question of 24-d manifold with an action of the Monster Group. | |
| Sep 15, 2015 at 14:09 | comment | added | Tom Mrowka | The plumbing should be of the $TS^4$ (disk bundle) so that the self intersection number of the S^4 is 2 (or -2) to get the $E_8$ matrix. That should take care of $\chi$ now. | |
| Sep 15, 2015 at 12:23 | comment | added | Chris Schommer-Pries | ( I meant $L$ is a bundle on $\mathbb{OP}^1$, not $\mathbb{OP}^2$) | |
| Sep 15, 2015 at 12:11 | comment | added | Chris Schommer-Pries | Also for others who might read this and be confused like I was, one key point is that for this bundle L on $\mathbb{OP}^2$, we have $e(L)$ is the generator but $p_2(L) = 6 e(L)$. This follows from the calculation that the map $S^8 \to BSO$ which is the generator in $\pi_8$ induces multiplication by 6 in cohomology (generated by $p_2$ on the RHS). This is also why on String manifolds there is a $p_2/6$ class. | |
| Sep 15, 2015 at 12:08 | comment | added | Chris Schommer-Pries | This is really really great, thank you! I have lots of questions. When you say the Milnor pluming, I take it you mean the pluming on the $E_8$-graph, yes? I am confused about two points. The first is: did you really mean 8 copies of $S^4 \times S^4$? or did you mean 7 copies? When I tried to compute the Euler characteristic of M I am off by 2, which could mean I made an arithmetic mistake, or that we really want 7 copies. More importantly, I see how the argument works if we know that $p_1(TM) = 0$. Can you remind me how we can see that? | |
| Sep 15, 2015 at 9:52 | history | edited | user80296 | CC BY-SA 3.0 |
improved grammar
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| Sep 15, 2015 at 9:37 | history | edited | user80296 | CC BY-SA 3.0 |
added 74 characters in body
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| Sep 15, 2015 at 9:30 | review | First posts | |||
| Sep 15, 2015 at 9:32 | |||||
| Sep 15, 2015 at 9:29 | history | answered | user80296 | CC BY-SA 3.0 |