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
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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 bP8=Z/28. I was hoping to use this manifold to prove that 240σ=0, but since the computation that bP8=Z/28 presupposes πst7=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 E8 and the number of S4×S4s. 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 p1 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 S4×S4 that are needed. If I am doing things right the first half of the construction (the 28 Milnor plumbings + cap) has χ=8∗28+2=226 while the connect sum of 8 S4×S4s has χ=8∗2+2=18. The connect sum of these two parts then has χ(M)=226+18−2=242. If you use 7 S4×S4s, you get 240 instead. | |
Sep 15, 2015 at 14:42 | comment | added | Tom Mrowka | To see that p1(TM) is zero note that it is enough to check this on the obvious generators for the 4-dimensional homology (the S4's). The tangent bundle of M on these spheres is the sum of two copies of TS4 and hence stably trivial so p1 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 TS4 (disk bundle) so that the self intersection number of the S^4 is 2 (or -2) to get the E8 matrix. That should take care of χ now. | |
Sep 15, 2015 at 12:23 | comment | added | Chris Schommer-Pries | ( I meant L is a bundle on OP1, not OP2) | |
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 OP2, we have e(L) is the generator but p2(L)=6e(L). This follows from the calculation that the map S8→BSO which is the generator in π8 induces multiplication by 6 in cohomology (generated by p2 on the RHS). This is also why on String manifolds there is a p2/6 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 E8-graph, yes? I am confused about two points. The first is: did you really mean 8 copies of S4×S4? 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 p1(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 |