Timeline for Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres?

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
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.
Jan 20, 2016 at 11:16	history	edited	Chris Schommer-Pries	CC BY-SA 3.0	fix error (8 to 7)
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
Sep 15, 2015 at 9:37	history	edited	user80296	CC BY-SA 3.0	added 74 characters in body
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

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