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 |
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| Sep 16, 2015 at 0:54 | comment | added | David Roberts♦ | @ChrisSchommer-Pries someone should do the Kummer approach without doing blowup, in the sense of working with the underlying differentiable stack of the orbifold (maybe this doesn't work! But I feel that it should). Then one could see how your case should work. Possibly one needs to work with a 2-stack rather than just a naive orbifold (for instance, the quotient of the torus by the action of a 2-group, for instance the nontrivial symmetric 2-group structure on $(\mathbb{Z}/2)//(\mathbb{Z}/2)$). | |
| Sep 15, 2015 at 15:21 | vote | accept | Chris Schommer-Pries | ||
| Sep 15, 2015 at 12:40 | comment | added | David Roberts♦ | Someone should mention G_2 manifolds at some point, surely... | |
| Sep 15, 2015 at 12:20 | comment | added | Chris Schommer-Pries | Then if we blow up those 256 points using $Y$, we get a manifold $M$ with $\chi(M) = 256 y - 128$. But then 240 is not an integer solution to this equation. So any Kummer-esque construction is going to have to be more involved. | |
| Sep 15, 2015 at 12:18 | comment | added | Chris Schommer-Pries | @DylanWilson I thought about that a little. If we just proceed naively and take the abelian group $T^8$ modulo $a \mapsto -a$ then we get a singular (orbifold) quotient with $2^8 = 256$ singular points. If we cut out balls in a neighborhood of each of these points we get a space $W$ with $\chi(W) = -256/2 = -128$ and $\partial W =$ 256 copies of $\mathbb{RP}^7$. I am not sure how to resolve the singularities. We must choose a manifold $Y$ with boundary $\mathbb{RP}^7$. Let $y = \chi(Y)$ be its Euler characteristic. (cont) | |
| Sep 15, 2015 at 9:29 | answer | added | user80296 | timeline score: 44 | |
| Sep 15, 2015 at 3:17 | comment | added | Sam Gunningham | In particular, the number 24 appears as the kissing number of the most dense lattice in 4-d. This 24 is surely the same 24, but I haven't thought enough about it to tell a coherent story (maybe such a story involves the binary tetrahedral group). Similarly 240 is the corresponding kissing number for a lattice in 8 dimensions... | |
| Sep 15, 2015 at 3:07 | comment | added | Sam Gunningham | Probably you are already aware of all this, but this book review written by John Baez contains a lot of interesting numerology that suggests an interesting answer to your question. math.ucr.edu/home/baez/octonions/conway_smith | |
| Sep 14, 2015 at 17:57 | comment | added | Dylan Wilson | Can you do a kummer-esque construction to the 8-torus and get anything nice? | |
| Sep 14, 2015 at 17:44 | comment | added | Chris Schommer-Pries | The Octionions $\mathbb{O}$, viewed as an 8-manifold, admit the kind of structure I have in mind, as does the 8-torus $T^8 = \mathbb{O}/ \Lambda$, or any (unstably) framed 8-manifold. Also the tautological rank eight bundle on $\mathbb{OP}^1 = S^8$ is also a vector bundle with this sort of structure. They are associated to "$GL_1(\mathbb{O})$-principal bundles" where $GL_1(\mathbb{O})$ is a "non-associative Lie group". This makes sense since each of these bundles admits a trivialization over a cover with no non-trivial triple intersections (so associativity doesn't factor into it). | |
| Sep 14, 2015 at 16:24 | comment | added | მამუკა ჯიბლაძე | @WillSawin Maybe one could utilize triality setup - there are three 8-dimensional spaces (presumably the tangent, "left" half-spinors and "right" half-spinors) with actions of these $I$, $J$, $K$, etc. jumping between them in such a way that it is impossible to come up with a compatible choice of a single system of isomorphisms between these three spaces which would produce associative actions. I realize this is more than vague, I just want to throw into the discussion the triality thing somehow :D | |
| Sep 14, 2015 at 15:52 | comment | added | Will Sawin | If $I,J,K,E,IE,JE,KE$ are maps from the tangent bundle to itself than their multiplication is necessarily associative. | |
| Sep 14, 2015 at 15:45 | history | asked | Chris Schommer-Pries | CC BY-SA 3.0 |