Timeline for Is there a lattice model of E8 manifold?

6 events

when toggle format	what		by	license	comment
Nov 3, 2012 at 8:42	answer	added	Matthias Kreck		timeline score: 11
Nov 2, 2012 at 3:57	comment	added	Misha		A possible solution is to use a finite simplicial complex $X$ homotopy-equivalent to the E8 manifold. I do not think anybody computed such a complex for E8.
Nov 2, 2012 at 2:10	comment	added	Will Sawin		Your "lattice model" is a description of the torus as a quotient of a simpler manifold, the plane, by a discrete group action without stabilizers, translation by a lattice. The E8 manifold has no such description, because it it simply connected. One can describe it as a 4-cube with the boundary glued together a certain way, but describing that way is difficult and probably can't be done combinatorially.
Nov 2, 2012 at 1:14	answer	added	Kevin Walker		timeline score: 7
Nov 1, 2012 at 23:24	comment	added	David Roberts♦		What does it mean for a combinatorial structure to have an intersection form? Presumably you need some form of Poincare duality. Indeed, 'combinatorial structure' is pretty much synonymous with simplicial complex. But the whole point of the E8 manifold is that it not homeomorphic to a simplicial complex. If you has some combinatorial structure which captured the topological properties, I'm sure you could arrive at a triangulation. This is certainly the case for the 'lattice model' for the torus you describe. But this is just intuition on my part.
Nov 1, 2012 at 22:46	history	asked	Physics Monkey	CC BY-SA 3.0