Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background

I'm using physics terminology because I'm not sure what the right mathematical terminology is, perhaps a simplicial complex?

I'm interested, for various physics reasons, in four manifolds and specifically in their intersection forms. I'm especially interested in the E8 manifold, but if I understand the situation correctly, this manifold is not smooth. I have even read statements that it cannot be triangulated in a certain sense. Obviously I can read the various definitions, but I don't have any intuition for them and I suspect getting the intuition would take me way too far afield. For my physics purposes I would have really liked this manifold to have a nice differential form cohomology, so I'm trying to understand what I can use instead.

Main question

As an example of a simpler structure that I could use, I would be happy with a lattice model of the E8 manifold. By "lattice model" I mean something like the way a large square lattice with periodic boundary conditions is a model of a torus. There is a discrete notion of points, links, and plaquettes so that the various topological properties are correctly captured. For example, I have in essence non-contractible loops and so forth.

Does something like this exist for the E8 manifold i.e. a discrete structure with the right intersection form, or is this impossible?

share|improve this question
    
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. –  David Roberts Nov 1 '12 at 23:24
3  
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. –  Will Sawin Nov 2 '12 at 2:10
    
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. –  Misha Nov 2 '12 at 3:57

2 Answers 2

It might be useful to look at a construction of the E_8 manifold. It is glued from 2 parts, one which is very simple and has a smooth structure and contains the intersection form, and the other one which is very mysterious. The simple one starts from the E_8 form and glues 8 copies of the tangent disc bundle of S^2 in a simple way so that the resulting manifold with boundary has intersection form E_8. The boundary is the Poincare homology sphere P. The mysterious part comes from Freedman's theory which implies that there is a compact contractible topological manifold with boundary P. The E_8-manifold is the union of the result of plumbing and the contractible topological manifold glued along P The second homology of the plumbed part maps isomorphically to the second homology of the E_8-manifold and so you can see the intersection form on the smooth part.

share|improve this answer
    
@Matthias, Is the Freedman's example smoothable? i.e. The closed 4manifold with poincare sphere $P$ as it's boundary. –  J. GE Nov 19 '12 at 22:16

I'm not certain I understand your question, but it sounds like you might be looking for a piecewise-linear (PL) structure on the E8 manifold. Roughly, this means than the manifold can be divided into 0-cells, 1-cells, ... 4-cells, and the local combinatorics of how the cells meet is tame/non-pathological/standard. If that's your question, then the answer is no: in dimension 4, a topological manifold has a PL structure if and only if it has a smooth structure.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.