Exotic $\mathbb{R}^4$ is a differentiable 4-manifold that is homeomorphic but not diffeomorphic to the Euclidean space $\mathbb{R}^4$ and there is a continuum of non-diffeomorphic differentiable structures of $\mathbb{R}^4$.

But are they (PL) triangulated 4-manifolds that would be homeomorphic to $\mathbb{R}^4$, meaning a topological space equipped with the “standard” PL structure obtained by a suitable subdivision of the standard cubulation, but not PL isomorphic to such $\mathbb{R}^4$?

Such discrete n-spaces are known from literature (cf. references to https://en.wikipedia.org/wiki/Discrete_exterior_calculus) and since discrete calculus parallels the smooth theory, I presume that exotic $\mathbb{R}^4$ should be valid also for such a discrete 4-space (?).

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    $\begingroup$ It is very unclear what a "discrete 4-space" can possibly mean. And what would a smooth structure on such a thing mean either. $\endgroup$ – Moishe Kohan Feb 18 at 4:43
  • $\begingroup$ When adding this kind of clarification that is essential to the post, please click on the tiny edit to improve the body of the post instead of commenting. $\endgroup$ – Lee David Chung Lin Feb 18 at 12:39
  • $\begingroup$ Anything you can do to improve the body of the post is welcome, because as it is currently written, the post is unclear. You will often find on this site that comments (such as the comment of @MoisheKohan) are hints for you to improve your post by editing it. $\endgroup$ – Lee Mosher Feb 18 at 21:06
  • $\begingroup$ Still totally unclear: What does it mean for the edge-set to form a lattice? What does it mean to satisfy discrete Stokes' theorem? What would it mean for such a graph to be "exotic"? For manifolds this means to be homeomorphic but not diffeomorphic to some standard target. What can this possibly mean for graphs? At this point, I do not think you have anything close to a question. $\endgroup$ – Moishe Kohan Feb 18 at 22:39

Actually, there is a (nontrivial) theorem somewhere in the vicinity of what you are trying to ask. The right objects to consider are triangulated manifolds (more precisely, PL triangulations, of manifolds, i.e. where links are PL spheres). (PL stands for "piecewise-linear," this is a generalization of the notion of a piecewise-linear function you might be familiar with.) Every triangulated manifold $M$ defines a graph (the 1-dimensional skeleton of the triangulation), but a triangulation actually contains much more information than that graph.

Every manifold admits infinitely many triangulations, provided that it admits one. Thus, the natural notion replacing diffeomorphism for smooth manifolds is the one of a PL homeomorphism. Equivalently, you can say that two triangulations are regarded as "the same" if they admit isomorphic subdivisions. Thus, one defines the notion of PL isomorphic triangulated manifolds. Now, one can ask:

Is there a (PL) triangulated manifold which is homeomorphic to ${\mathbb R}^4$ (equipped with the "standard" PL structure, obtained by a suitable subdivision of the standard cubulation) as a topological space but is not PL isomorphic to such ${\mathbb R}^4$.

The answer to this is indeed positive and the proof uses the result about the existence of exotic differentiable structures on ${\mathbb R}^4$. The proof boils down to nontrivial a theorem (due to Kirby and Siebenmann) that in dimensions $\le 6$ the categories PL and DIFF are naturally isomorphic. In particular, if $M, M'$ are smooth 4-dimensional manifolds which are homeomorphic but not diffeomorphic then they can be PL triangulated so that the resulting PL manifolds are not PL isomorphic. From this, it follows that for every exotic smooth ${\mathbb R}^4$ there exists an exotic PL triangulated ${\mathbb R}^4$.

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  • $\begingroup$ Would you please advise why would a triangulation of a manifold M, defining a graph, contain (strictly speaking) much more information than that graph? $\endgroup$ – Szymon Feb 19 at 14:17
  • $\begingroup$ @Szymon: Consider the complete graph $G$ on 3 vertices. It could be a triangulation of the circle. Or, if you add the 2-dimensional simplex to this graph (whose boundary is the graph $G$), you obtain a triangulation of the 2-dimensional disk. On the other hand, I do not have examples of non-isomorphic triangulations of manifolds of the same dimension, which have isomorphic 1-dimensional skeleta. $\endgroup$ – Moishe Kohan Feb 19 at 15:24
  • $\begingroup$ Indeed, but this 2-dimensional simplex (triangle) would define this (closed?) disc and would contain all the information necessary to reproduce it. It is only the fact that a disc and a triangle are visually different that would support an argument that a triangulation contains more information than the underlying graph. Am I right? $\endgroup$ – Szymon Feb 19 at 16:12
  • $\begingroup$ @Szymon: My point is that 1-skeleton of a simplicial complex contains less information than the entire simplicial complex, that's all. In the example I gave, you have two complexes with the same 1-dimensional skeleton but of different dimension. $\endgroup$ – Moishe Kohan Feb 19 at 16:32
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    $\begingroup$ As I said, if you have any noncompact manifold (and $R^4$ is noncompact and any spaces homeomorphic to it is noncompact as well), you need infinitely many simplices to triangulate it. As a simple example, consider the real line $R$. Each simplex in the triangulation of $R$ is either a vertex or a compact interval. You need infinitely many of these to cover the line. $\endgroup$ – Moishe Kohan Mar 5 at 1:27

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