Discrete exotic 4-manifolds. 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 (?).   
 A: 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 ${\mathbb R}^4$. 
