4
$\begingroup$

I'm confused about various definitions of triangulations and piece-wise linearity. I read, for example, on wikipedia

"..the question of whether all topological manifolds have triangulations is an open problem, though it is known that some do not have piecewise-linear triangulations "

Could you please check whether I understand a couple of concepts correctly?

(1) A triangulation of any topological space $X$ is just a homeomorphism $K\to X$ of a simplicial complex $K$ and $X$. (However, in the context of polyhedra in $\mathbb{R}^n$, "triangulation of a polyhedron" usually refers to subdivision of the polyhedron to geometric simplices in $\mathbb{R}^n$.)

(2) Each smooth manifold has a triangulation (even a piece-wise smooth triangulation).

(3) A piecewise linear map from a simplicial complex $K$ to $\mathbb{R}^n$ is a map such that for some subdivision $K'$ of $K$, $f$ is (affine) linear on each simplex of $K'$.

(4) A PL manifold is a manifold such that the transition maps between charts $h_U(U\cap V)\to h_V(U\cap V)$ are piece-wise linear; for this to make sense, we need to subdivide an open set in $\mathbb{R}^n$ into an infinite number of simplices. (?)

(5) Most common manifolds, such as spheres $S^n\subseteq\mathbb{R}^{n+1}$ are PL manifolds. All smooth manifolds are PL manifolds; this means that there exists a PL atlas generating the same topology as the original smooth atlas (?).

(6) In the wikipedia sentence at the beginning, do they mean that it is an open problem whether each topological manifold admits a triangulation, but it is known that not each topological manifold is a PL manifold (in other words, there might not be a PL atlas generating the topology of the manifold)? Or does it say something else?

(7) A "combinatorial manifold" (see, for example, Bryant, PL topology) is a simplicial complex such that the link of $p$-simplex is either the boundary of a $(n-p)$-simplex, or an $(n-p-1)$-simplex. Each homeomorphism between a combinatorial manifold and a topological manifold $M$ determinaes a PL structure on $M$.(?)

(8) In Rourke-Sanderson (Introduction fo PL topology), they have the notion of polyhedra in $\mathbb{R}^n$. Cell complexes define a polyhedra and any cell complex can be subdivided into a simplicial complex consisting of geometric simplices in $\mathbb{R}^n$. In this context, the intersection of any two cell complexes is a cell complex again and any two simplicial subdivisions of a cell complex have a common refinement.

(9) On the other hand, two triangulations of a manifold do not have a common sub-triangulation in general (this is the Hauptvermutung). While in RS, they have a positive results for triangulations of polyhedra in $\mathbb{R}^n$, a similar construction might be impossible for triangulations of manifolds which can be given by "wild" (although homeomorphisms) maps $K\to X$.

Thanks for the feedback.

$\endgroup$

1 Answer 1

4
$\begingroup$

Part 4 is wrong: A PL manifold is a topological manifold together with a PL structure, which is an atlas with PL transition maps. No need for infinite subdivisions (but different transition maps require different subdivisions of course.

Part 5 is also somewhat oddly stated. You already explained that each smooth manifold admits a smooth triangulation, which is moreover combinatorial in the sense of part 7.

Part 6 is correct but outdated: it is known since 80s that there are topological 4-manifolds which are not triangulable. Last year Manolescu proved the same in all dimensions.

For part 7 you can just say that the categories of PL manifolds and combinatorial manifolds are isomorphic.

$\endgroup$
2
  • $\begingroup$ Thanks a lot! Just for part 4: aren't charts of an atlas maps from open to open sets -- and if yes, how can you finitely subdivide an open set? $\endgroup$ Jun 26, 2014 at 4:06
  • $\begingroup$ @PeterFranek: In the PL setting you use charts with nonempty interior so that the image in $R^n$ is a polyhedron. $\endgroup$ Jun 26, 2014 at 14:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .