29
votes
Can one determine the dimension of a manifold given its 1-skeleton?
You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five ...
25
votes
Accepted
Critical dimensions D for "smooth manifolds iff triangulable manifolds"
All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand. The ...
24
votes
Accepted
Acute triangles in "obtuse" polygons?
Take a very obtuse isosceles triangle and chop its acute angles.
17
votes
Accepted
Triangulation with simplices of same volume
I don't know a reference, but I think that we can use a theorem of Moser to get what you want.
Start with any triangulation $\mathcal{T}$ of $M$, whose cardinal is denoted by $k$ and denote by $\...
14
votes
Are there invariants of cell complexes similar to the Euler characteristic?
In a certain restricted setting the answer is that the Euler characteristic is the only such invariant. This is not a complete answer to the question since more general things are allowed. However, it ...
11
votes
Accepted
Are triangulations of compact manifolds PL homeomorphic?
Pachner‘s Theorem is about PL-homeomorphic manifolds, while the Hauptvermutung is asking for a PL-homeomorphism between manifolds which a priori are only homeomorphic.
11
votes
The number of $2$-simplices and the number of $1$-simplices in a $4$-dimensional simplicial complex
Perhaps the best results in this direction are given by Klee's Dehn-Somerville equations:
Klee, V., A combinatorial analogue of Poincar\'e's duality theorem, Can. J. Math. 16, 517-531 (1964). ZBL0134....
11
votes
Accepted
When is a triangulation of sphere two-colorable?
As Fedor Petrov mentions in the comments, a necessary and sufficient condition is that each vertex has even degree. Here is a proof.
Let $T^*$ be the dual graph of the triangulation. That is, the ...
11
votes
Bordism for oriented triangulable manifolds without smooth differentiable structures
I am not sure whether this answers your question in full, but I can suggest an idea. It is based on triangulated, oriented, labelled manifolds. You can drop the orientation and then move from ...
10
votes
Presentations of exotic 4-manifolds
I guess that this is as explicit and low-tech as it gets: if $X$ is a K3 surface (i.e. a non-singular quartic hypersurface in $\mathbb{CP}^3$, with the complex orientation), then $X \# \overline{\...
8
votes
Accepted
Is every triangulation of a Euclidean ball by convex tetrahedra shellable?
I haven't had time to check through the details of the construction, but the example B_3_9_18 found in the proof of Theorem 2 here by Frank Lutz appears to be embeddable in 3-space.
Frank specializes ...
7
votes
Can triangulations (or some related combinatorial structure) distinguish smooth structures on $RP^4$?
$\newcommand{\RP}{\mathbb{RP}}\newcommand{\C}{\mathbb C}\newcommand{\cC}{\mathcal C}$Here's a TFT-style argument
for why it should be possible in principle to use an invariant of triangulations to ...
7
votes
Minimum weight triangulation of lattice points in a circle
I would like to propose a suggestion for finding some asymptotic bound, I think it should be $$\frac{1}{2}\cdot (2+\sqrt{2})\cdot \pi r^2.$$
Namely, the ratio of this number to the actual weight will ...
7
votes
Accepted
Is every (not necessarily PL-) triangulation of a manifold pure, non-branching and strongly-connected?
Suppose that $M$ is a connected $d$-dimensional topological manifold without boundary. (We make the last assumption to simplify matters.) Let $\Delta$ be the given pseudo-triangulation. So the ...
7
votes
Accepted
Refining a triangulation
Suppose that $S$ is a closed, connected surface with negative Euler characteristic. Suppose that $T$ is a triangulation of $S$.
Define "refine" to mean "replace each triangle by four ...
7
votes
Ideal triangulations of $3$-manifolds with "cusps" of genus $\ge 2$
A nice class of examples are the (generalized) triangulations with only one edge. The manifolds obtained by removing an open neighborhood of the vertex have totally geodesic boundary (or a cusp in the ...
7
votes
Three-dimensional triangulations with fixed number of vertices
Here are some observations that might be useful: not an answer, but too long for a comment.
The Euler characteristic of $S^3$ is zero (this holds for any compact three-manifold without boundary). So ...
7
votes
Minimum number of common edges of triangulations
Modifying Alex Ravsky's excellent construction, we can get rid of the $O(k)$ shared horizontal and vertical edges. Four new vertices are placed far enough from the original figure, suitably aligned ...
6
votes
Finite union of closed convex sets is triangulable?
This is not true even for unions of two compact convex sets.
To construct a suitable counterexample, consider the unit disk $$D=\{z\in\mathbb C:|z|\le 1\}$$in the complex plane $\mathbb C$. By $\...
6
votes
Accepted
Properties a triangulation must have in order to describe a manifold
From the comments:
Suppose that $T$ is a triangulation. If all vertex links are PL $(n-1)$-dimensional spheres then the realisation space of $T$ is a PL manifold and thus a topological manifold. (In ...
6
votes
Accepted
Triangulation of a simplex
You are looking for the edgewise subdivision:
Edelsbrunner, H.; Grayson, D. R., Edgewise subdivision of a simplex, Discrete Comput. Geom. 24, No. 4, 707-719 (2000). ZBL0968.51016.
The basic idea is to ...
6
votes
Accepted
Does every triangulable manifold have a vertex-transitive triangulation?
There exists many closed connected hyperbolic 3-manifolds $M$ with trivial symmetry group, and hence trivial mapping class group. $M$ cannot be homeomorphic to a simplicial complex $\tau$ which admits ...
6
votes
Does every triangulable manifold have a vertex-transitive triangulation?
This is to supplement Ian's answer and get examples in all dimensions $\ge 3$.
Let $M={\mathbb H}^n/\Gamma$ be a compact hyperbolic $n$-manifold; suppose that $f\in Homeo(M)$ is a homeomorphism of ...
6
votes
Minimum number of common edges of triangulations
By Theorem 1 from [DGM], for each $n\ge 9$, there exists a geometric thickness-two graph with $n$ vertices and $6n − 19$ edges.
When we partition this graph into two straight-edged planar graphs and ...
5
votes
Eberhard-type theorems for Fisk triangulations?
This is only a partial answer.
First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring ...
5
votes
Accepted
Minimum weight triangulation of lattice points in a circle
The examples shown below for $\ $ $r=4,\ 5,\ $ and $\ 6\ $ illustrate the idea described in Dmitri's answer. Some modifications in the interior plus the choices of edges near the boundary of the ...
5
votes
The number of $2$-simplices and the number of $1$-simplices in a $4$-dimensional simplicial complex
The answer is yes for pure abstract simplicial complexes. Thus we consider a finite set $X$ and a system of subsets $S$ in $X$, such that $S$ is closed under taking subsets, every $s \in S$ has at ...
5
votes
Presentations of exotic 4-manifolds
If you're still interested in this, I've been working on triangulations of exotic 4-manifolds for some time now. I've implemented an algorithm to produce triangulations of 4-manifolds from Kirby ...
5
votes
Accepted
Is every triangulation the projection of a convex hull
I think you are asking whether every triangulation is a regular triangulation. The answer is no. A common example of a non-regular triangulation is this:
5
votes
Minimum number of common edges of triangulations
This question was considered in On representations of some thickness-two graphs by Hutchinson, Shermer, and Vince. They define a graph to be doubly linear if it is the union of two straight-edged ...
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