# Local complexity of triangulations

According to the introduction in

Cooper, D.; Thurston, W. P., Triangulating 3-manifolds using 5 vertex link types, Topology 27, No. 1, 23-25 (1988). ZBL0656.57004.

It is known that, for any dimension $$n$$, there is a finite set of link types such that every $$n$$-manifold has a triangulation in which the link of each vertex is in this set.

(I assume the statement is about PL manifolds.)

What is a proof of or a reference to this known result?

Edit. There is a related discussion by Florian Frick here, but he only gets a bound on the valence of "ridges" (codimension 2 faces). I do not see how to generalize his sketch to faces of higher codimension. (Actually, I cannot follow his sketch.)

• Cooper is alive and I believe responds to emails. Ask him? And let us know the answer, please! – Sam Nead Dec 26 '20 at 15:00
• @SamNead, I do believe my answer is an answer! Cooper is a colleague of mine and I have asked him whether what he intended is similar. – Fedya Dec 26 '20 at 18:46
• @Fedya: You probably mean “the answer”, since it is “an answer” regardless. – Moishe Kohan Dec 26 '20 at 21:02
• I mean that it answers the question, not that it's what Daryl would have intended in 1988. – Fedya Dec 27 '20 at 2:40
• Sorry, sorry. I meant (and should have written) "And let us know the outcome, please!" I mean, don't share anything private, but do let us know whatever history you find out... I am giving up now. Internetting is hard. – Sam Nead Dec 27 '20 at 13:05

I think that there's a fairly useless argument using Riemannian geometry for smoothable PL manifolds (so in dimensions at most $$7$$ this should work). Assume that $$M$$ is a PL manifold which admits a compatible smooth structure (but note that the smooth structure need not be unique). Take a Riemannian metric on the manifold, and rescale it so that the curvature is nearly zero, and the injectivity radius is large everywhere. We may also assume that each point is contained in a large normal neighborhood, say for some radius $$R>>1$$. Then take a net of points in the manifold, a maximal collection of points with distance $$> \epsilon$$ between every pair, $$\epsilon <<1$$. Then the $$2\epsilon$$ balls about these points cover the manifold. Now take the Delaunay triangulation with respect to this covering. Because the metric is nearly flat on the scale $$2\epsilon$$, this will give a triangulation of bounded degree.

There are many details that need to be checked for this argument to work, but I expect that this is the sort of construction Cooper-Thurston may have had in mind. Maybe to convince oneself that this might work, the rescaled limit about any point for a Riemannian metric (the asymptotic cone) will be Euclidean, and the net will be locally finite, hence the Gromov-Hausdorff limits of such triangulations will be $$\epsilon$$-net Delaunay triangulations of Euclidean space, which have bounded degree. Each edge will be between vertices of distance at most $$2\epsilon$$, so the $$3\epsilon$$ ball about every vertex will contain the disjointly embedded epsilon balls about the vertex and its neighbors. Hence the degree ($$+1$$) will be at most $$3^d$$ where $$d$$ is the dimension.

• This sounds plausible for smooth manifolds (and maybe this is what Thurston had in mind, especially since the construction of triangulations via Delaunay construction appears in "Word Processing in Groups"). However, there are PL manifolds which are not smoothable (actually, your link mentions such examples). – Moishe Kohan Dec 23 '20 at 19:30
• Yes, this only works up to dimension 7 in general I guess. – Ian Agol Dec 23 '20 at 20:25

Here's a stab at a proof that works for non-smoothable manifolds, which seems to be related to Florian Frick's sketch. I'm going to call manifolds with at most $$L$$ simplices incident to a vertex "of geometry bounded by $$L$$" or just "of bounded geometry". The operator $$*$$ denotes the join.

First: Lemma. Every PL sphere of geometry bounded by $$L$$ can be filled with a PL ball of geometry bounded by $$C(L)$$.

Proof of lemma. Use the Delaunay triangulation argument in Ian Agol's answer. Smooth out the sphere, take a Riemannian filling, and take a Delaunay triangulation of the filling such that the Delaunay triangulation on the boundary is a subdivision of the original triangulation. You can use a specific sequence of subdivisions of each simplex $$\Delta^i$$ depending only on the scale $$\varepsilon$$. This can be done since Delaunay nets are constructed greedily. Then use some fixed extensions of these subdivisions to $$\Delta^i \times [0,1]$$ to interpolate between the original boundary and the new, subdivided boundary.

Now take a PL manifold $$M^d$$ equipped with an arbitrary PL triangulation.

1. Barycentrically subdivide once. The resulting complex has a vertex for each face of $$M$$. The link of the vertex corresponding to a $$k$$-face $$\sigma$$ looks like $$\partial\Delta^{d-k} * \operatorname{BarySub}(\operatorname{link}(\sigma))$$. The star of the vertex corresponding to every $$d$$-face and $$(d-1)$$-face already has geometry bounded by $$C(d)$$.
2. Now look at the star of each vertex corresponding to a $$(d-2)$$-face. This looks like $$\partial\Delta^{d-2} * CS$$ where $$S$$ is a circle with many edges. Using the lemma, we can replace the interior of this subcomplex by $$\partial\Delta^{d-2} * D(S)$$, where $$D(S)$$ is a bounded geometry filling of $$S$$. Now the subcomplex $$D(S)$$ has geometry bounded by $$C(d,2)$$.
3. If we now look at the star of each vertex corresponding to a $$(d-3)$$-face, it looks like $$\partial\Delta^{d-3} * C\Sigma$$ where $$\Sigma$$ is a triangulated 2-sphere. This surface is a union of patches which are the $$D(S)$$'s from the previous step, therefore it has bounded geometry. Using the lemma, we can replace $$C\Sigma$$ with a 3-ball of geometry bounded by $$C(d,3)$$.
4. Keep going like this. At the $$k$$th stage, the $$k$$-skeleton of the dual CW complex to the original triangulation is subdivided into a triangulation with geometry bounded by $$C(d,k)$$. At the end of the procedure, the resulting manifold has bounded geometry.
• Thank you, I will try to fill in the details. – Moishe Kohan Dec 25 '20 at 12:11
• I just made the argument somewhat more explicit, hope this helps. – Fedya Dec 26 '20 at 18:28