Return to Answer

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 depending only on the scale $\varepsilon$. This can be done since Delaunay nets are constructed greedily. Then use some finite simplexwise procedure 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.

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.

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, using the Delaunay triangulation argument in Ian Agol's answer, every PL sphere of geometry bounded by $L$ can be filled with a PL ball of geometry bounded by $C(L)$. The procedure is: smooth it out, take a Riemannian filling, take a Delaunay triangulation of the filling, and then use some finite procedure to interpolate between the original boundary and the new boundary which may have gotten subdivided. (This part of the argument is just a sketch, but only deals with smoothable manifolds.)

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. 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. 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.

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 depending only on the scale $\varepsilon$. This can be done since Delaunay nets are constructed greedily. Then use some finite simplexwise procedure 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.

Here's a stab at a proof that works for non-smoothable manifolds. 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, using the Delaunay triangulation argument in Ian Agol's answer, every PL sphere of geometry bounded by $L$ can be filled with a PL ball of geometry bounded by $C(L)$. The procedure is: smooth it out, take a Riemannian filling, take a Delaunay triangulation of the filling, and then use some finite procedure to interpolate between the original boundary and the new boundary which may have gotten subdivided. (This part of the argument is just a sketch, but only deals with smoothable manifolds.)

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. 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 this subcomplex 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 surface. This surface is a union of patches which are the $D(S)$'s from the previous step, therefore it has bounded geometry. We can replace $C\Sigma$ with a 3-ball of geometry bounded by $C(d,3)$.
4. Keep going like this. At the end of the procedure, the resulting manifold has bounded geometry.

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, using the Delaunay triangulation argument in Ian Agol's answer, every PL sphere of geometry bounded by $L$ can be filled with a PL ball of geometry bounded by $C(L)$. The procedure is: smooth it out, take a Riemannian filling, take a Delaunay triangulation of the filling, and then use some finite procedure to interpolate between the original boundary and the new boundary which may have gotten subdivided. (This part of the argument is just a sketch, but only deals with smoothable manifolds.)

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. 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. 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.