In euclidean n-space, it's easy to show that given a set $S$ of radius $< r$, the $a$-neighbourhood of $S$ is a ball, for any $a \geq 2r$.

Proof: Let $S$ be contained in $B_r(y)$, $y \in \mathbb{R}^n$. Note that if $a \ge 2r$ then $ B_r(y) \subset Nbd_a(S)$. Let $z\in Nbd_a(S) \backslash B_r(y)$. Consider the triangle with vertices $z$, $y$ and $s$ with $s\in S$. The length of the edge $yz$ is greater than $r$ which is greater than the length of the edge $ys$. It follows that the angle at $z$ is less than $\pi/2$ (less than $\pi/3$, in fact), which means that points on the edge $yz$ near $z$ are closer to $s$ than $z$ is, which implies that these points are also in $Nbd_a(S)$. Hence $Nbd_a(S)$ is star-shaped with respect to $y$.

I'd like a result for a metric $PL$ manifold, of the form:

Theorem: For an metric $PL$ manifold $M$, there is some $\epsilon > 0$ such that every subset $S$ of radius $r$ and all $a$ with $2r \leq a \leq \epsilon$, the neighbourhood $Nbd_a(S)$ is homeomorphic to a ball.

Can someone provide a proof?

  • $\begingroup$ The corresponding theorem for compact Riemannian manifolds just asks that $\epsilon$ be a bit smaller than the injectivity radius, and that on $\epsilon$ balls the geodesic coordinates don't distort angles much. $\endgroup$ – Scott Morrison Jul 12 '10 at 22:24
  • $\begingroup$ Is there any particular reason that PL should be important here, or is it just that in the situation you care about, you have a PL manifold? $\endgroup$ – Dan Ramras Jul 12 '10 at 23:37
  • $\begingroup$ I can think of more than one thing that you might mean by "metric PL manifold". $\endgroup$ – Tom Goodwillie Jul 13 '10 at 0:42
  • $\begingroup$ @Tom, I had in mind a simplicial manifold, with each simplex inheriting a metric from a linear embedding in R^n, although I don't really mind polyhedra more complicated that simplices. $\endgroup$ – Scott Morrison Jul 13 '10 at 2:19
  • $\begingroup$ @Dan, this will be used as a lemma inside a larger argument, where for reasons of technical convenience we've already decided to work in the PL setting. In the current draft, in fact, we nonchalantly give the Riemannian argument. $\endgroup$ – Scott Morrison Jul 13 '10 at 2:22

If I understand the question right, the answer is no. Make a triangulated $2$-manifold with Euclidean metrics on the simplices, such that the total angle around some vertex is very small. Let $S$ consist of two points, both of which are the same small distance $D$ from that vertex and (subject to that) as far from each other as possible. If $2r$ is the distance between the points then for a pretty big range of values of $a$ the union of $a$-balls centered at these two points is topologically an annulus.

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  • $\begingroup$ Hmmm, sure. What if I asked that there's some $\epsilon$, as above, and also some $k>0$, so the condition holds for $rk \leq a \leq \epsilon$? I think that would also suffice for my application. $\endgroup$ – Scott Morrison Jul 13 '10 at 4:13
  • $\begingroup$ In that same example, no matter what $a$ and $r$ are, there will always be a set (in fact, a point) of radius $<r$ whose $a$-neighborhood is topologically an annulus. $\endgroup$ – Tom Goodwillie Jul 13 '10 at 4:55
  • $\begingroup$ So essentially Tom's example is when your "Riemannian manifold" has a conic singularity and no lower bound on the injectivity radius. $\endgroup$ – Willie Wong Jul 13 '10 at 10:36
  • $\begingroup$ Hence replacing 2r with kr for some constant k. The idea is that k would be determined by the "narrowest" cone point. $\endgroup$ – Scott Morrison Jul 13 '10 at 15:45
  • $\begingroup$ Let me make sure of the quantifiers. You seem to want a single $\epsilon$ and $k$ such that for every $r$, for every $a$ such that $rk\le a\le\epsilon$, the $a$-nbhd of a set having radius $r$ is a ball? But this fails in the cone example: go close enough to the singularity so that a "ball" of radius $a$ hits itself, and choose $r<a/k$, and let your set be a point. $\endgroup$ – Tom Goodwillie Jul 13 '10 at 17:03

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