# Smoothness of boundary of $r$-neighborhood of convex core

The boundary of an $$r$$-neighborhood of the convex core of hyperbolic $$n$$-manifold is smooth, by page 73 of Hyperbolic Manifolds and Kleinian Groups. The authors does not provide a proof for this fact. This does not look very obvious as the $$r$$-neighborhood of a convex set in a Riemannian manifold may not be smooth. One example is that: the convex hull of two intersecting geodesic segment in hyperbolic 2-space, one going from (0,1) to $$\infty$$, one is a quarter circle from $$(0,1)$$ to $$(1,0)$$. Its $$r$$-neighborhood has 3 parts. By observing the curvature of each part we can see that its boundary is not smooth.

If it is not smooth, what is the maximal possible regularity of the boundary. This paper https://link.springer.com/article/10.1007/BF02921327 talks about how to perturb the boundary of a cusp and tube in a controlled way so that it becomes smooth. Would similar argument work in this case?

If it is true by some special property of the boundary of the convex core for hyperbolic manifolds, I would also wonder whether this is true for negatively curved manifold with pinched sectional curvature $$\kappa \in [−b,−1]$$, where $$b > 1$$. The classical paper Geometrical finiteness with variable negative curvature by Bowditch does not seem to discuss the regularity issues of the boundary of the $$r$$-neighborhood of the convex core. Any reference for this issue, preferably with a proof would help. In general the $$r$$-neighborhood of a convex set with piecewise smooth boundary is not smooth.

Just coming across this 3 years too late, but thought I'd mention for posterity that I don't think the boundary of the r-nbhd is smooth. It's not so hard to prove it's $$C^{1,1}$$. Intuitively, it's $$C^1$$ because every point on the boundary of the r-nbhd is on the boundary of a r-ball centered in the convex core. That entire r-ball is contained in the r-nbhd of the convex core, and the r-nbhd is also convex, so you have a support plane, and hence you have a well defined tangent plane at that point. It's also not hard to show that these tangent planes vary lipschitzly. Walter, "Some analytical properties of geodesically convex sets", is a good reference in a more general Riemannian setting.
You're probably not going to do much better than $$C^{1,1}$$, though. For instance, in $$3$$-dimensions, if the bending lamination of a convex core boundary component has a closed leaf, locally the r-nbhd of the convex core is just going to look like the r-nbhd of the intersection of a pair of half-spaces, and its boundary won't be $$C^2$$, sort of like in the example you suggested. Probably the boundary of the r-nbhd is $$C^2$$ exactly when the convex core has totally geodesic boundary, which is unusual.
• The $C^{1,1}$ claim is written down in Section 2 of arxiv.org/abs/1908.09814. One extra comment is that the boundary of an $r$-neighborhood $N_r(C)$ of a convex set $C$ in a Hadamard manifold has reach $\ge r$. To show that $r$-ball rolls freely inside $N_r(C)$ note that each point $x$ of $\partial N_r(C)$ lies on the boundary the $r$-ball centered at the nearest point projection of $x$ to $C$, and that ball lies in $N_r(C)$. Aug 25, 2022 at 15:09
• Oh, very nice, I hadn't seen that reference! It's true that it's a little hard to extract the $C^{1,1}$ claim from the Walter paper, since he really proves that it's $C^2$ a.e. I was imagining you also get local bounds on the second derivatives by relating them to first derivatives of the closest point projection, as in Walter's previous paper, which gives a local lipschitz bound on first derivatives. Aug 25, 2022 at 15:36
• Almost everywhere is not good enough. I think, the $C^{1,1}$ conclusion goes back to Federer. Aug 25, 2022 at 16:27
• Hmm, let me see. In Walter's previous paper ("On the Metric Projection onto Convex Sets in Riemannian Spaces") he proves in Thm 2 that the closest point projection to a convex set C is locally lipschitz. Then in Cor 1 he writes the gradient of the distance function to C in terms of the closest point projection, in a way that shows that this gradient is also locally lipschitz. So presumably that means that the boundary of the r-nbhd is the preimage of a regular value under a $C^{1,1}$ map, and hence $C^{1,1}$? Aug 25, 2022 at 19:17
• I think showing that gradient is locally Lipschitz is enough to conclude that the map is $C^{1,1}$. I find Federer's paper hard to read but I think the point is that being of positive reach is invariant under diffeormorphisms, so the kind of Riemannian structure we use is irrelevant, see the corollary on p.57 in Bangerts' "Sets of positive reach". And positive reach is equivalent to $C^{1,1}$. Aug 25, 2022 at 20:01