# Gromov Hausdorff distance to tubular neighborhood

Let $$M$$ be a compact path metric space in $$\mathbb{R}^d$$, and for $$\sigma>0$$, $$M_\sigma:=\{y\in\mathbb{R}^d:\min_{x\in M}\|x-y\|\leq\sigma\}$$ the $$\sigma$$-tube around $$X$$ in $$\mathbb{R}^d$$. I consider both $$M$$ and $$M_\sigma$$ metric spaces with respect to the shortest path metric (geodesic, not necessarily Euclidean distances) induced by $$\mathbb{R}^d$$, with possibly distinct intrinsic dimensions. We are furthermore given constants $$s,\epsilon>0$$, such that for $$x,y\in M$$, $$\|x-y\|. Is there any bound we can provide on the Gromov Hausdorff distance $$d_{GH}(M, M_\sigma)$$ in terms of $$s,\epsilon$$, and the diameter of $$M$$, when $$\sigma$$ is sufficiently small?

The tubular neigborhood can significantly alter the metric, e.g., the tubular neighborhood of a nearly closed circle can suddenly include the circle itself. However, I suspect that such information would be encoded by $$s$$ and $$\epsilon$$, and that for $$\sigma$$ sufficiently small (according to these parameters), the path from $$x$$ to $$y$$ in $$M_\sigma$$ travels `near' the path from the (not necessarily unique) projections of $$x$$ and $$y$$ on $$M$$, and the length of these paths will then be similar.

I could believe that similar problems have been investigated before, but I don't find any helpful references. It would be great if someone could point out some possible directions on this problem.

• Your intuition that such problems have been investigated before is certainly correct. You could try looking up the notions of "quasi-isometry" and "coarse embedding".
– HJRW
Commented Aug 9, 2020 at 13:44
• More specifically, I don't quite understand the question. It's clear that $๐_{๐บ๐ป}(๐,๐_\sigma)$ is bounded above by $\sigma$, so for $\sigma$ sufficiently small compared to $s,\epsilon$ and the diameter of $M$, that appears to provide the bound you are looking for. But presumably I misunderstood something...
– HJRW
Commented Aug 9, 2020 at 13:48
• It is certainly easy to show that there exists a correspondence under which $d_{M_\sigma}$ is not much larger than $d_M$ (more specifically, at most 2$\sigma$). However, showing that $d_M$ is not much larger than $d_{M_\sigma}$ is another story. Commented Aug 10, 2020 at 12:47
• Consider a circle $C$ with radius $r$ and a very small angle $\theta$. Now discard any segment in the circle defined by $\theta$. This is now a space homeomorphic to the line, with diameter $r(2\pi - \theta)$. When we let $\sigma = r\sin(\theta / 2)$, i.e., the length of the chord corresponding to the discarded segmented, then $M_\sigma$ includes the entire circle $C$, and its diameter (distance between the two furthest points) is now closer to half the original diameter of $M$. Hence, $d_{GH}(M, M_{\sigma})$ is then at least close $\pi r$. Commented Aug 10, 2020 at 12:51
• However, when $\sigma$ is less then two times the length of the chord defined by $r\sin(\theta/2)$, $M_{\sigma}$ would not include the entire circle $C$, and their metrics of $M$ and $M_\sigma$ should now be much closer. The roles of $s$ and $\epsilon$ are here fulfilled by the length of a chord and the length of a segment in the circle, respectively. Commented Aug 10, 2020 at 12:54

I think I have figured this out. More specifically, it should hold that $$d_{GH}(M, M_\sigma) \leq \max\left\{2\sigma, \left(\frac{\epsilon}{s-2\sigma}-1\right)(\mathrm{diam}(M)+2\sigma)+\epsilon\right\},$$ whenever $$\sigma < s/2$$.

Sketch of the proof:

Define the correspondence $$C$$ as $$(x,y)\in C\leftrightarrow y\in \overline{B}_{\mathbb{R}^d}(x,\sigma)$$ Clearly it holds that $$(x,y),(x',y')\in C$$ implies that $$d_{M_\sigma}(y,y')\leq d_M(x,x')+2\sigma.$$ For the more difficult direction, take $$0<\delta\leq s-2\sigma$$ and split up the path from $$y$$ to $$y'$$ in $$M_\sigma$$ into $$k$$ parts of length at most length $$s - 2\sigma - \delta$$. This can be done with $$k\leq \frac{d_{M_\sigma}(y,y')}{s - 2\sigma - \delta}+1$$ segments. Each of these segments corresponds to a segment to a segment in $$M$$ with length at most $$\epsilon$$. We find that $$d_M(x, x')\leq \left(\frac{d_{M_\sigma}(y,y')}{s - 2\sigma - \delta}+1\right)\epsilon$$. Now subtract $$d_{M_\sigma}(y,y')$$ from both sides, and bound $$d_{M_\sigma}(y,y')$$ in the right hand side by $$\mathrm{diam}(M_\sigma)\leq\mathrm{diam}(M)+2\sigma$$. Finally, let $$\delta\rightarrow 0$$.

With the clarification given in the comments, what you are asking is if there is a bound on $$d_{GH}(M, M_\sigma)$$ which tends to $$0$$ as $$\sigma\to 0$$. This question has negative answer. An example is given by $$M$$ which is the comb space:

The thing is that for each $$\sigma=1/n$$, $$M$$ contains a pair of points $$p=(0,1), q=(\frac{1}{n},1)$$ such that $$d_{M_\sigma}(p,q)=1/n,$$ while $$d_M(p,q)= 2+ \frac{1}{n}$$. The existence of these pairs of points prevents the GH-convergence $$M_\sigma\to M$$. (The space satisfies other conditions in your question: $$diam(M)=3$$, one can take $$s=\sqrt{2}, \epsilon=3$$.)

One way to define the GH distance is via distortion of bisurjective correspondences:

Definition. Let $$A, B$$ be compact metric spaces and $$R\subset A\times B$$ be a bisurjective correspondence meaning that its projection to both $$A$$ and $$B$$ is surjective: For every $$a\in A$$ there is $$(a,b)\in R$$ and for every $$b\in B$$ there is $$(a,b)\in R$$. Define the distortion of $$R$$ by:
$$dis(R):= \sup_{(a,b), (a',b')\in R} |d(a, a')- d(b,b')|.$$ Then $$d_{GH}(A,B)= \frac{1}{2}\inf_R dis(R)$$ where the infimum is taken over all bisurjective correspondences as above. Up to a uniform factor (which is irrelevant for our purposes), $$d_{G}$$ can be defined using $$\epsilon$$-surjective maps: $$\inf \{dis(f), f: A\to B \ \hbox{is \epsilon-surjective}\},$$ where $$dis(f)= \sup \{ |d(f(a), f(a'))- d(a,a')| : a, a'\in A\}$$ and $$f$$ is $$\epsilon$$-surjective if each $$b\in B$$ is within distance $$\epsilon$$ from some $$f(a)$$.

In other words, if $$dis(f)\le \epsilon$$ then $$f$$ is a $$(1,\epsilon)$$-quasiisometry: $$d(a,a')-\epsilon \le d(f(a), f(a'))\le d(a,a')+\epsilon,$$ Therefore, a sequence of compact metric spaces $$M_n$$ converges to a metric space $$M$$ if and only if there is a sequence of $$(1,\epsilon_n)$$-quasiisometries $$f_n: M_n\to M,$$ which are $$\epsilon_n$$-surjective and $$\lim_{n\to\infty}\epsilon_n=0$$.

What you get in your setting is different: The inclusion map $$f: M\to M_\sigma$$ defines (when $$\sigma\le s/3$$) a quasi-isometry $$M\to M_\sigma$$:

$$M$$ is $$\epsilon$$-dense in $$M_\sigma$$ and $$f$$ satisfies (for all $$a, a'\in M$$) $$\frac{\sigma}{\epsilon}d_M(a, a') - \sigma\le d_{M_\sigma}(f(a), f(a'))\le d_M(a, a').$$ The multiplicative (Lipschitz) factor $$\frac{\sigma}{\epsilon}\ne 1$$ in the LHS makes all the difference. This is the difference between the GH distance and quasi-isometries mentioned by in Henry's comment. The attempt to estimate (from above) the GH distance made in your post will also result in a map with such multiplicative factor $$\ne 1$$ and that just is not good enough. One can define a measure of closeness between compact metric spaces using quasi-isometries instead of the GH-distance. I do not know if it is useful for anything. If you are content with, say, $$C^2$$-smooth compact submanifolds $$M$$ instead of general compact subspaces then, indeed, you get GH-convergence $$M_\sigma\to M$$.

For more on the topic, see this question and

Burago, D.; Burago, Yu.; Ivanov, S., A course in metric geometry, Graduate Studies in Mathematics. 33. Providence, RI: American Mathematical Society (AMS). xiv, 415 p. (2001). ZBL0981.51016.

There are several other questions one can ask along the lines of your post. The more interesting of these is:

Do not fix the dimension of the ambient Euclidean space, but assume that the extrinsic diameter of $$M$$ is $$\le D$$. Is there a uniform upper bound on $$\liminf_{\sigma\to 0+} d_{GH}(M, M_\sigma)$$ in terms of $$s, \epsilon$$ and $$D$$? This question also has a negative answer but examples are harder; they use the comb space as one of the building blocks.

The reason to use the extrinsic diameter is that if the intrinsic diameter is bounded by $$D$$ then, trivially, $$d_{GH}(M, M_\sigma)\le D+\sigma,$$ which you find uninteresting. If the extrinsic diameter of $$M$$ is bounded by $$D$$ and the ambient dimension $$n$$ is fixed, one again obtains an upper bound on the intrinsic diameter of $$M$$ in terms of $$D$$ and $$n$$.

Edit. Here is a correct phrasing of your question:

1. Suppose that $$M\subset {\mathbb R}^n$$ is a rectifiably-connected subset, such that, when equipped with the intrinsic path-metric $$d_M$$, $$M$$ is compact. Does it follow that the family of neighborhoods $$M_\sigma$$ of $$M$$ (also equipped with the intrinsic path-metrics) converge to $$M$$ in the GH topology?
1. Suppose that $$M$$ is a compact connected $$C^1$$-smooth submanifold in $${\mathbb R}^n$$. Is $$M$$ still compact with respect to its intrinsic path-metric?
1. Suppose that $$M$$ is a compact connected $$C^2$$-smooth submanifold in $${\mathbb R}^n$$. Can one estimate $$d_{GH}(M, M_\sigma)$$ in terms of intrinsic and extrinsic differential-geometric invariants of $$M$$?

Now, this question has positive answer:

1. Consider the identity embeddings $$f_\sigma: M\to M_\sigma$$. Then each $$f_\sigma$$ is $$\sigma$$-surjective and 1-Lipschitz. Thus (see the interpretation of GH distance above in terms of maps), we just need to prove that $$\lim\sup_{\sigma\to 0+} \sup_{p,q\in M} |d_M(p,q)- d_{M_\sigma}(p,q)|=0.$$ A proof is by contradiction: If this limit is $$\delta>0$$, then (by compactness!) there are sequences $$p_i, q_i\in M$$ converging to $$p, q\in M$$ (with respect to the topology given by its path-metric) such that $$\lim_{i\to\infty} (d_{M_{1/i}}(p_i,q_i) - d_M(p_i,q_i))=\delta.$$ Let $$c_i: [0,1]\to M_{1/i}$$ be nearly geodesic paths connecting $$p_i$$ to $$q_i$$. These paths can be taken uniformly Lipschitz (with respect to the Euclidean metric) since the diameter of $$M_{1/i}$$ is $$\le diam(M)+ 2$$. By applying Arzela-Ascoli theorem combined with the Lebesgue dominant convergence theorem, we obtain a limit path $$c$$ in $$M$$ connecting $$p$$ to $$q$$ whose length is $$\le d_M(p, q)-\delta$$. A contradiction.

2. For $$C^2$$-smooth submanifolds, it is a classical fact proven in pretty much every Riemannian geometry textbook that for a $$C^2$$-smooth Riemannian metric, the manifold topology agrees with the topology given by the Riemannian distance function. For a $$C^1$$-smooth submanifold, you can argue instead as follows. It suffices to show that $$(M, d_M)$$ is sequentially compact. By the compactness of $$M$$ (with the subspace topology), it suffices to show that if $$p_i\to p$$ in the subspace topology of $$M$$, then $$d_M(p_i, p)\to 0$$. Writing the induced Riemannian metric in local $$C^1$$-coordinates, it becomes merely continuous but this is enough. (Actually, one needs even less than continuity.) The proof now becomes just a calculus computation:
$$\lim_{i\to\infty} \int_{0}^{\epsilon_i} \sqrt{g(c_i'(t), c_i'(t))}dt \le \lim_{i\to\infty} K \epsilon_i =0,$$ where $$g$$ is a continuous Riemannian metric on a domain in $${\mathbb R}^k$$, $$c_i: [0, \epsilon_i]\to {\mathbb R^k}$$ are arc-length parameterizations of line segments (emanating from the origin) of length $$\epsilon_i$$, satisfying $$\epsilon_i\to 0$$. The constant $$K$$ is an upper bound on the $$g$$-norm of unit vectors in $${\mathbb R}^k$$ near the origin. (Hence, all what you need is that, in the local coordinate, the metric $$g$$ is measurable and locally bounded on unit vectors, where unit is understood with respect the Euclidean norm.)

3. An estimate for $$C^2$$-smooth compact submanifolds can be given in terms of the 2-nd fundamental form (you need it for submanifolds of arbitrary codimension):

If $$\sigma$$ is sufficiently small (less than the normal injectivity radius of $$M$$ in $${\mathbb R^n}$$), you have a well-defined nearest-point projection $$r_\sigma: M_\sigma\to M$$. You need is to estimate the Lipschitz constant $$L$$ of $$r_\sigma$$. The estimate is essentially the same as the one for the circle example: $$L^{-1} \ge 1- \sigma C,$$
where, up to some multiplicative constant depending only on the dimension $$n$$, $$C$$ is the supremum-norm of the 2nd fundamental form of $$M$$. (In the circle example, $$1/C$$ is the radius of the circle.) Thus, for $$p, q\in M$$, you have $$0\le d_M(p, q)- d_{M_\sigma}(p,q)\le CD\sigma.$$ Thus (up to a uniform multiplicative constant depending only on $$n$$), $$d_{GH}(M, M_\sigma)\le CD\sigma,$$ if $$\sigma$$ is less than the normal injectivity radius of $$M$$.

• Thanks for your clear clarification. You are correct that $d_{GH}$ does not converge to 0 under my stated assumptions, for your given example. As you might have supsected, I'm indeed content with more restrictive spaces, such as $C^2$ smooth compact submanifolds. In matter of fact, for my purpose, though these are not the exact spaces I'm working with, it would be sufficient to assume $M$ is a smooth curve in $\mathbb{R}^d$. Do you know any references where your stated fact that GH convergence is achieved for these spaces is proven? I'll look into some of your current references as well. Commented Aug 15, 2020 at 13:44