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 $x,y\in M$, $\|x-y\|<s\implies d_M(x,y)<\epsilon$. 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.

  • $\begingroup$ 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". $\endgroup$
    – HJRW
    Commented Aug 9, 2020 at 13:44
  • 1
    $\begingroup$ 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... $\endgroup$
    – HJRW
    Commented Aug 9, 2020 at 13:48
  • $\begingroup$ 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. $\endgroup$
    – rvdaele
    Commented Aug 10, 2020 at 12:47
  • $\begingroup$ 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 $\pi r$. $\endgroup$
    – rvdaele
    Commented Aug 10, 2020 at 12:51
  • $\begingroup$ 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. $\endgroup$
    – rvdaele
    Commented Aug 10, 2020 at 12:54

2 Answers 2


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:

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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 $M$.

  • $\begingroup$ 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. $\endgroup$
    – rvdaele
    Commented Aug 15, 2020 at 13:44

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