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This question might be very easy, but I am little confused by the Gromov-Hausdorff convergence.

My situation is the following: I have a fixed set $X$ which is finite or countable; on it I have locally finite metrics $d_n$ and $d$. For the application that I have in mind, I've found that a good notion of convergence of the sequence of spaces $(X,d_n)$ to the space $(X,d)$ would be described by the uniform convergence of $d_n$ to $d$. Therefore, I am wondering if this convergence is equivalent to Gromov-Hausdorff's convergence.

Question: Are the following statements equivalent:

  1. $d_n\rightarrow d$ uniformly
  2. For any point $x\in X$, the sequence of pointed locally compact metric spaces $(X,d_n,x)$ converges to $(X,d,x)$ in Gromov-Hausdorff sense?

It would sound strange for my intuition if they turn out to be different, but I am in trouble to write down a proof, basically because I am quite new in the definition of Gromov-Hausdorff convergence and I am pretty confused/scared by all these isometric embeddings one should consider.

Thank you in advance for any help,

Valerio

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2 Answers 2

No: 2 does not imply 1.

Let $X$ be the set $\mathbb{Z} \times \mathbb{Z}$. Define a metric $d$ on $X$ by $$ d((x, y), (x', y')) = 2|x-x'| + |y-y'| $$ and define a metric $e$ on $X$ by $$ e((x, y), (x', y')) = |x-x'| + 2|y-y'|. $$ Then $d \neq e$, but for each $x \in X$, the pointed metric spaces $(X, d, x)$ and $(X, e, x)$ are isometric in a basepoint-preserving way (rotate by 90 degrees about $x$).

Now consider the sequence $$ d, e, d, e, \ldots $$ of metrics on $X$. It does not converge uniformly (or even pointwise). However, the sequence $$ (X, d, x), (X, e, x), (X, d, x), (X, e, x), \ldots $$ of pointed metric spaces does converge in the Gromov-Hausdorff sense: since they're all pointedly isometric, the distance between each element of the sequence and the next is always zero.

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thank you very much! –  Valerio Capraro Dec 5 '11 at 15:55
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1) does imply 2). An alternative equivalent definition of Hausdorff-Gromov distance is as follows. A correspondence between two sets $X, Y$ is a subset of $X\times Y$ which intersects each horizontal and each vertical fiber. Then, given two metric spaces $(X,d)$, $(Y,e)$, their Hausdorff-Gromov distance is

$$ \frac12 \inf_R \max_{(x,y), (x',y')\in R} |d(x,x')-e(y,y')|, $$

where the infimum is over all correspondences (This is Theorem 7.3.25 in the book of Burago-Burago-Ivanov).

When the underlying space is the same, the diagonal correspondence gives that the Hausdorff-Gromov distance between $(X,d)$ and $(X,e)$ is at most half the uniform distance between the metrics $d$ and $e$ (likewise for pointed metric spaces). In particular, 1) implies 2).

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perfect! Thank you very much. –  Valerio Capraro Dec 5 '11 at 13:47
    
At least in the diagonal case, it is totally trivial to construct a space giving that distance. The idea is that you make two copies of the space, one with each metric, and allow "jumping" between corresponding points, adding a fixed distance to those paths. The formula tells you how to chose the fixed distance such that it is never efficient to jump and then jump back, meaning that the new distances are the same as the old. –  Will Sawin Dec 5 '11 at 16:46
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