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There is the question, because when we consider the Gromov-Hausdorff distance, we must fix the metric, so we use the natural metric induced from the embedding $\mathbb{S}_n \to \mathbb{R}^{n+1}$. Is it possible for us to compute the Gromov-Hausdorff distance $d_{G-H}(\mathbb{S}_n,\mathbb{S}_m)$ for two different spheres $\mathbb{S}_n$ and $\mathbb{S}_m$, $m\neq n$?

For example if we want to calculate $d_{G-H}(\mathbb{S}_2,\mathbb{S}_3)=\inf_{M,f,g}d_{M}(\mathbb{S}_2,\mathbb{S}_3)$, where $M$ ranges over all possible metric space and $f:\mathbb{S}_2\to M$ and $g:\mathbb{S}_3\to M$ range over all possible isometric (distance-preserving) embeddings.

At least we can embed $\mathbb{S}_2$,$\mathbb{S}_3$ into $\mathbb{R}^3$ in a canonical way. This will lead to a upper bound: $d_{G-H}(\mathbb{S}_2,\mathbb{S}_3)\leq \sqrt{2}$. And in general case we have $d_{G-H}(\mathbb{S}_m,\mathbb{S}_n)\leq d_{G-H}(point,S_m)+d_{G-H}(point,S_n)\leq 2,\forall 0\leq n\leq m$. But it is difficult to get a lower bound control for me. Because we need to take the inf in all possible metric spaces $M$. Especially I conjecture $d_{G-H}(\mathbb{S}_m,\mathbb{S}_n)\geq \lambda_{m,n}\frac{m-n}{m},\forall 0\leq n\leq m$, where $\liminf_{m,n\to \infty}\lambda_{m,n}>0$.

I only know the knowledge of Gromov-Hausdorff from Peterson's Riemann Geometry. Unfortunately there is not enough information to compute the Gromov-Hausdorff distance, so this problem may be very stupid, I will appreciate any pointer.

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    $\begingroup$ Please check your question for typos. $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '17 at 3:58
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    $\begingroup$ @Mark Sapir,Appreciate for help!I am reading the article you point out,it seems this article mainly focus on investigating the Gromov-Hausdorff limit space of a sequence of hyperbolic group equipped with modified G-H metric defined in 2.A with some special condition to ensure the limit space exists.and take a sequences corvarage to the limit space,the hyperbolic property and some other thing is stayed by the process of take limit. $\endgroup$ – Hu xiyu Oct 28 '17 at 4:26
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    $\begingroup$ Your conjecture would imply that the GH distance is unbounded. But it's clearly bounded, since the GH distance of any sphere to a point is equal to 2 (when the sphere is endowed with the restriction of Euclidean distance, as you seem to assume, or $\pi$ when endowed with geodesic distance) and hence the GH distance between any two spheres is $\le 4$. $\endgroup$ – YCor Oct 28 '17 at 7:29
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    $\begingroup$ Clearly from standard embeddings we get $d_{GH}(S_n,S_m)\le\sqrt{2}$ for all $n,m\ge 0$. Would it be reasonable to simply conjecture that it's an equality whenever $n\neq m$? $\endgroup$ – YCor Oct 28 '17 at 7:59
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    $\begingroup$ And in addition there is the obvious remark (I didn't first notice) that $d(S_n,S_m)\le 1$ for all $n,m$ (form a metric space $S_n\sqcup S_m$ with the given distance on each component and points in different components being at distance 1. This works for any two metric spaces of diameter $\le 2$. So my previous comment should rather ask whether $d_{GH}(S_n,S_m)=1$ for all $n\neq m$. $\endgroup$ – YCor Oct 28 '17 at 10:08

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