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Question 1. Let $\epsilon > 0$ and $V > 0$. Is there always a complete connected Riemannian manifold $M$ with $$ \operatorname{diam} M < \epsilon\quad\text{ (small diameter)} \quad \text{and} \quad \operatorname{vol}M > V\quad\text{(large volume)}? $$ In other words, can we construct worlds with room for arbitrarily many planets, but where any two planets are arbitrarily close to each other?

Note that $M$ has to be compact by Hopf-Rinow. I'm almost certain that the answer is 'yes', but I'm having a hard time finding an explicit sequence $M_1, M_2, M_3,\ldots$ such that $\operatorname{diam}(M_n)$ goes to zero while $\operatorname{vol}(M_n)$ goes to infinity.

Question 2. (Fixed dimension) Is there a sequence of complete connected Riemannian manifolds $M_1, M_2, M_3, \ldots$ with $$ \dim M_n = m, \quad \lim_{n \to \infty}\operatorname{diam} M_n = 0, \quad \text{and} \quad \lim_{n \to \infty}\operatorname{vol}M_n = \infty $$ for some fixed $m \in \mathbb{N}$.

For this one, I have less intuition, but I'm more inclined towards 'no'. Note that for $m = 0, 1$, this is clearly impossible. For $m = 2$ my intuition is very strongly in favour of 'no'.

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    $\begingroup$ By itself, diameter does not control volume. See e.g. mathoverflow.net/a/314653/1573. $\endgroup$
    – Igor Belegradek
    Commented Aug 22, 2021 at 16:14
  • $\begingroup$ @Igor So a the connected sum of very many spheres of ever smaller radius would answer the question? $\endgroup$
    – Harry Wilson
    Commented Aug 22, 2021 at 17:43
  • $\begingroup$ @HarryWilson: yes, the union of a large number of unit spheres connected by thin nearly cylindrical tubes answers the question, which is a special case (not even a duplicate) of the one I linked to. $\endgroup$
    – Igor Belegradek
    Commented Aug 22, 2021 at 17:48

1 Answer 1

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The answer to both questions is positive, even in dimension 2. Take a round sphere of diameter $\epsilon$, and make many, say $N$ little holes in it. Then take $N$ spheres of diameter $\epsilon$ and make one little hole in each. Then glue these $N$ spheres to the first sphere along the boundaries of the holes. The diameter of the resulting surface is about $3\epsilon$ while the volume (area) is approximately $(N+1)\epsilon$. Now make $\epsilon$ as small as you wish, while $N>V/\epsilon$.

Here is another construction. Take a round sphere $S$ of diameter $\epsilon$, think of it as the Riemann sphere, and consider the $N$-fold ramified covering $S_1\to S,\;z\mapsto z^n$. Here $S_2$ is also a sphere, and equip it with the pullback of the metric from $S$. Then the solume of $S_1$ will be $N\epsilon$ while diameter is at most $2\epsilon$. If a smooth metric is desirable, approximate it with a smooth one.

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  • $\begingroup$ Are your upper bounds of the diameter that obvious? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 23, 2021 at 6:03
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    $\begingroup$ For the first example, yes. Any two points $x,y$ must be contained in two spheres $S_x,S_y$. Suppose $S_x,S_y$ meet the central sphere at $p_x,p_y$, then $dist(x,y)\le dist(x,p_x)+dist(p_x,p_y)+dist(p_y,y)\le 3\epsilon$. $\endgroup$
    – user7868
    Commented Aug 23, 2021 at 7:56
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    $\begingroup$ caveat on the ramified covering : the pullback metric is not smooth ! (Though it can be smoothed without changing the diameter and the volume too much). $\endgroup$
    – Thomas Richard
    Commented Aug 23, 2021 at 9:23
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    $\begingroup$ @Thomas Richard: so approximate it with a smooth one. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 23, 2021 at 12:04
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    $\begingroup$ @მამუკა ჯიბლაძე: for the second example it is obvious too. In the complex coordinate, go straight to $0$ and then straight to the destination. This takes at most $2\epsilon$. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 23, 2021 at 12:08

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