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Let $M$ be a closed Riemannian manifold. Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$. Is it true that $$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$

• A complete solution is given by S. Ivanov [it can not be marked as accepted due to software limitations].

• This is a modification of a problem of A. Nabutovsky. Here is yet related question about universal covers.

• You can reformulate it for compact length metric space --- no difference.

• The answer is YES if the cover is regular (but that is not as easy as one might think).

• The estimate $\mathop{diam}\tilde M\le 2(m-1)\cdot 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.

• We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.

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Let $M$ be a closed Riemannian manifold. Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$. Is it true that $$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$

• A complete solution is given by S. Ivanov [it can not be marked as accepted due to software limitations].

• This is a modification of a problem of A. Nabutovsky. Here is yet related question about universal covers.

• You can reformulate it for compact length metric space --- no difference.

• The answer is YES if the cover is regular (but that is not as easy as one might think).

• The estimate $\mathop{diam}\tilde M\le 2(m-1)\cdot \mathop{diam} M$ for $m>1$ is trivial.

• We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.

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