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edited Oct 29 2010 at 3:57 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
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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edited Mar 3 2010 at 18:44 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
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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edited Dec 30 2009 at 20:30 |
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edited Dec 12 2009 at 20:46 |
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edited Dec 12 2009 at 19:02 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
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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edited Dec 11 2009 at 20:32 |
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edited Dec 11 2009 at 17:56 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
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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edited Dec 7 2009 at 1:06 |
Diameter of m-fold cover(
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edited Dec 6 2009 at 3:08 |
Diameter of m-fold cover(
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
This is a modification of a problem of A. Nabutovsky. 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 $\mathop{diam}\tilde M\le 2m\cdot 2(m-1)\cdot \mathop{diam} M$ 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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edited Dec 4 2009 at 18:52 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$
Comments:
This is a modification of a question problem of A. Nabutovsky. 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 $\mathop{diam}\tilde M\le 2m\cdot \mathop{diam} M$ is trivial. We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.
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edited Dec 4 2009 at 18:31 |
Diameter of m-fold Riemannian cover
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edited Dec 4 2009 at 3:33 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?$$\ \ \ \ \ \ \ (*)$$
Comments:
This is a modification of a question of A. Nabutovsky. You can reformulate it for compact length metric space --- no difference. The answer is YES if the cover is regular (but that is not as easy as one might think). The estimate $\mathop{diam}\tilde M\le 2m\cdot \mathop{diam} M$ is trivial. The estimate We have equality in $\mathop{diam}\tilde M\le m\cdot \mathop{diam} M$ is exact (*)$ for covers of $S^1$ and for many some covers of figure-eight.
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edited Dec 4 2009 at 2:46 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam}\tilde mathop{diam} M\ ?$$
Comments:
This is a modification of a question of A. Nabutovsky. You can reformulate it for compact length metric space --- no difference. The answer is YES if the cover is regular (but that is not as easy as one might think). The estimate $\mathop{diam}\tilde M\le 2m\cdot \mathop{diam}\tilde mathop{diam} M$ is trivial. The estimate $\mathop{diam}\tilde M\le m\cdot \mathop{diam}\tilde mathop{diam} M$ is exact for covers of $S^1$ and for many covers of figure-eight
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m-fold Riemannian cover
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam}\tilde M\ ?$$
Comments:
This is a modification of a question of A. Nabutovsky. 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 $\mathop{diam}\tilde M\le 2m\cdot \mathop{diam}\tilde M$ is trivial. The estimate $\mathop{diam}\tilde M\le m\cdot \mathop{diam}\tilde M$ is exact for covers of $S^1$ and for many covers of figure-eight
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