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](https://mathoverflow.net/questions/8534/diameter-of-universal-cover) 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**](http://en.wikipedia.org/wiki/Covering_space#Deck_transformation_group.2C_regular_covers) (but that is not as easy as one might think).
* The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.
* We have equality in $(\*)$ for covers of $S^1$ and for some covers of figure-eight.



Diameter of m-fold cover
I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.


Split $\gamma$ into $m$ arcs $a\_1,\dots,a\_n$ of equal length: $\gamma=a\_1a\_2\dots a\_m$, $L(a\_i)=L(\gamma)/m>diam(M)$. Let $b\_i$ be a shortest path in $M$ connecting the endpoints of $a\_i$. Note that $L(b\_i)\le diam(M)< L(a\_i)$. I want to replace some of the components $a\_i$ of the path $\gamma$ by their "shortcuts" $b\_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.


To switch from $a\_i$ to $b\_i$, you left-multiply $\gamma$ by a loop $l\_i:=a\_1a\_2\dots a\_{i-1}b\_i(a\_1a\_2\dots a\_i)^{-1}$. More precisely, if you replace the arcs $a\_{i\_1},a\_{i\_2},\dots,a\_{i\_k}$, where $i\_1< i\_2<\dots< i\_k$, by their shortcuts, the resulting path is homotopic to the product $l\_{i\_1}l\_{i\_2}\dots l\_{i\_k}\gamma$.
So it suffices to find a product $l\_{i\_1}l\_{i\_2}\dots l\_{i\_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi\_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.


Now consider the following $m+1$ elements of $\pi\_1(M,p)$: $s\_0=e$, $s\_1=l\_1$, $s\_2=l\_1l\_2$, $s\_3=l\_1l\_2l\_3$, ..., $s\_m=l\_1l\_2\dots l\_m$. Two of them, say $s\_i$ and $s\_j$ where $i< j$, are in the same left coset. Then $s\_i^{-1}s\_j=l\_{i+1}l\_{i+2}\dots l\_j\in H$ and we are done.


