MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I recently came across the following question while working on some problems on manifolds with lower Ricci curvature bounds.

Given $n$ does there exist a large $R>0$ with the following property:

Suppose $M^n$ is a closed Riemannian manifold of diameter $=1$ such that for any $p$ in the universal cover $\tilde M$ the ball $B(p,R)$ is contained in a homeomorphic copy of $\mathbb R^n$ which in turn is contained in $B(p,R+\frac{1}{10})$. Then $M$ is aspherical.

In our situation we had extra geometric assumptions which allowed us to prove asphericity but I've been wondering if the above holds as is. I suspect not but I could not construct a counterexample.

share|cite|improve this question
Welcome to Math overflow! – Igor Rivin Oct 4 '11 at 20:34
Thanks! Igor Belegradek and Anton Petrunin mentioned this site to me and it looked like a good place to ask this kind of question. – Vitali Kapovitch Oct 4 '11 at 20:55

Sorry, I realized that this is not an answer. I am constructing a Riemannian 3-manifold $M$ with small diameter and nontrivial $\pi_2 M$ such that for any point $p$ in the univesal cover $\widetilde M$ there is a sequence of open embeddings $$B_R(p)\hookrightarrow\mathbb R^3\hookrightarrow B_{10\cdot R}(p),$$ and its composition coinsides with the inclusionn $B_R(p)\hookrightarrow B_{10\cdot R}(p)$.

I hope that it still might be interesting.

Take a the surface of an $(2{\cdot}R+\tfrac1{100})$-long and $\varepsilon$-thin cylinder $C$ with caps in $\mathbb R^3$ (further $C$ is called sausage). Think of it as a surface of revolution around $X$-axis. Idetify points on $C$ along the folloing equivalence relation $$x\sim y\ \ \ \text{if}\ \ \ x-y=(\tfrac12,\varepsilon,0).$$

This way you obtain a $2$-dimensional CW-complex, say $W=C/\sim$ with $\pi_1 W=\mathbb Z$ and nontrivial $\pi_2 W$. If you equip $W$ with the induced intrinsic metric then then $\mathop{\rm diam} W\approx \tfrac12$ and any $R$-ball in the universal cover $\widetilde W$ is contractible in a ball of radius $R+\tfrac1{10}$. (A rough reason: $\widetilde W$ glued from a sequence of sausages. If a ball of radius $R$ intersects one sausage then it can not contain it all, but the ball of radius $R+\tfrac1{10}$ with the same center containa at least one of the ends, which makes possible to shrink the intrsection to a point.)

Now $W$ can be embedded into $\mathbb R^3$, it seems that thickening and then doubling produces a $3$-dimensional manifold $M$ with the property described above. (Fortunately or unfortunately, any ball in $\widetilde M$ contains a closed curve such that to shrink it one has to go about $R$-far out of the ball.)

share|cite|improve this answer
Hi Tosha, This looks interesting but I don't quite understand the example. Perhaps I'm misunderstanding it but it seems to me that your space $Z$ is obtained from a long thin $S^2$ by gluing a long interval at the top to one at the bottom with a shift of $1/2$. Then $\tilde Z$ looks to be an infinite string of $S^2$'s where consecutive spheres are glued to each other by long intervals and $\pi_1(Z)=\mathbb Z$ acts by translations. Isn't this right? The $R$-balls in this space don't look contractible to me. – Vitali Kapovitch Oct 5 '11 at 3:10
Ups, now it is corrected. – Anton Petrunin Oct 5 '11 at 3:55
Ups again --- this is not an answer. – Anton Petrunin Oct 5 '11 at 7:10
Yes, I can see that in this example it holds that for any $p\in \tilde W$ we have that $B_R(p)$ is contractible in $B_{10R}(p)$ but it does not work for $B_{R+1/10}(p)$. This is weaker than I wanted but maybe the example can be modified? Also, I can't visualize what happens after you thicken and double W to make it a manifold. Can you clarify that part, please? P.S. when I was typing my original question I could see a preview of LaTex formatting while I was typing That was very convenient. Is there something like this for comments? – Vitali Kapovitch Oct 5 '11 at 14:59
@Vitali, for $W$ everything is OK (but cotractibility instead of $\mathbb R^n$). I get into problem when I pass to doubling. – Anton Petrunin Oct 5 '11 at 15:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.