# Is the following a sufficient condition for asphericity?

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.

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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.)

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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