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The first time I heard about the asymptotic cone, I ingenuously thought "Well... the asymptotic cone of $\mathbb Z^2$ minus the origin is $\mathbb R^2$ minus the origin". At that point somebody said me "Are you crazy? The asymptotic cone is always a geodesic space! The asymptotic cone of $\mathbb Z^2$ minus one point is $\mathbb R^2$".

This is seems very strange to my (certainly questionable) intuition. Trying to imagine myself moving further and further from $\mathbb Z^2$ minus the origin, I would say the hole gets smaller and smaller... why does it disappear? I would say that it gets infinitesimal...

Question: Is there any similar notion of asymptotic cone, taking into account, in some sense, the holes?

Why I am interested in this?

First of all because I am human being and, even if I am male, I am quite curious. Second, because I think that there might be some interesting relation with the A-theory of graphs. For instance, the fundamental group of $\mathbb Z^2$ minus one point is equal to $\mathbb Z$ in A-theory. The $A$-theory is really constructed as the discretization of the standard homotopy theory and then, conversely, it is possible that the homotopy theory of the continuous version of a discrete object (as the asymptotic cone in my intuition should be) is related to the A-theory of the object itself. Third because, if the classical asymptotic cone is useful in many cases, maybe a more precise notion as the one I am trying to imagine, can be even more useful... (or maybe not, since such a notion cannot be a quasi-isometric invariant).

Thanks in advance,

Valerio

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4  
Answering your title question: To get to the other side. –  Will Jagy May 10 '12 at 18:18
    
I thought it is because our state no longer has money for this, so having ultralimits that fill holes all by themselves is our contribution to the road repair. In view of this, the idea of making ultralimits that do not fill the holes, strikes me as counterproductive. –  Misha May 11 '12 at 3:30

2 Answers 2

when you take Gromov-Hausdorff limits you have to restrict to complete spaces if you want a well-defined notion because otherwise the limits are not unique. After all the Gromov-Hausdorff distance between $\mathbb R^2$ and $\mathbb R^2\backslash \{pt\}$ is zero. so no such constructions can see infinitesimal holes.

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Thanks a lot! However I don't think this is the point, since $\mathbb R^2$ and $\mathbb R^2$ minus a ball are quasi-isometric and then they have the same asymptotic cone. I think I have understood my mistake: I was imaging $\mathbb Z^2$ drawn on a blackboard and I was moving at in infinity orthogonally. This is the mistake, one has to imagine to move at infinity in some sense inside the space. Now I understand the definition: if I want to construct a subset of $\mathbb Z^2$ having isoperimetric cone with a hole I probably should take out larger and larger squares further and further away. –  Valerio Capraro May 10 '12 at 18:00
    
I don't understand the remark about $R^2$ and $R^2$ minus a point being quasi-isometric. I don't see it being relevant. my point is that G-H convergence can't tell apart limit spaces which have zero G-H distance between them. so if you don't restrict to complete limit spaces you have no notion of a well defined limit space. –  Vitali Kapovitch May 10 '12 at 22:38
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Vitali, asymptotic cones of a space $X$ is always complete, even if $X$ is incomplete, see e.g. Bridson-Haefliger, lemma I.5.53. The point is that not all asymptotic cones are GH limits, e.g. rescaling of a hyperbolic plane by as small constant has no GH limit as volume grows too fast, but its utralimit (i.e. asymptotic cone in this case) exists no matter what, and is a metric tree. –  Igor Belegradek May 11 '12 at 2:50
    
Igor, I realize this. As you say it's a rather trivial consequence of the definition that an ultralimit of any sequence of metric spaces is always complete. but as far as I could understand from the original question that was not the point of the confusion of the OP. In the example he gave rescaled sequence of $Z^2$'s converges to $R^2$ on the nose without any ultralimits. –  Vitali Kapovitch May 11 '12 at 3:38

Although asymptotic cones are not Gromov-Hausdorff limits of spaces in general, they are always complete spaces. See the survey Druţu, Cornelia Quasi-isometry invariants and asymptotic cones. Internat. J. Algebra Comput. 12 (2002), no. 1-2, 99–135. Also if two spaces are quasi-isometric, then the asymptotic cones (corresponding to the same ultrafilters and the same scaling constants) are bi-Lipschitz equivalent, hence homeomorphic.

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