I am currently trying to understand "Curvature bounded below: a definition a la Berg--Nikolaev" by Nina Lebedeva and Anton Petrunin.

They start with a complete, intrinsic metric and space $X$ and say that one can assume that $X$ is geodesic (a shortest path exists between two arbitrary two points) otherwise one passes to the Ultraproduct of X. My question is what is the Ultraproduct of $X$ and why is it geodesic?

Wikipedia says that for metric spaces the Ultraproduct is somewhat more special and one considers the Ultralimit... I found something about Ultralimits in Metric spaces of Non-positive curvature in particular the following 3 results:

  1. Every Ultralimit of metric spaces is complete.
  2. The Ultralimit of a sequence of metric spaces is a length space if every metric space in the sequence was a length space.
  3. The Ultralimit of a sequence of metric spaces is a geodesic space if every metric space in the sequence was a geodesic space

However they do not explain, why I am allowed to consider a geodesic space if I start with just an intrinsic one.

Thanks in advance!


If you have a sequence of paths $\gamma_i$ (between $p$ and $q$) of length converging to the infimum then the equivalence class of the sequence $\langle \gamma_i:i\in\mathbb N\rangle$ will give a minimizing path in the ultrapower.

  • $\begingroup$ Exactly: in the 2nd construction that I mentioned. $\endgroup$ – Moishe Kohan May 2 '16 at 13:48

I do not know for sure, but here are two possible interpretations. Let $(X,d)$ be a metric space, $\omega$ a nonprincipal ultrafilter on ${\mathbb N}$. The standard definition of the ultraproduct $$ X^*:= \prod_{n\in {\mathbb N}} X/\omega $$ of $X$ as a set is the quotient of $$ \prod_{n\in {\mathbb N}} X $$ by the following equivalence relation: $(x_n)\sim (y_n)$ iff these two sequences are equal on a subset contained in $\omega$. Now, you need to define a metric on this ultraproduct. The most natural thing to do is to define a metric with values in the field of nonstandard real numbers ${}^*{\mathbb R}$: $$ d^*((x_n), (y_n))= [d(x_n, y_n)] $$ where $[t_n]$ is the nonstandard real number represented by the sequence $(t_n)$ of real numbers. For such a nonstandard metric one can define all the usual concepts like completeness, geodesic property, etc.; it is easy to see that the ultraproduct $(X^*, d^*)$ is a geodesic metric space in this sense. Suppose, however, that you want to have a traditional metric, taking values in real numbers. You need a projection from ${}^*{\mathbb R}_+$ to $[0, \infty]$ and this is what the ultralimit of sequences accomplishes: $$ lim_{\omega}: {}^*{\mathbb R}_+ \to [0, \infty]. $$ Then the only natural thing I can think of is the ultralimit of the constant sequence $(X, d, x)$ of pointed metric spaces as the ultraproduct of $(X,d)$: It picks up a certain subset of $(X^*, d^*)$ and takes its further quotient, using the map $lim_\omega$.

Then the results that you quoted show that such ultraproduct is again complete and geodesic.

Hope it helps. I will add the tags "nonstandard analysis", "model-theory" and "logic" to your question, maybe logicians on this site can provide further insight.

  • $\begingroup$ It may be helpful to clarify your notation to indicate that it is the product that is being factored "modulo the ultrafilter" rather than $X$ itself. $\endgroup$ – Mikhail Katz May 2 '16 at 9:29
  • $\begingroup$ @Mikhail Katz: You are right of course, but this abuse of the notation is standard. $\endgroup$ – Moishe Kohan May 2 '16 at 13:59
  • $\begingroup$ I just meant adding parentheses, not anything more involved (like clarifying that you are factoring by an ideal rather than by an ultrafilter). $\endgroup$ – Mikhail Katz May 2 '16 at 14:37

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