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I've found lots of (more or less precise) definitions of the Alexandrov curvature, but I'm mainly interested in that of "Alexandrov curvature bounded below". Could anyone give me that or give me a reference?

Thanks in advance, Valerio

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From "A.D. Alexandrov spaces with curvature bounded below", Burago, Y. and Gromov, M. and Perel'man, G., Russian Mathematical Surveys, 47, 1992, p.5:

A locally complete space $Μ$ with intrinsic metric is called a space with curvature $\ge k$ if in some neighbourhood $U_x$ of each point $x \in M$ the following condition is satisfied: (D) for any four (distinct) points $(a, b, c, d)$ in $U_x$ we have the inequality $\tilde \angle bac + \tilde \angle bad + \tilde \angle cad \le 2\pi$. If the space $Μ$ is a one-dimensional manifold and $k > 0$, then we require in addition that diam $M$ does not exceed $\pi/\sqrt{k}$.

(Here $\tilde \angle pqr$ is the angle at the vertex $\tilde q$ of the triangle $\tilde \triangle pqr$ on the two-dimensional "$k$-plane" of curvature $k$, which has side lengths $|pq|$, $|qr|$, $|rq|$.)

They say in the Introduction, p.1,

We are talking, roughly speaking, about spaces with an intrinsic metric, for which the conclusion of Toponogov's angle comparison theorem is true (although only in the small).

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  • $\begingroup$ Thanks very much. So it is well possible that a Banach space has curvature strictly bigger than 0, for example $l_2^p$ with $p>2$, isn't it? I intuitively thought that Banach spaces were flat in any sense! $\endgroup$ Apr 12, 2011 at 17:43
  • $\begingroup$ @Valerio, If Banach space has curvature $\ge 0$ or $\le 0$ then it is a Hilbert space. $\endgroup$ Apr 12, 2011 at 18:10
  • $\begingroup$ Many thanks, Anton, for the advice. I've actually founded the book on the web. I'm going to read that as soon as possible, since my research is just going towards the metric geometry. And I have to say that a love this kind of things! $\endgroup$ Apr 12, 2011 at 19:33
  • $\begingroup$ I would suggest to start with the last chapter in "Metric Geometry" of Burago--Burago--Ivanov. There is also an easy to read introduction to curvature bounded below written by Shiohama, but it is hard to find. – $\endgroup$ Apr 13, 2011 at 0:16

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