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I have a question regarding the definition of CAT(0) spaces. I am using the following definition:

$X$ complete metric space is CAT(0) if $\forall z,y \in X$, $\exists m \in X$ such that $\forall x \in X$,

$$ d(x,m)^2\leq \cfrac{d(x,y)^2 + d(x,z)^2 }2 - \cfrac{d(y,z)^2}4$$

What is the geometric meaning of this inequality? I guess there is one but I can't find it.

Thanks for your help

User
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1 Answers1

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This is not the definition the theory began with; the comparison of geodesic triangles came first. The four-point condition you see has been distilled from that, even shedding the geodesicity assumption. Since a complete CAT(0) space is geodesic, I recommend drawing geodesics between the points you have. The picture will look like this:

parallelogram

I added an (imagined) point $W$ on the continuation of geodesic $XM$ to emphasize that the condition is a form of the Parallelogram Law. Indeed, multiplying both sides by $4$ an moving $d(y,z)^2$ to the left, you will see the parallelogram law with inequality sign.

What is the meaning of inequality replacing equality? It allows the curvature to be negative, not necessarily zero. (With equality we'd have a flat CAT(0) space.) Negative curvature means geodesic triangles are thin, and therefore the midpoint $m$ can be closer to $x$ than it would be in a flat space:

curvature

  • Thank you for your answer. I now see the meaning. Is it possible that the triangle has edges that curves out instead of inside the triangle in a CAT(0) space? – User Jul 14 '14 at 16:14
  • @user132951 "Curving in/out" are just intuitive concepts, they do not have a precise meaning in a metric space, because we don't have "straight lines* to compare to. Actually geodesics as on the picture are "straight lines". But if you state the question in precise terms: could a triangle in CAT(0) be "fat" instead of "thin", then the answer is no. This is exactly what the original definition of CAT(0) is. –  Jul 14 '14 at 16:35