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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 $\forall x \in X$,

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

- Geometric interpretation of connection forms, torsion forms, curvature forms, etc
- How can a $C^1$-continuous surface have infinite curvature?
- Curvature of planar implicit curves
- maximum curvature of 2D Cubic Bezier
- Curves with constant curvature and constant torsion
- Centre of the circle

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

Thanks for your help

- Relations between curvature and area of simple closed plane curves.
- Why is the Riemann curvature tensor the technical expression of curvature?
- How to show the volume preserving mean curvature flow preserve volume?
- Is a uniquely geodesic space contractible? II
- Distinguishing the Cylinder from a “full-twist” Möbius strip
- Shortest Path and Minimum Curvature Path - implementation
- Are there simple examples of Riemannian manifolds with zero curvature and nonzero torsion
- Integral of a curve with respect to its curvature?
- Injectivity radius of Exponential and curvature
- Curves with constant curvature and constant torsion

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:

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:

- Imaginary-Order Derivative
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- How to solve this equation for x? $0 = (x+k)e^{-(x+k)^2}+(x-k)e^{-(x-k)^2}$
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