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I want to prove some Alexandrov space M is parabolic cone X x R.Since Alex has no Riemannian metric,so how to do?Is there any (triangle) formula about the relation of distance of two points in M and distance of the projection points in X?

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The metric of warp products $F\times_f B$ is fiber independent. It means that if $(x,a),(y,b)\in F\times_f B$ and $(x',a),(y',b)\in F'\times_f B$ then $$|x-y|_F=|x'-y'|_F \ \ \ \Rightarrow\ \ \ |(x,a)-(y,b)| _{F\times_f B} = |(x',a)-(y',b)| _{F\times_f B}$$ If I remember right, this statement is due to S. Alexander and R. Bishop.

If you apply this to parabolic cone, you get a nice formula for distances, which depent only on $|x-y|_F$ and two real values $a$ and $b$.

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@Anton Petrunin:Nice formula.But I don't know how to get the formula.BGp(Alex 1) gives the elliptic cone formula,but no parabolic cone. – jiangsaiyin Oct 29 '12 at 13:14

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