# Convex spherical neighborhood in Alexandrov areas Answer

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Convex spherical neighborhood in Alexandrov areas

There is a variational gauge to learn whether or not geodesic balls are convex in finite dimensional launch Alexandrov area $$(X,d)$$. Otherwise, if $$X$$ is compact, then my respond says nothing.

If $$X$$ is related, let $$M^*=M^*(x_0)$$ breathe the clique of noncompact maximal minimizing geodesic rays. For $$lambda in M^*$$, let $$h_lambda(x)$$ breathe exclusive horofunction centred at $$lambda$$ and satisfying $$h_lambda(x_0)=0$$. Then $$b(x,y):=sup_{lambda in M^*} |h_lambda(x)-h_lambda(y)|$$ defines a probably degenerate metric on $$b:Xtimes Xto [0,+infty)$$ satisfying $$b(x,y)leq d(x,y)$$ for each $$x,y in X$$. From the $$d$$-geodesic convexity of $$h_lambda$$, we discover the $$b$$-metric balls of capricious radius are $$d$$-geodesically convex subsets of $$X$$. The $$d$$ geodesic convexity of the horofunction $$h_lambda$$ is well-known property of Alexandrov metrics $$d$$.

If $$(X,d)$$ has the property that equality is obtained all over the place $$b(x,y)=d(x,y)$$ we resolve all $$d$$-metric balls in $$X$$ are $$d$$-geodesically convex.

So in your illustration, we necessity additional data on the clique of innumerable minimizing rays $$M^*=M^*(x_0)$$ at a basepoint $$x_0$$. In monotonous phrases, for each pair of factors $$x,y$$, you necessity learn whether or not there exists a horosphere $$H_lambda$$ which separates $$x,y$$. In such illustration $$b$$ is nondegenerate distance duty with $$b(x,y)=0$$ iff $$x=y$$.

If equality is attained all over the place and $$b=d$$, then $$b$$ is Alexandrov. But if the inequality is strict $$b , then I do not know whether or not the metric is once more Alexandrov, and this seems fascinating query.

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