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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 $b<d$ , then I do not know whether or not the metric is once more Alexandrov, and this seems fascinating query.

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