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Let $M$ be an Alexandrov space with curvature $\geqslant -1$. Then we have the following theorem which is often used to perturb a regular point to points we want.

Let $g_0$ be a $\lambda$-superharmonic function which has a minimum $x_0$ strictly inside of an open domain $\Omega\subset M$. Suppose that $x_0$ is regular ($T_{x_0}M =\mathbb{R}^n$) and $g=(g_1,...,g_n)$ be a coordinate system (almost isometry) on this domain of concave functions. Consider the map $q$ $$ x=(x_1,...,x_n)\mapsto \min_{z\in M} g_0(z)+x_1g_1(z)+...+x_n g_n(z)=g_0(z)+\langle x,g\rangle $$ Then there is a $\epsilon>0$ such that $q$ is well defined on $(0,\epsilon)^n$ and $\delta(x)$-expanding map, i.e. $|q(x)q(y)|\geqslant \delta(x)|xy|$

It's well known that we can find an explosion $(d(p_1, \cdot),...,d(p_n, \cdot):M \mapsto \mathbb{R}^n$, it's an almost isometry in an open neighborhood of $x_0$. We know that $d(p_i,\cdot)$ are semi-concave. How can we find a coordinate $g_i$ such that $g_i$ are all concave? And if we choose $g_i=d(p_i,\cdot)$ in the above theorem, can we get the same result?

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The needed construction was given by Perelman and it is described in many places, for example 7.11 in my "Semiconcave functions...".

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  • $\begingroup$ But what I need is that the n stricly concave function $f_i$ forms an almost isometry. $\endgroup$ – mafan Nov 1 '15 at 10:30
  • $\begingroup$ @mafan: Take you favorite distance chart and modify the distance function using the construction. $\endgroup$ – Anton Petrunin Nov 1 '15 at 10:44

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