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$g_{ij}(x)\in L^\infty(\mathbb{R}^n, M^{n\times n})$ is a metric in $\mathbb{R}^n$ satisfying $\lambda |x|^2\leq g_{ij}x^ix^j\leq \Lambda |x|^2$($\lambda>0$&$\Lambda>0$)

Can we find a Lipschitz homorphism $x=f(y)$ of $\mathbb{R}^n$ with $f(0)=0$, s.t. the new metric $h_{kl}(y)=g_{ij}(f(y))\frac{\partial f^i}{\partial y^k}\frac{\partial f^j}{\partial y^l}$ satisfying Gauss lemma, That is： $$h_{kl}(y)y^k=y^l$$

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  • $\begingroup$ The answer is no. In dimension 2, you can find a homeomorphism, but in general it is not Lipschitz; it is quasiconformal. But in dimension $>2$ you cannot find any such homeomorphism (the system of PDE which it has to satisfy is overdetermined). $\endgroup$
    – Alexandre Eremenko
    Commented Dec 9, 2019 at 13:21
  • $\begingroup$ I think you are talking about uniformization theorem, did it? $\endgroup$
    – Yuchen Bi
    Commented Dec 10, 2019 at 4:55

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