I want prove that the metric $g=dx^2+\cosh^2(x)dy^2$ is complete with constant curvature $-1$ on $\mathbb{R}^2$, for that I considered the following function (suggested in an article) $f:(\mathbb{R}^2,g)\to (\mathbb{R}^2,g_{-1})$, where $g_1=dx^2+e^{2x}dy^2$, defined by $$f(x,y)=(-y+\ln(\cosh x),e^y\tanh x).$$ I gotten to show that this is a diffeomorphism with inverse $$f^{-1}(x,y)=\Big(\sinh^{-1}(ye^x),\ln(\sqrt{e^{-2x}+y^2})\Big),$$ I need to know explicitly how the mentioned function $f$ preserves the metric. This is the Danilo Blanusa's article: Über die Einbettung hyperbolischer Räume in euklidische Räume I'm trying to read in detail this article about the immersion of $\mathbb{H}^2$ in $\mathbb{R}^6$ in which I found this metric, I have not been able to find the other article where it mentions the function mentioned. enter image description here

  • $\begingroup$ Is $g$ the euclidean metric? If yes, then it cannot be an isometry as the curvature is different. $\endgroup$ Jul 7, 2021 at 9:20
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    $\begingroup$ @ArcticChar $g$ is given at the top of the question and in the title $\endgroup$
    – Didier
    Jul 7, 2021 at 9:22
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    $\begingroup$ What does I can't get that mean? (1) You know the conclusion is false? (2) Something prevents you from checking the condition for being an isometry? $\endgroup$ Jul 7, 2021 at 12:13
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    $\begingroup$ Your definition should include an explanation of the notation $(\mathbb H^2,g_{-1})$. Presumably that notation refers to the hyperbolic plane, however there are several different models of the hyperbolic plane in common use (the upper half plane; the Poincaré disc; the Beltrami-Klein disc) and you should say exactly which one you mean, including an explanation of the enigmatic $g_{-1}$. $\endgroup$
    – Lee Mosher
    Jul 7, 2021 at 14:12
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    $\begingroup$ Let me suggest, then, that you add some context to your problem, in particular what article you found this statement in. $\endgroup$
    – Lee Mosher
    Jul 7, 2021 at 20:25

1 Answer 1


Maybe i'm wrong with this but i would appreciate your comment please: Let $$f(x,y)=(\underbrace{-y+\ln(\cosh x)}_{u},\underbrace{e^y\tanh x}_{v})$$ as $du=\tanh x\,dx-dy$ and $dv=e^y\cosh^{-2}x\,dx+e^y\tanh x\,dy$ then we've got $du^2=\tanh^2 x\,dx^2-2\tanh x\,dxdy+dy^2$ and $dv^2=e^{2y}\cosh^{-4}x\,dx^2+2e^{2y}\cosh^{-2}x\tanh x\,dxdy+e^{2y}\tanh^2 x\,dy^2$ $$ \begin{array}{ccl} f^*g_{-1}&=&du^2+e^{2u}dv^2\\ &=&\tanh^2 x\,dx^2-2\tanh x\,dxdy+dy^2 +e^{-2y}\cosh^2x \Big(e^{2y}\cosh^{-4}x\,dx^2+2e^{2y}\cosh^{-2}x\tanh x\,dxdy+e^{2y}\tanh^2 x\,dy^2\Big)\\ &=&dx^2+(1+\sinh^2x)dy^2\\ &=&dx^2+\cosh^2x dy^2\\ &=&g \end{array} $$ Then $g$ is complete. What do you think?

  • $\begingroup$ (+1) Looks good! $\endgroup$ Jul 8, 2021 at 12:16

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