# The metric $g=dx^2+\cosh^2(x)dy^2$ is complete

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.

• Is $g$ the euclidean metric? If yes, then it cannot be an isometry as the curvature is different. Jul 7, 2021 at 9:20
• @ArcticChar $g$ is given at the top of the question and in the title Jul 7, 2021 at 9:22
• 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? Jul 7, 2021 at 12:13
• 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 $g_{-1}$. Jul 7, 2021 at 14:12
• Let me suggest, then, that you add some context to your problem, in particular what article you found this statement in. Jul 7, 2021 at 20:25

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?