6
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ 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
  • 2
    $\begingroup$ @ArcticChar $g$ is given at the top of the question and in the title $\endgroup$
    – Didier
    Jul 7, 2021 at 9:22
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

2
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.