I'm analyzing the following isometric immersion of $(\mathbb H^2,g\_D)$ in $(\ell^2,g\_\infty)$ given by $f(x,y)=(x\_1,x\_2,\dots,x\_{2m-1},x\_{2m},\dots)$ with
\begin{align}\label{5.1}
 x\_{2m-1}=\color{red}{2}\operatorname{Re}\frac{(x+iy)^m}{\sqrt{m}},\quad x\_{2m}=\color{red}{2}\operatorname{Im}\frac{(x+iy)^m}{\sqrt{m}},\quad m=1,2,\dots
\end{align}
I tried to check that it really is an isometric immersion, but I cannot calculate $f^\*g\_{\mathbb R^\infty}=g$, some metric $g$, or give it shape, I have tried to do it by means of its polar representation but I have gotten confused without reaching anything concrete. Any ideas how to attack this problem?


[Here](https://www.e-periodica.ch/cntmng?type=pdf&pid=com-001:1932:4::24) I leave the original document.



Isometric immersion of $\mathbb H^2$ into $\mathbb R^\infty$ built by Bieberbach
After a few tries I got the following:


Instead of taking real variable I take complex variable, that is let $z\_m=\dfrac{\color{red}{2}z^m}{\sqrt{m}}$, donde $z\_m=x\_{2m-1}+ix\_{2m}$. Then $dz\_m=\color{red}{2}\sqrt{m}z^{m-1}dz$, thus
\begin{align\*}
\varphi^\*g\_\infty&=\sum\_{m=1}^\infty dx\_{m}^2\\
&=\sum\_{m=1}^\infty (dx\_{2m-1}^2+dx\_{2m}^2)\\
&=\sum\_{m=1}^\infty |dz\_m|^2\\
&=\sum\_{m=1}^{\infty}\color{red}{4}m|z|^{2(m-1)}|dz|^2\\[2mm]
&=\color{red}{4}|dz|^2\sum\_{m=1}^{\infty}m|z|^{2(m-1)}\\[2mm]
&=\color{red}{4}\frac{|dz|^2}{(1-|z|^2)^2}\\[2mm]
&=\color{red}{4}\frac{dx^2+dy^2}{(1-(x^2+y^2))^2}\\
&=g\_D
\end{align\*}


