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 I leave the original document.
Best Answer
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*}