# Isometric immersion of $\mathbb H^2$ into $\mathbb R^\infty$ built by Bieberbach

I'm analyzing the following isometric immersion of $$\mathbb H^2$$ in $$\mathbb R^\infty$$ 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.

My attempt was: $$\color{red}{[\rm{Updated}]}$$

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*}

• What do you denote by $\mathbf{R}^\infty$? I understand it consists of certain sequences, but with which condition and which norm/distance, and I don't know what you denote by $dx_m^2$.
– YCor
Oct 27, 2021 at 7:26
• From the context, I think, ${\mathbb R}^\infty$ means $\ell_2$ and $dx_i$ denotes the linear functional on $\ell_2$ defined by $dx_i((x_1, x_2,....))=x_i$. Your notation $f^*g_{\mathbb R^\infty}=\displaystyle\sum_{m=1}^\infty dx_m^2$ is horrifying. What you mean is that you define the standard inner product on $\ell^2$, treat it as a Riemannian metric $g$ and take its pull-back by $f$. Oct 27, 2021 at 18:11
• Cross-posted at MSE: math.stackexchange.com/questions/4288146/…. As a general rule, you should avoid such simultaneous cross-posting in order to avoid duplication of efforts. Post your question on one site (say, MSE, wait a week or so and if there is no satisfactory answer, make a note of cross-posting and post on the other site (say, MO). Oct 27, 2021 at 18:14
• @MoisheKohan I understand, I'm sorry. Oct 27, 2021 at 19:04
• Apart from the typo in the last line of your computation (should be just $(1-x^2-y^2)^{2}$), this is the standard formula for the hyperbolic metric on the unit disk. Oct 28, 2021 at 0:45