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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?

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    $\begingroup$ 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 $dx_m^2$. $\endgroup$
    – YCor
    Commented Oct 27, 2021 at 7:26
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    $\begingroup$ 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 $f$. $\endgroup$
    – Moishe Kohan
    Commented Oct 27, 2021 at 18:11
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    $\begingroup$ 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). $\endgroup$
    – Moishe Kohan
    Commented Oct 27, 2021 at 18:14
  • $\begingroup$ @MoisheKohan I understand, I'm sorry. $\endgroup$
    – Zaragosa
    Commented Oct 27, 2021 at 19:04
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    $\begingroup$ 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. $\endgroup$
    – Moishe Kohan
    Commented Oct 28, 2021 at 0:45

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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*}

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