Local isometric immersion from $\mathbb H^{n}$ into $\mathbb R^{2n-1}$ I found this local isometric immersion from $\mathbb H^{n}$ into $\mathbb R^{2n-1}$, given by Schur (1886) en Über die Deformation der Räume constanten Riemannschen Krümmungsmaasses as follows, $(1\leq k\leq n-1)$:
\begin{align*}
x_{2k-1}&=\frac{a^2}{z_n}\cos \frac{z_k}{a}\\
 x_{2k}&=\frac{a^2}{z_n}\sin \frac{z_k}{a}\\
 x_{2n-1}&=a\int^{z_n}\frac{\sqrt{z_n^2-(n-1)a^2}}{z_n^2}dz_n 
\end{align*}
but i'm trying to prove that

*

*Is a local isometric immersion.


Here, taking $\phi:\mathbb H^n\to \mathbb R^{2n-1}$ given by $\phi(z_1,\dots,z_n)=(x_1,\dots,x_{2n-1})$ I imagine that $\phi^*g_{\mathbb R^{2n-1}}=g_{\mathbb H^n}$ which would prove that is a isometric immersion, but the conditions for $x_{2n-1}$ to be well defined make it only a local immersion.



*It's have a constant curvature $K\equiv -1/a^2$.


This is where I have some problems, would it be a consequence of the above?



*Any ideas to prove that image $\phi(z_1,\dots,z_n)$ is not a complete surface?

I started to see this example as a coincidence but I was thinking a bit about what happens in $\mathbb R^3$: there are $3$ types of smooth surfaces of revolution with negative constant curvature given by $x(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$, this is clear when solving
$$K=-\frac{f''(v)}{f(v)}.$$
Is there something similar in $\mathbb R^{2n-1}$, how many surfaces with these characteristics exist? is there a differential equation as in $\mathbb R^3$?
 A: First, the calculation is easier, if you do a change of coordinates. Let
\begin{align*}
  \theta_k &= \frac{z_k}{a}\\
  \tau &= \frac{a}{z}\\
  t &= \frac{a}{z_n}.
\end{align*}
Then
\begin{align*}
  x_{2k-1} &= at\cos\theta_k\\
  x_{2k} &= at\sin\theta_k\\
  x_{2n-1} &= a\int_{z=a\sqrt{n-1}}^{z=z_n} \frac{\sqrt{z^2-a^2(n-1)}}{z^2}\,dz\\
           &= \int_{z=a\sqrt{n-1}}^{z=z_n}\frac{a}{z}\sqrt{1 - \frac{a^2(n-1)}{z^2}}\,dz\\
           &= -a\int_{\tau=\frac{1}{\sqrt{n-1}}}^{\tau=t}\sqrt{\tau^{-2}-(n-1)}\,d\tau
\end{align*}
Differentiating, we get
\begin{align*}
  dx_{2k-1} &= a(\cos\theta_k)\,dt - at(\sin\theta_k)\,d\theta_k\\
  dx_{2k} &= a(\sin\theta_k)\,dt + at(\cos\theta_k)\,d\theta_k\\
  dx_{2n-1} &= -a\sqrt{t^{-2}-(n-1)}\,dt.
\end{align*}
Since
$$
  dx_{2k-1}^2 + dx_{2k}^2 = a^2(dt^2 + t^2\,d\theta_k^2),
$$
the metric is
\begin{align*}
  g &= dx_1^2 + \cdots dx_{2n-1}^2\\
    &= a^2((n-1)\,dt^2 + a^2t^2|d\theta|^2 + a^2(t^{-2}-(n-1))\,dt^2\\
    &= a^2(t^{-2}\,dt^2 + t^2|d\theta|^2).
\end{align*}
It is not hard to show that this is the hyperbolic metric, and the level sets of $t$ are horospheres.
One way is to compute the sectional curvature using the orthonormal frame of $1$-forms given by
\begin{align*}
  \omega^k &= at\,d\theta_k,\ 1 \le k \le n-1\\
  \omega^n &= at^{-1}\,dt.
\end{align*}
Differentiating, we get
\begin{align*}
  d\omega^k &= a\,dt\wedge d\theta_k\\
            &= -t\,d\theta_k\wedge t^{-1}\,dt\\
            &= -t\,d\theta_k\wedge\omega^n\\
  d\omega^n &= 0
\end{align*}
Therefore, the connection $1$-forms are
\begin{align*}
  \omega^k_j &= 0\\
  \omega^k_n &= t\,d\theta_k
\end{align*}
The curvature $2$-forms are
\begin{align*}
  \Omega^k_j &= d\omega^k_j + \omega^k_i\wedge\omega^i_j + \omega^k_n\wedge\omega^n_j\\
             &= -t^2\theta^k\wedge\theta^j\\
             &= -a^{-2}\omega^k\wedge\omega^j\\
  \Omega^k_n &= d\omega^k_n + \omega^k_j\wedge\omega^j_n\\
             &= dt\wedge d\theta_k \\
             &= -a^{-2}(at\,d\theta_k)\wedge at^{-1}dt\\
             &= -a^{-2}\omega^k\wedge\omega^n.
\end{align*}
This shows that the $n$-dimensional submanifold has constant sectional curvature $-a^{-2}$.
For each $0 \le c < \infty$, the level set $x_{2n-1} = c$ is a flat $(n-1)$-dimensional horotorus. The map $(\theta_1, \dots, \theta_{n-1}, t) \mapsto (x_1, \dots, x_{2n-1})$ is an embedding if $0 < t < \frac{1}{\sqrt{n-1}}$ but not when $t = \frac{1}{\sqrt{n-1}}$ (and $x_{2n-1} = 0$). The submanifold is therefore a manifold with boundary (the torus at $x_{2n-1} = 0$) and a cusp as $x_{2n-1} \rightarrow -\infty$.
A: Here's another way to do this (the calculations below are for the hyperbolic metric where the sectional curvature is equal to $-1$. The letter $a$ below represents a function and not the constant $a$ used in the question):
Start by checking that a metric of the form
$$
g = da^2 + b^2(dz_1^2 + \cdots dz_{n-1}^2),
$$
where $a$ and $b$ are functions of $z_n$ only, is hyperbolic if
$$
(a')^2 = \left(\frac{b'}{b}\right)^2.
$$
Next, check that the immersion given by
\begin{align*}
x^{2k-1} &= r(z_n)\cos z_k\\
x^{2k} &= r(z_n)\sin z_k\\
x^{2n-1} &= h(z_n),
\end{align*}
where $1 \le k \le n-1$, has the induced metric
$$
g = ((h')^2 + (n-1)(r')^2)\,dz_n^2 + r^2(dz_1^2 + \cdots + dz_n^2)
$$
Therefore, it is a hyperbolic metric if
$$
(h')^2 + (n-1)(r')^2 = \left(\frac{r'}{r}\right)^2.
$$
It is now straightforward to check that the metric defined in the question satisfies this equation.
