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 $\mathbb R^3$?
Best Answer
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 $x_{2n-1} \rightarrow -\infty$.