# 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

1. 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.

1. 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?

1. 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 $$\mathbb R^3$$?

• Edit to include your computations. Sep 23, 2021 at 14:40
• One way to do this is to write the Euclidean metric $g_E = (dx^1)^2 + \cdots (dx^{2n-1})^2$ in terms of $dz^1, \dots, dz^n$, use what you get to define an orthonormal frame of $1$-forms, and compute the connection $1$-forms and curvature $2$-forms. A beautiful paper on isometric embeddings of hyperbolic $n$-space in $\mathbb{R}^{2n-1}$ can be found here: annals.math.princeton.edu/1980/111-3/p05 Sep 26, 2021 at 19:54
• @Deane Thanks for the link, the article is very good Sep 28, 2021 at 0:20

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 $$x_{2n-1} \rightarrow -\infty$$.

• Thank you so much, your answer to my first two questions has been very satisfactory, I was trying to do it with the Christoffel's symbols, I would be infinitely grateful if you could also do it with the Christoffel's symbols even if it was in great strides. Thank you very much again. Sep 28, 2021 at 0:20
• If you compute $g = dx_1^2 + \cdots + dx_{2n-1}^2$ using the original definitions in terms of $z_1, \dots, z_n$, you get a diagonal metric, and it should not be difficult to compute the Christoffel symbols and the curvature. Perhaps you could edit your question and show how far you got. Sep 28, 2021 at 13:37

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.

• Interesting thank you very much for taking the time Oct 3, 2021 at 19:04