# A 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) in 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*}

I'm trying to prove the following statements:

1. It 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 has a constant curvature $$K\equiv -1/a^2$$.

This is where I have some problems: is this a consequence of the above result? I'm trying with Christoffel's symbols.

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 $$\mathbb R^3$$?

• I'm not sure what you are asking. By direct calculation, $$\phi^*({dx_1}^2 + \cdots + {dx_{2n-1}}^2) = (a/z_n)^2\,\bigl({dz_1}^2 + \cdots + {dz_{n}}^2\bigr)$$ when $(z_n/a)^2>(n{-}1)$, so this answers Q1 and Q2 once you know that the formula on the RHS is a metric of sectional curvature $K=-1/a^2$. As for Q3, the metric on the RHS is defined for $z_n>0$ and is complete on that half-space, as it is homogeneous (translation in the first $n{-}1$ coordinates and scaling in $z_i$). Clearly, it is not complete on the half-space $z_n>a\sqrt{n{-}1}$, so the image of $\phi$ cannot be complete either. Oct 2, 2021 at 9:39
• @RobertBryant Thank you very much for your answer, you can do the detail I ask you please. This problem has driven me crazy for many days now, I am making several errors in the calculation of the sectional curvature by means of the Christoffel symbols. I understand what you say but when I do the detail of the (not) completeness and the rest I get a little confused. If I got to the direct calculation you mention, at first I thought that that might be enough for the sectional curvature to be $-1/a^2$ but I think that's not the case. Oct 2, 2021 at 20:44
• The integral in the question is badly stated, since it has $z_n$ both as a limit and as the argument of integration, and since it is not obvious what lower limit makes it converge. I'd prefer giving the result of the integral, using $a\sqrt{n-1}$ as the lower limit: $$- \sqrt {u} + \log (\frac {1 + \sqrt {u}} {\sqrt {1 - u}})\ \text {with}\ u = 1 - \frac {(n - 1) a^2} {x_n^2}$$ Oct 4, 2021 at 15:32
• @MattF. Choose non-zero real numbers $a_i,\; 1\leq i\leq n-1$, such that $\sum_i a_i^2=1$. The immersion $$f\colon D:=\{(x_1,\dots,x_n)\in\mathbb R^n\;|\; x_n<0\}\to \mathbb R^{2n-1},\; (x_1,\dots,x_n)\mapsto (y_1,\dots,y_{2n-1})$$ defined by $$y_{2i-1}=a_i e^{x_n}\cos(x_i/a_i), y_{2i}=a_i e^{x_n}\sin(x_i/a_i),\; 1\leq i\leq n-1,\; y_{2n-1}=\int_0^{x_n} \sqrt{1-e^{2u}} du$$ induces on $D$ a non complete metric of constant negative curvature. Oct 4, 2021 at 20:32

The isometric immersion that you describe above is the higher dimensional pseudosphere. Now, concerning your final question, I presume that you need to search about isometric immersions of the hyperbolic space $$\mathbb H^n$$ by means of a warped product representation (of $$\mathbb H^n$$) into the Euclidean space.

Now, some additional things that you might be interested to:

1. There are many (explicit in some cases) local isometric immersions from $$\mathbb H^n$$ to $$\mathbb R^{2n-1}$$. These can be constructed by using either the Ribaucour or the Bäcklund transformation (for instance, see the papers by Dajczer-Tojeiro and Tenenblat-Terng).
2. Local isometric immersions of the hyperbolic plane $$\mathbb H^2$$ into $$\mathbb R^3$$ imply "local" solutions, that is, solutions that are not defined on the whole $$\mathbb R^2$$, of the sine-Gordon equation and vice versa. Therefore, it follows from Hilbert's theorem that there is no "global" solution, that is, a solution defined on the whole plane $$\mathbb R^2$$, of the sine-Gordon equation. Just like in the case of dimension two, the same also happens in the higher dimensional case where now you will end up with a system of PDES (see for instance Dajczer-Tojeiro). We can have local solutions to this system but we don't know if there exists any global. The existence of a global solution would imply the existence of a global isometric immersion of $$\mathbb H^n$$ into $$\mathbb R^{2n-1}$$, which would give a non affirmative answer to the major still open problem (in submanifolds) up to this day, which is the following conjectured extension of Hilbert's theorem:

There is no global isometric immersion from $$\mathbb H^n$$ to $$\mathbb R^{2n-1}$$

However, the above holds true in some very special cases. For instance:

• If the immersion is also minimal (the mean curvature vanishes) (see Moore).

I should also mention here that $$\mathbb H^2$$ admits no minimal immersion in any Euclidean space. (for a proof of this fact see either "Lectures on minimal submanifolds" by Lawson, or Bryant, or Di Scala).

• (weaker) If the immersion has also bounded mean curvature (see here)
• (even weaker) If also the length of the mean curvature of the immersion does not go to infinity too fast, that is, exponentially fast (see here)

I also recommend the following: