On $\mathbb{R}^4$ with coordinates $x_1,x_2,x_3,x_4$ consider the riemannian metric $g:=\displaystyle{\frac{dx_1^2+dx_2^2+dx_3^2+dx_4^2}{x_1^2+x_2^2}}$ defined on $X:=\{x_1^2+x_2^2\neq 0\}$ and call $d$ the intrinsic induced metric.
How can I compute explicitly $d(p_1,p_2)$ for any $p_1,p_2\in X$?
Suppose for example $p_1=(1,0,1,0)$ and $p_2=(1,1,2,2)$.
I computed the geodesic equations
$$\ddot \gamma_1=\frac{2\gamma_2\dot\gamma_1\dot\gamma_2+\dot\gamma_1(\dot\gamma_1^2-\dot\gamma_2^2-\dot\gamma_3^2-\dot\gamma_4^2)}{\gamma_1^2+\gamma_2^2}$$
$$\ddot \gamma_2=\frac{2\gamma_1\dot\gamma_1\dot\gamma_2-\dot\gamma_2(\dot\gamma_1^2-\dot\gamma_2^2+\dot\gamma_3^2+\dot\gamma_4^2)}{\gamma_1^2+\gamma_2^2}$$
$$\ddot \gamma_3=\frac{2(\gamma_1\dot\gamma_1+\gamma_2\dot\gamma_2)\dot\gamma_3}{\gamma_1^2+\gamma_2^2}$$
$$\ddot \gamma_4=\frac{2(\gamma_1\dot\gamma_1+\gamma_2\dot\gamma_2)\dot\gamma_4}{\gamma_1^2+\gamma_2^2}$$
which have the following solutions:
$$\gamma_1=\frac{k\operatorname{sech}(kt+d)\cos(at+b)}{\sqrt{A^2+B^2}}$$
$$\gamma_2=\frac{k\operatorname{sech}(kt+d)\sin(at+b)}{\sqrt{A^2+B^2}}$$
$$\gamma_3=\frac{Ak\operatorname{tanh}(kt+d)}{A^2+B^2}+c_1$$
$$\gamma_4=\frac{Bk\operatorname{tanh}(kt+d)}{A^2+B^2}+c_1$$
where $k,d,a,b,A,B,c_1,c_2$ are constants, $\operatorname{sech}$ is the hyperbolic secant and $\operatorname{tanh}$ is the hyperbolic tangent.
The distance $d(p_1,p_2)$ is such that there exists a geodesic $\gamma=(\gamma_1,\gamma_2,\gamma_3,\gamma_4):[0,d(p_1,p_2)]\rightarrow X$ of the previous form with $\gamma(0)=p_1,\gamma(d(p_1,p_2))=p_2$.
It's still not clear to me what is the easiest way to compute $d(p_1,p_2)$: should I find constants $k,d,a,b,A,B,c_1,c_2$ such that there exists a geodesic $\gamma$ and a constant $d(p_1,p_2)$ with $\gamma(0)=(1,0,1,0)$ and $\gamma((d(p_1,p_2))=(1,1,1,2)$? This seems extremely complicated and not directly solvable. Can you suggest me another way to proceed?
Best Answer
Not that you are doing anything wrong, it is just that this is way too complicated and it is unclear if you will ever arrive to an answer within a reasonable timeframe if you follow that direction. Here are the steps to solve this efficiently:
Suppose that $M=M_1\times M_2$ is the direct product of two Riemannian manifolds with the product metric, where $M_1$ is uniquely geodesic: Any two points are connected by a unique (up to an affine reparameterization) geodesic. Show that the distance function on $M$ is given by the Pythagorean formula (maybe you already proved this in your class/textbook you are using): $$ d^2((x_1,x_2), (y_1, y_2))= d^2(x_1, y_1) + d^2(x_2, y_2). $$ (Use the fact that the LC connection on $M$ is the sum of two connections on the factors to show that geodesics in $M$ are given by $c(t)= (c_1(t), c_2(t))$, where $c_1, c_2$ are geodesics in $M_1, M_2$.)
Compute (or copy from somewhere, say, wikipedia) a formula for the distance function in the upper half space model of the hyperbolic 3-space; the best one I know uses cross-ratios.
Show that your space is isometric to the Riemannian direct product $$ {\mathbb H}^3 \times S^1. $$ Hint: Consider first the subspace of your space given by $x_2=0, x_1>0$; then rotate this half-space around the 2-plane $x_1=x_2=0$. The circle $S^1$ will be parameterized by the angle of rotation. Equivalently, write your metric in cylindrical coordinates on $X$.