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:





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?

  • $\begingroup$ As far as I know there's not much that can be done to simplify this problem - just be sure to use all the symmetry you have available. For example rotations in the $12$ and $34$ planes, translations in the $34$ directions should allow you to reduce the number of unknowns. If you compute the speed of your geodesic you should get $\sqrt{a^2 + k^2}$, so you can make the problem a little more concrete by requiring the endpoints to be $0$ and $1$, and the distance to be found will then be $\sqrt{a^2 + k^2}$. $\endgroup$ – Anthony Carapetis Jan 2 '17 at 8:46
  • $\begingroup$ You are following a wrong lead. Hint: First consider the case of the Riemannian direct product of the real line and the hyperbolic 3-space. Do you know how to compute distances here? Now, relate my example and yours. $\endgroup$ – Moishe Kohan Jan 4 '17 at 12:28
  • $\begingroup$ @Moishe Cohen: are you saying my method is wrong or just too complicated? In the hyperbolic 3-space $\mathcal{H}$ (with coordinates $x_1,x_2,x_3$) the geodesics are the semicircles orthogonal to the plane $x_3=0$. So geodesics in $\mathcal{H}\times \mathbb{R}$ are couples of semicircles and segments. But I can't really see how $\mathcal{H}\times \mathbb{R}$ and $(\mathbb{R}^4,g)$ relate (they don't seem to be isometric): could you please elaborate a little more? $\endgroup$ – user372511 Jan 5 '17 at 12:38

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:

  1. 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 $M_1, M_2$.)

  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.

  3. 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 $X$.

  • $\begingroup$ Thank you! You've been very clear, I still have to wait 20 hours to award the bounty, I'll do it as soon as I can. In the mean time is the formula you're talking about the one at the bottom of this wikipedia page en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model ? $\endgroup$ – user372511 Jan 5 '17 at 16:22
  • $\begingroup$ @user372511: I actually had in mind the one in "distance function" part, but the other distance formula is OK as well, it is just less transparent. $\endgroup$ – Moishe Kohan Jan 5 '17 at 16:34

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