I am writing something for the journal of the university on the Lorentz space, and I want to prove that the following definition of the hyperbolic distance on the upper sheet of the hyperboloid $v \cdot v=-1$ is a metric:

The hyperbolic distance between $u$ and $v$ is the only positive number $\eta (u,v)$ such that $$\cosh(\eta(u,v))=-u \cdot v. $$

The first properties are easy, but all the proofs of the triangle inequality that I have seen seem complicated. In Ratcliffe's Foundations of Hyperbolic Manifolds the lorentzian cross product is used along with many of its properties, and I would like not having to introduce the cross product just for this proof.

I have thought about showing that the hyperbolic angle is the same as the lorentzian arc-length distance, and then showing that the metric given by the arc length satisfies the triangle inequality, but this seems even more tedious. Is there a shorter and nicer proof of this?

  • $\begingroup$ Aren't the geodesics just the intersections of the hyperboloid with 2-planes through the origin? I would focus on that and showing the unit speed parametrization for any geodesic, wih special attention to those through the vertex (0,0,1). These should mirror a similar calculation for the unit sphere. I think this approach is given in entirety in Spivak's five volumes. For the triangle inequality, insist that one point be (0,0,1) and the second (a,0,b) or the like...I see you do not restrict the dimension, may not matter. $\endgroup$
    – Will Jagy
    Apr 16, 2010 at 19:10
  • $\begingroup$ Another way to proceed is to start with the Riemannian metric and then obtain the hyperbolic distance via integration along a geodesic. There is a lovely introduction to the various models of hyperbolic space by Cannon, Floyd, Kenyon, and Parry at citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – Sam Nead
    Apr 16, 2010 at 20:00

2 Answers 2


We need to check $\eta(u,v)+\eta(v,w)\ge\eta(u,w)$. Introduce coordinates $x,y,z$ so that the form is $x^2+y^2-z^2$.

First, verify that there is a Lorentz map sending $v$ to $(0,0,1)$. Since it is an isometry, we may now assume that $v=(0,0,1)$. This is the main idea. For added convenience, you may also rotate the $xy$-plane so that the $y$-coordinate of $u$ equals 0.

Next, observe that the formula yields equality in the case when the projections of $u$ and $w$ to the $xy$-plane are endpoints of a segment containing (0,0). This is straigtforward if you write $u=(\sinh a,0,\cosh a)$ and $w=(-\sinh b,0,\cosh b)$ where $a,b\ge 0$.

Finally, rotate $w$ around the $z$-axis until it comes to a position as above. The product $u\cdot w$ grows down (it equals contant plus the scalar product of the $xy$-parts, since $z$-coordinates are fixed). Hence $\eta(u,w)$ grows up while the two other distances stay, q.e.d.

Of course, for writing purposes the last step is just an application of Cauchy-Schwarz for the scalar product in $\mathbb R^2$.

This was about two-dimensional hyperbolic plane, in higher dimensions just insert more coordinates (they will not actually show up in formulae).

  • 3
    $\begingroup$ The spelling is Cauchy-Schwarz, from Hermann Schwarz. If you are interested in Fourier transform and distributions instead, then you probably mean Laurent Schwartz. $\endgroup$ Oct 29, 2018 at 12:23

The inequality we want to prove is $$ \left|\frac{u-v}{1-\overline{u}v}\right| \le \left|\frac{u-a}{1-\overline{u}a}\right| + \left|\frac{a-v}{1-\overline{a}v}\right| $$ for $u,v,a$ in the open unit disk ${\mathbf D}$. Let $f$ be the map $$ f(\zeta)=\frac{\zeta-a}{1-\overline{a}\zeta} \quad (\zeta \in {\mathbf D}). $$ It is easily seen that $f$ maps ${\mathbf D}$ into itself. Indeed, if $\zeta,a \in {\mathbf D}$, then $(1-|\zeta|^2)(1-|a|^2)> 0$, it follows that $|\zeta|^2+|a|^2 < 1+ |a|^2 |\zeta|^2$, and after adding $-\overline{a}\zeta-a\overline{\zeta}$ on both sides, we get $ |\zeta-a|^2 < |1-\overline{a}\zeta|^2$, that is $f(z)=\frac{\zeta-a}{1-\overline{a}\zeta} \in {\mathbf D}$. A straightforward computation shows that $$ \left|\frac{f(\zeta)-f(\xi)}{1-\overline{f(\zeta)}f(\xi)}\right| =\left|\frac{\zeta-\xi}{1-\overline{\zeta}\xi}\right|. $$ Thus, in the inequality we want to prove we may replace $u,v,a$ by $f(u)=:z$, $f(v)=:w$, $f(a)=0$, respectively, and we are left with proving that $$ \left|\frac{z-w}{1-\overline{z}w}\right| \le |z|+|w|, $$ that is, $$ |z-w| \le |z|\,|1-\overline{z}w|+|w|\,|1-\overline{z}w| $$ for $z,w \in {\mathbf D}$. Since $|1-\overline{z}w|=|1-\overline{w}z|$, this is equivalent to the inequality $$ |z-w| \le |z-|z|^2 w|+|w-|w|^2z|. $$ Denote the points $z$ and $w$ in the complex plane by $Z$ and $W$, and let $O$ denote the origin. The point $|z|^2w$ is a point $P$ on the line segment between $O$ and $W$, and $z-|z|^2w$ is the vector from $P$ to $Z$. It follows that $|z-|z|^2w|$ equals the (Euclidean) length $|PZ|$ of the line segment between $P$ and $Z$. Analogously, $|w-|w|^2z|=|QW|$ for some point $Q$ on the line segment between $O$ and $Z$. Let $S$ be the point of intersection of the line segments $PZ$ and $QW$. We then have $$ |z-w| = |ZW| \le |ZS|+|SW| \le |PZ|+|QW| = |z-|z|^2 w|+|w-|w|^2z|, $$ which is what we wanted to prove.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.