In Is the shortest path in flat hyperbolic space straight relative to Euclidean space?

I answered by refering to the Triangle Inequality (https://en.wikipedia.org/wiki/Triangle_inequality , Euclid's elements, book 1 proposition 20 )

The triangle inequality is a theorem of neutral geometry (so valid in both euclidean and hyperbolic geometry )

Still others objected, that it is NOT a theorem of hyperbolic geometry.

Why am I wrong?

In which geometries does the Triangle Inequality holds and in which does it fail?

  • $\begingroup$ come on...they said it exactly, for time-like vectors (length<0) the triangle inequality is in the other direction. $\endgroup$ – Troy Woo Jun 18 '15 at 8:52
  • $\begingroup$ @Troy I think OP maybe isn't familiar with the concept of causal character.. $\endgroup$ – Ivo Terek Jun 18 '15 at 10:33
  • $\begingroup$ @IvoTerek I guess... $\endgroup$ – Troy Woo Jun 18 '15 at 11:37
  • 4
    $\begingroup$ The term "hyperbolic space" was used incorrectly in the post you linked to. The manifold $\mathbb R^2$ with the metric $ds^2$ defined there is $2$-dimensional Minkowski space. Hyperbolic space is a Riemannian manifold of constant negative sectional curvature, and the triangle inequality holds there, as it does on every connected Riemannian manifold with the Riemannian distance function. $\endgroup$ – Jack Lee Jun 18 '15 at 18:23

I think that confusion arose because the OP of the other question used the term "hyperbolic plane" loosely, and not in a precise way. Mainly, we denote by $\Bbb L^2$ the pair $(\Bbb R^2, {\rm d}s^2)$, where $${\rm d}s^2 = -{\rm d}x^2 + {\rm d}y^2.$$

This is another way of saying that the scalar product (which is not positive definite anymore) between two vectors is: $$\langle {\bf x}, {\bf y} \rangle = -x_1y_1 + x_2y_2$$

Given a non zero vector $\bf x$, we say that:

  • $\bf x$ is spacelike if $\langle {\bf x}, {\bf x} \rangle > 0 $;
  • $\bf x$ is timelike if $\langle {\bf x}, {\bf x} \rangle < 0 $;
  • $\bf x$ is lightlike if $\langle {\bf x}, {\bf x} \rangle = 0 $;

We define $\|{\bf x}\| = \sqrt{|\langle {\bf x},{\bf x}\rangle|}$, but this is not a norm - almost everything fails.

If ${\bf x},{\bf y}$ are spacelike vectors spanning a spacelike plane (i.e., no linear combination of them is lightlike or spacelike, for higher dimensions), we have the Cauchy-Schwarz inequality and the triangle inequality, as usual: $$|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\|{\bf y}\|, \quad \|{\bf x}+{\bf y}\| \leq \|{\bf x}\| + \|{\bf y}\|.$$ If ${\bf x},{\bf y}$ are timelike vectors, we have the Cauchy-Schwarz inequality and the triangle inequality, reversed: $$|\langle {\bf x},{\bf y}\rangle| \color{red}{\geq} \|{\bf x}\|\|{\bf y}\|, \quad \|{\bf x}+{\bf y}\| \color{red}{\geq} \|{\bf x}\| + \|{\bf y}\|.$$

These strange versions are used in special relativity, for example, in explaining the twins' paradox.

Also, you have a lot of non-intuitive results such as: two lightlike vectors are parallel if and only if they are orthogonal, etc.

However, if you consider $\Bbb L^3$ (you can guess what it is, right?), and look at: $${\Bbb H}^2(1) = \{ {\bf x} \in \Bbb L^3 \mid \langle {\bf x}, {\bf x} \rangle = -1, ~ x_3 > 0\},$$ we'll have that the restriction of $\langle \cdot, \cdot \rangle$ to the tangent planes of this set (which is a two-sheeted hyperboloid, and we take the upper sheet) will be a positive definite inner product as we know and love. This set is the hyperbolic plane. $\Bbb L^2$, the $2$-dimensional Lorentz-Minkowski space isn't. Geometry in Lorentz spaces is more screwed up than geometry in hyperbolic spaces.


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