# Does distance in hyperbolic space satisfy such properties which Euclidean distance have?

In Euclidean space $E^n$, the distance between two points $x, y$ is just $|x-y|$, and for each fixed $x_0$, the image $y\to\nabla_x|x_0-y|$ is $S^{n-1}$, so it satisfies

(1)$\operatorname{rank}(\frac{\partial^2}{\partial_x\partial_y}d_{\mathbb{R}^n}(x,y))=n-1$

(2) $\nabla_xd_{\mathbb{R}^n}(x,y)\subset T^*_x\mathbb{R}^n$ has non-vanishing Gaussian curvature.

First, I want to know the explicit expression of $d_{\mathbb{H}^n}(x, y)$ in the hyperboloid model, and to see if they also satisfy the above two properties.

What is $\nabla_x$? How do you define it in hyperbolic case? What notion of Hessian do you use in the hyperbolic case? – studiosus Mar 9 '14 at 4:43
If $\langle x, y\rangle$ denotes the Lorentz inner product, the hyperbolic distance between $x$ and $y$ is $\cosh^{-1}|\langle x, y\rangle|$. – Andrew D. Hwang Mar 10 '14 at 1:38
@IncnisMrsi: $\nabla$ has several different meanings in Riemannian geometry (covariant derivative is one of them, of course). However, for a Riemannian manifold $M$, the notation $\nabla_x$, where $x\in M$, is meaningless. Hence, my request for a clarification (which OP so far did not provide.) Voting to close. If you know what OP had in mind, you are welcome to edit the question. – studiosus Nov 6 '14 at 17:29
The enigmatic “$∇_x$, where $x∈M$” notation is a hint that suggests to differentiate the expression by $x$, instead of by $y$, or by both, or by neither ☺ – Incnis Mrsi Nov 6 '14 at 17:37