Spheres in hyperbolic spaces

Question

I've been playing too much HyperRogue and it's given me ideas. I'm currently working on a text-based RPG, and there's a part of it where the PC is immersed in a hyperbolic space (notwithstanding whether they could even survive in such an environment), and they're supposed, at a moment, to look at the local Sun.

Now, here's my question. Or rather, my questions :

• Is a ball (as defined in math $\lbrace x \in ℍ^3 : ||x-C|| < r \rbrace$ ) even possible in a hyperbolic 3-space? (Answered by Lubin : Yes)
• If so, what does it look like? More precisely:
  • If I'm on the floor of a planet in a hyperbolic space and I look at the sun, would it look like our Sun or would it look different?

Thanks a lot

Answer 1

Here is at least a partial answer to your query (a full answer would actually attempt to include pictures):

Let's work in the hyperboloid model, which for hyperbolic $3$-space will look like $\mathbb{H}^3=\{(t,x,y,z)\in\mathbb{R}^4|t^2-x^2-y^2-z^2=1 \text{ and } t>0\}$. (In fact, one can see from this that $t\geq 1$.) The distance between two such points $\vec{v}_1=(t_1,x_1,y_1,z_1)$ and $\vec{v}_2=(t_2,x_2,y_2,z_2)$ is given by $d(\vec{v}_1,\vec{v}_2)=\text{arcosh}(t_1t_2-x_1x_2-y_1y_2-z_1z_2)$. So, the ball of radius $r\geq0$ centered at $\vec{v}_0=(t_0,x_0,y_0,z_0)$ is given by: $$B_r(\vec{v}_0)=\{(t,x,y,z)\in\mathbb{H}^3|\text{arcosh}(t_0t-x_0x-y_0y-z_0z)\leq r\}$$ Since $r\geq 0$* and $\cosh$ is a monotone increasing function for nonnegative inputs, one can also re-write this as: $$B_r(\vec{v}_0)=\{(t,x,y,z)\in\mathbb{H}^3|t_0t-x_0x-y_0y-z_0z\leq \cosh(r)\}$$ Now, let's choose a particularly nice $\vec{v}_0$ to center our ball around: $\vec{v}_0=(1,0,0,0)$. Then we will have: $$B_r(1,0,0,0)=\{(t,x,y,z)\in\mathbb{H}^3|t\leq \cosh(r)\}$$ Any cross-section of this which is perpendicular to the $t$-axis will be equal to the set of points $\{(t,x,y,z)\in\mathbb{R}^4|x^2+y^2+z^2\leq t^2-1\}$ for a fixed value of $t\geq 1$, i.e. each cross section $t$ fixed of the ball of radius $r$ centered at $(1,0,0,0)$ will look like a $3$-sphere of radius $\sqrt{t^2-1}$. So, in a way, the ball looks sort of `cone-like'. (Think of the analogous situation in two dimensions.)

"But what about centering around other points $\vec{v}_0$?!" I hear you cry, shaking your fist at the MathStackExchange Heavens. Well, friend, have I got good news for you.** It is sufficient to consider $\vec{v}_0$ of the form $\vec{v}_0=(t_0,\sqrt{t_0^2-1},0,0)$ because the subgroup $SO(3)$ inside of the isometry group $SO(1,3)$ of $\mathbb{H}^3$ acts transitively on the set $\{(t,x,y,z)\in\mathbb{H}^3|x^2+y^2+z^2=c^2\}$ for some fixed $c\in\mathbb{R}$. "Okay," you say, scratching your head in perplexity, "What does $B_{r}(t_0,\sqrt{t_0^2-1},0,0)$ look like?"

Fix some $m\in\mathbb{R}$ with $0\leq m\leq\cosh(r)$, and look at the set $\{(t,x)\in\mathbb{R}^2|0\leq t^2-x^2\leq 1 \text{ and } t_0t-x_0x=m\}$, where $x_0=\sqrt{t_0^2-1}$. This is just the equation of a line, and as we move along the line (and now including the $y$ and $z$ coordinates, with $t^2-x^2-1=y^2+z^2$), we see that we trace a cone. Of course, the reason that we have done this, is that we have constructed a cross-section of our ball $B_{r}(t_0,\sqrt{t_0^2-1},0,0)$, and from this we can see that we get another cone-like object (whatever the name for the $3$-dimensional version of a cone is).

So, to summarize, balls (of positive radius) in hyperbolic space look like cones.

*An interesting thing here is that the radius of the ball could just as well be taken to be negative instead, in which case, following a similar process, we would find $t\geq\cosh(r)$ instead.

**Yes.

Answer 2

You should have some intuitions about hyperbolic circles from playing HyperRogue. If you are standing close to a big circle, it is impossible to tell whether it is a circle of radius 20, a circle of radius 2000, a horocycle (a limit circle of infinite radius), or a equidistant curve of a big radius -- these shapes look very similar in the Poincaré model.

One important subtlety: if you are actually in the space, you will obviously not see things as in the Poincaré model. However, the angular sizes of objects, which correspond to perceived sizes, will be correct (assuming that the eye is in the center of the model), and the properties of hyperbolic space make it difficult to tell the distance to faraway objects, so the Poincaré model is a good approximation. Check out the Hyperbolic VR to do some experiments yourself.

Now, spheres/horospheres/equidistant surfaces behave just like circles/horocycles/equidistant curves, but in one extra dimension -- in the Poincaré disk model, rotate a circle around an axis $AB$ where $A$ is the center of the model and $B$ is the center of the circle, and you get a sphere in the Poincaré ball model. Thus, a faraway sphere will look like a disk, just like in our world.

One important point: if you are standing in distance $d$ from a big circle/horocycle/equidistant surface, its angular size is on the order of $\exp(-d)$. If you go 1 absolute unit away, you perceive it as about $e$ times smaller. You also see roughly the same, small, part of a big sphere from the point further away (a bit bigger, but just a bit -- it is not the case that you can see almost half of the sphere if you are far away from it, as happens in the Euclidean world). (Again, you can do 2D experiments in HyperRogue here.)

And in the hyperbolic geometry, any direction which is not "towards" the sun is away from it. Thus, if the spherical sun appears in the zenith in the given point of the hyperbolic plane, and you go one step in any direction, the sun will be $e$ times smaller, and (even if it was very far away) it will no longer be in zenith (2D experiments yet again). There have been some discussions about the hyperbolic sun on the HyperRogue forum, an interesting possibility is that there are actually many suns, and if you walk away, you simply see another sun above you.