94
$\begingroup$

I had this idea to build a model of Earth in Minecraft. In this game, everything is built on a 2D plane of infinite length and width. But, I wanted to make a world such that someone exploring it could think that they could possibly be walking on a very large sphere. (Stretching or shrinking of different places is OK.)

What I first thought about doing was building a finite rectangular model of the world as like a mercator projection, and tessellating this model infinitely throughout the plane.

enter image description here

Someone starting in the US could swim eastwards in a straight line across the Atlantic, walk across Africa and Asia, continue through the Pacific and return to the US. This would certainly create a sense of 3D-ness. However, if you travel north from the North Pole, you would wind up immediately at the South Pole. That wouldn't be right.

After thinking about it, I hypothesized that an explorer of this model might conclude that they were walking on a donut-shaped world, since that would be the shape of a map where the left was looped around to the right (making a cylinder), and then the top was looped to the bottom. For some reason, by simply tessellating the map, I was creating a hole in the world.

Anyway, to solve this issue, I thought about where one ends up after travelling north from various parts of the world. Going north from Canada, and continuing to go in that direction, you end up in Russia and you face south. The opposite is true as well: going north from Russia, you end up in Canada pointing south. Thus, I started to modify the tessellation to properly connect opposing parts of Earth at the poles.

When going north of a map of Earth, the next (duplicate) map would have to be rotated 180 degrees to reflect the fact that one facing south after traversing the north pole. This was OK. However, to properly connect everything, the map also had to be flipped about the vertical axis. On a globe, if Alice starts east of Bob and they together walk North and cross the North Pole, Alice still remains east of Bob. So, going north from a map, the next map must be flipped to preserve the east/west directions that would have been otherwise rotated into the wrong direction.

enter image description here

Now the situation is hopeless. After an explorer walks across the North Pole in this Minecraft world, he finds himself in a mirrored world. If the world were completely flat, it would feel as if walking North will take you from the outside of a 3D object to its inside.

Although I now think that it is impossible to trick an explorer walking on infinite plane into thinking he is on a sphere-like world, a part of me remains unconvinced. Is it really impossible? Also, how come a naive tessellation introduces a hole? And finally, if an explorer were to roam the world where crossing a pole flips everything, what would he conclude the shape of the world to be?

$\endgroup$
  • 1
    $\begingroup$ This isn’t really about mathematics I would say. However, did you consider working with command blocks that immediately teleport the player to the other side after „walking over the edge of the world“? $\endgroup$ – sampleuser Mar 29 at 9:21
  • $\begingroup$ Hi @sampleuser, thanks for your reply. I can teleport them as they cross the edges, but they would still see the edge of the world and conclude they were on a finitely sized rectangle. Besides the game, I want to know if I can theoretically trick someone on a plane into thinking they are on a sphere! $\endgroup$ – 986 Mar 29 at 9:27
  • 70
    $\begingroup$ @sampleuser this question is absolutely, 100% about mathematics. Specifically, it's about topology. $\endgroup$ – Nathaniel Mar 29 at 17:41
  • 25
    $\begingroup$ Put an ice wall around Antarctica, and have NASA patrol it and arrest anyone who gets too close. $\endgroup$ – Acccumulation Mar 29 at 21:18
  • 6
    $\begingroup$ That being said, your idea has actually been proposed by Merilees back in 1972.The so-called "double Fourier sphere method" finds its use in approximating functions defined on the surface of the sphere, e.g. in the Spherefun extension of the Chebfun project. See this for more details. $\endgroup$ – J. M. isn't a mathematician Mar 30 at 13:17
69
$\begingroup$

What you want to do is not possible because there is no flat sphere. That is, there is no way to put a metric on a topological sphere such that the curvature is everywhere zero. This can be shown using the Gauss-Bonnet theorem: the global curvature (by which I mean the integral of the curvature on the whole sphere) is equal to ($2\pi$ times) the Euler characteristic, which for a sphere is $2$ (and not $0$).

On the other hand, it is very well-known to gamers that there are flat tori: you just teleport on the other side when you hit a wall. This is illustrated by the fact that the Euler characteristic of a torus is $0$, so there can be a flat metric on a torus (and indeed you can define one by expressing the torus as a quotient of the plane).

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Thanks for the answer. I suspect then that the second picture in my question could make an explorer think they were on a Klein bottle, which apparently has Euler characteristic 0, and travelling on which one would switch from outside to inside. On the other hand, if only in that picture, a player could pass through the plane as they walked across a pole, then it would be just like a sphere! $\endgroup$ – 986 Mar 29 at 10:05
  • 3
    $\begingroup$ Exactly, the Klein bottle is the other compact surface (without boundary) on which you can play a 2D video game, but it's not orientable, which makes things weird (but possibly fun too!). $\endgroup$ – Captain Lama Mar 29 at 14:15
  • 3
    $\begingroup$ @Captain Isn't the projective plane another possiblity? (And the game Obduction uses it: the map is circular, and moving past the edge puts you at the opposite point on the circle.) $\endgroup$ – aschepler Mar 29 at 17:49
  • 14
    $\begingroup$ @Acccumulation The whole point of the Gauss-Bonnet theorem is that (global) curvature is actually in some sense a topological property. But in the setting of the question, in a game such as Minecraft (or any similar context), there is a metric. The question is "can we make some kind of tesselation such that someone living on it could believe they live on a sphere?". And you won't be able to do that. Any solution will have some kind of singularity that characters will notice. (Unless I missed something, of course, in which case I'm happy to get an explanation). $\endgroup$ – Captain Lama Mar 29 at 21:17
  • 3
    $\begingroup$ Another way to think of the Gauss-Bonnet theorem is that you cannot gift-wrap a ball unless you either crease or tear the wrapping paper, since the sphere and the plane do not have the same Gaussian curvature. $\endgroup$ – J. M. isn't a mathematician Mar 29 at 23:36
91
$\begingroup$

Although you can't make a sphere from a plane, there are map projections that tessellate "naturally" (and place the tricky singular points in the ocean where people tend not to notice them). You can't, for topological reasons, avoid the points at the corners, but this kind of map does avoid some of the problems of mirroring and is continuous except at those corner points.

Most well known is the "Peirce quincuncial" projection. Wikipedia has an image showing the projection quincuncial projection

Image by Strebe - Own work, CC BY-SA 3.0

| cite | improve this answer | |
$\endgroup$
  • 10
    $\begingroup$ I wrote a Mathematica function for the quincuncial projection here. $\endgroup$ – J. M. isn't a mathematician Mar 29 at 23:34
  • 7
    $\begingroup$ The dark gridlines at 45° (the equator) threw me off, the above image is actually only 4 repeated tiles. Each tile is a square with each corner in Antarctica. wiki page: en.wikipedia.org/wiki/Peirce_quincuncial_projection $\endgroup$ – Burnsba Mar 30 at 13:37
  • 1
    $\begingroup$ As far as I can understand, within one tile, say the upper left quarter of the image, the circle of equator is mapped to a square (or "diamond", ◇). In this example, the lines of longitude (white) are at 0° (dark), 15°E, 30°E, 45°E, …, 180°, 175°W, 160°W, 145°W, …, 15°W. However, the locations where the equator bends (the corners of the ◇) are at other longitudes, namely at 70°E, 160°E, 110°W, and 20°W. When tiling, think of a chessboard. Every time you place a tile on a black field of the chessboard, first rotate it by 180 degrees. Each corner where the equator bends, meets itself in an X. $\endgroup$ – Jeppe Stig Nielsen Mar 30 at 15:12
  • $\begingroup$ @Burnsba that's correct; you need four copies of the square with the projection to get something that tiles. $\endgroup$ – J. M. isn't a mathematician Mar 31 at 13:09
  • 2
    $\begingroup$ Might this explain the Bermuda Triangle? $\endgroup$ – TonyK Mar 31 at 16:30
16
$\begingroup$

A mathematical model

Assume you managed to trick the player into thinking they are on a sphere while they are really walking on an infinite plane. What would the world have to look like?

First of all, whenever the player is standing at some point $x$ on the flat world, they are deceived to think they really are at some point $i(x)$ on the imaginary spherical world. In other words, the player's imagination creates a mapping $i : \mathbb{R}^2 \to S^2$.

The assumption

As another answer points out, it is impossible for $i$ to be a local isometry because of the difference in curvatures of the plane and the sphere. Another easy argument is that on the sphere there is a triangle with three right angles, while on the plane clearly there is not. But we can relax our expectations and only demand that $i$ be a rough local isometry. What do I mean by that?

Our player is just a human and as such, they can't really distinguish between $1$ meter and $99$ centimeters, they also can't see very far away. Thus we assume that for each sufficiently close points $x, y \in \mathbb{R}^2$ the following equality up to a small margin $\varepsilon$ between distances on the plane and on the sphere holds: $$(1-\varepsilon) \cdot d_{\mathbb{R}^2}(x, y) \leqslant d_{S^2} \big( i(x), i(y) \big) \leqslant (1+\varepsilon) \cdot d_{\mathbb{R}^2}(x, y).$$

A solution

It can be proven (though it's quite technical) that under this assumption $e : \mathbb{R}^2 \to S^2$ must be a covering map. But $S^2$ is simply connected, so it follows that $\mathbb{R}^2$ is homeomorphic to $S^2$, which is a contradiction. Hence a function with the properties stated above does not exist.

Which means what you are trying to do - is impossible.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ And the "Pierce quincuncial" projection described in another answer shows that it's quite possible to get the sphere as a branched cover of the plane, and putting the branch points in the oceans is a clever choice. Of course, this doesn't get you a local isometry, but it comes very close to getting you a local homeomorphism, which is nothing to sneeze at. $\endgroup$ – John Hughes Mar 30 at 12:51
  • 1
    $\begingroup$ ...and as also noted, there is in fact a whole family of map projections based on the doubly-periodic elliptic functions, which allows the construction of a branched cover. $\endgroup$ – J. M. isn't a mathematician Mar 30 at 13:07
15
$\begingroup$

Unlike the other answers to this question, I claim that it's possible to trick an explorer on an infinite plane into thinking he's on a sphere. In fact, I'm about to trick you, by providing just a screenshot of a work-in-progress video game project I'm currently working on, together with a bunch of other excellent internet strangers.

The previous answers have all shown that it isn't really possible to do what you want to satisfaction in euclidean space. So instead let's do it in hyperbolic space. In hyperbolic space, planes are hyperbolic planes, they have negative curvature, and interesting surfaces with zero curvature are spheres of infinite radius called horospheres (in hyperbolic space, this is not the same as a plane!) Horospheres work intrinsically just like euclidean space, but they look curved when embedded into hyperbolic space.

Here's a screenshot of a horosphere with some terrain:

Horosphere

Looks pretty round, right? Well, it is indeed round. However, it's not spherical, even if you ignore the altitude variations. It's a horosphere. The surface of this "planet" is actually Euclidean. You can draw a square grid onto the surface without distorting it.

To further confuse you, here's a different view, this time with inverted terrain (air is now inside the horosphere instead of outside of it).

Inverted horosphere

Looks much bigger on the inside, I know, but it's the exact same horosphere as before! From this perspective it should be easier to tell that the horosphere is infinite, but seen in action this can still convince explorers from the Euclidean world that they're inside a relatively small sphere if they don't happen to look up.

It's also possible to achieve a very different effect: If you place terrain along a hyperbolic plane, it will look perfectly flat at ground level:

Hyperbolic plane from ground level

However, if you look at it from just a few blocks up, it begins to seem like a pretty small planet:

Hyperbolic plane from above

None of these screenshots involve cheap rendering tricks, this is exactly what hyperbolic space actually looks like.

| cite | improve this answer | |
$\endgroup$
  • 6
    $\begingroup$ Hello! I'm one of those internet strangers Magma alluded to; I'm actually responsible for a great deal of the procedurally generated terrain that occupies that hyperbolic space above. (with a great deal of help from Magma, of course.) For further information, come check out our wiki or join us in the Hyperrogue Discord server, where we have a channel dedicated to development of this game. We'll be happy to answer more questions about hyperbolic space and its intersection with voxels. $\endgroup$ – Cedric Hutchings Mar 30 at 23:36
6
$\begingroup$

Not really a full answer, but elaboration on the OP's "I was creating a hole in the world" observation: It's been known since antiquity that one can make a stereographic projection of a plane onto sphere, missing one single point. This is used notably in complex analysis in the construction of the Riemann sphere; by taking the complex plane and adding a single "point at infinity", one has a structure equivalent to a sphere.

Riemann sphere

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

I'd like to add another visual which complements James K's answer.

What you need is this:

A plain white square pillow. The pillow is made from two squares of fabric whose edges are sewn together.

Topologically, the above pillow is just a sphere. It would be pretty easy to take a globe and put it on this pillow; you'd need to stretch and shrink things a bit, but not too severely.

Once you've done that, it's not too hard to cover the plane with copies of the pillow. Start with an infinite pile of globe pillows, and use a seam ripper to separate each pillow into the top half and the bottom half. Place one top half in the center of an infinite table, then surround it with 4 bottom halves, so that their edges match up the same way as on the original pillow. Then surround these bottom halves with 8 top halves, matched appropriately, and so on forever.

As your explorer walks across the quilt that you've created, they will have a very hard time distinguishing the quilt from the original pillow. They'll only notice a difference if they come across one of the corners, and if they're astute enough to somehow notice that, standing at the corner, they are now immediately surrounded by two copies of the world, not one.

To see what this would actually look like, look at the image in James K's answer again. One side of the pillow would have the northern hemisphere on it, and the other side would have the southern hemisphere. The equator would, of course, lie along the seam of the pillow. The orientation of the map has been chosen so that one corner is in the middle of the Atlantic Ocean, one is in the middle of the Indian Ocean, and the other two are in the Pacific Ocean.

That image shows that if you take two copies of the northern hemisphere and two copies of the southern hemisphere, you can form them into a square which tiles the plane in the usual way. The two copies of each hemisphere are 180 degree rotations of each other.

| cite | improve this answer | |
$\endgroup$

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