Imagine that you're a flatlander walking in your world. How could you be able to distinguish between your world being a sphere versus a torus? I can't see the difference from this point of view.

If you are interested, this question arose while I was watching this video about the shape of space by Jeff Weeks.

Najib Idrissi
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    Maybe if you had one or two long ropes with you ? – jibe Jul 02 '14 at 15:05
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    Could you clarify whether you mean an *exact* sphere, or anything topologically equivalent to a sphere. Some solutions (e.g. Steven's) will only work if you're on a geometrical sphere, and will break on topologically-identical spaces. – cloudfeet Jul 02 '14 at 17:33
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    There should be some sort of answer involving 3 90 degree turns, right? – Wossname Jul 02 '14 at 21:52
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    I think the simplest answer would be: look up in the sky =P. But without using that observation, I'd get a surveyor to find out that the points on the torus have zero curvature while a sphere has constant positive curvature. This is if we are talking about what we literally call a sphere and a torus and not something topologically homeomorphic too either. –  Jul 03 '14 at 03:13
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    @Bryan: The flatlanders don't have an 'up' for the sky to be in. – dotancohen Jul 03 '14 at 10:13
  • @Bryan If you can see part of the (torus) planet in the sky, that would mean that the gravity field of it will "pull you up", right? and even before you see it, the gravity force will be like if you are going downhill, which will be scary – Display Name Jul 03 '14 at 10:33
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    @SargeBorsch If we were accounting for gravity, I don't think we'd be having this worry for the planet would most definitely be a sphere (if it's sufficiently large). –  Jul 03 '14 at 12:00
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    Actually supposedly there is no physical law that forbids [the creation of toroidal planets](http://io9.com/what-would-the-earth-be-like-if-it-was-the-shape-of-a-d-1515700296). – Bennett Gardiner Jul 03 '14 at 12:29
  • I asked a similar question [here](http://math.stackexchange.com/questions/743034/topological-surface-thought-experiment). One answerer suggested using persistent homology. – Viktor Vaughn Jul 03 '14 at 14:34
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    @BennettGardiner awesome reading, thanks for the link! – Display Name Jul 03 '14 at 15:08
  • @Bennet Gardiner: Till someone stands on both his feet, ours is a toric planet, too :) – Jack D'Aurizio Jul 03 '14 at 23:00
  • @dotancohen: Even if they could look up, people living on the outer parts of Planet Doughnut wouldn't have a view of the hole. – Dan Jul 04 '14 at 08:05
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    http://math.stackexchange.com/questions/102376/flatlander-on-torus – TROLLHUNTER Jul 09 '14 at 04:54
  • Unless you (or someone) can define precisely what it means to "walk on a surface", then this is a soft question. – Najib Idrissi Nov 03 '16 at 10:06

20 Answers20


Get a (two-dimensional) dog and a very long (one-dimensional) leash. Send your dog out exploring, letting the leash play out. When the dog returns, try to pull in the leash. (Meaning, you try to reel in the loop with you and the dog staying put.) On a sphere, the leash can always be pulled in; on a torus, sometimes it can't be.

(See homotopy.)

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    This seems cruel to two-dimensional dogs – Ben Grossmann Jul 02 '14 at 15:07
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    If you are on a sphere, you cannot know if every loop is homotopically trivial or if you did not search enough to get a non-trivial one. – Seirios Jul 02 '14 at 15:11
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    @Seirios: Yes, at best this establishes semidecidability — and that's assuming we have some effective way of enumerating some dense collection of paths on the torus, which is not at all clear if we're not using metric properties. –  Jul 02 '14 at 15:42
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    @Omnomnomnom: Dogs like going for walks. –  Jul 02 '14 at 15:42
  • What kind of path would the dog have to follow for pulling in the leash to be impossible? It seems to me that only friction could prevent it. – Brilliand Jul 02 '14 at 16:04
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    @Brilliand, he walks all the way around the torus (going through the hole). Pulling on the leash would have to pull the dog back around the torus, something that can't happen on a sphere. (If the dog walked around the equator of a sphere, the leash could slip over one of the poles when you pulled it back in.) – cjm Jul 02 '14 at 16:25
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    That works if you're on an exact sphere. However, the surface of an hourglass is topologically identical to the surface of a sphere - but if your dog walks around the narrow "neck" then it could be impossible to pulled in, even though you're essentially on a sphere. – cloudfeet Jul 02 '14 at 17:30
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    I'm curious - where did you learn the dog leash metaphor for homotopy? Or did you invent it? I think I did a long time ago in The Spinor Spanner (http://links.jstor.org/sici?sici=0002-9890%28197311%2980%3A9%3C977%3ATSS%3E2.0.CO%3B2-M) but maybe I was channeling an earlier source. – Ethan Bolker Jul 02 '14 at 17:32
  • @cjm So the leash would be able to pull the dog back, it's just that you would see the dog moving away from you at first? I interpreted the answer as saying that retrieving the dog by pulling the leash would be impossible. It doesn't seem like this test could be effective if the dog was out of sight when you started pulling the leash. – Brilliand Jul 02 '14 at 17:39
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    @Brilliand, that's why the test says "When the dog returns." But this test isn't without its issues, as cloudfeet pointed out. Also, it could take quite a while for the dog to walk all the way around. – cjm Jul 02 '14 at 17:44
  • @Brilliand, I've made a small edit which I hope will more clearly express the idea of pulling in the leash. –  Jul 02 '14 at 17:55
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    @cloudfeet: Good point. How about if we assume I can jiggle the leash to get it over that kind of obstacle? And by "jiggle" I of course mean "make arbitrarily large deformations [but still keeping the endpoints fixed]". –  Jul 02 '14 at 17:56
  • @EthanBolker: I came up with it independently, but of course I'm happy to concede priority. (As I recall, it was when I learned that $\pi_1(\mathbb{PR}^2)=\mathbb{Z}/2\mathbb{Z}$, was confused, and sought to express my confusion in the most concrete possible terms, because it was really a violation of physical intuition. So I made up a story of sending my dog out on a loop, twice.) –  Jul 02 '14 at 18:15
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    @Steven: In the case of an hourglass, that "jiggle" would need to cover half the distance of your world. You can more systematically search (the "loop testing" step 2 in my answer), but you're still probabilistic in finding loops in the first place (your dog makes a random walk). – cloudfeet Jul 02 '14 at 18:41
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    @Steven: I think our approaches are not completely dissimilar, just I defined the "jiggle" more thoroughly (equivalent to my step 2), and I aimed for determinism (instead of a random-walk dog). – cloudfeet Jul 02 '14 at 18:43
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    good intuitive answer ! Thanks (+1) – Julien__ Jul 03 '14 at 04:38
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    how do you tie a two dimensional leash to a two dimensional dog? – Martijn Jul 03 '14 at 15:53
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    @Martijn Staple it to his tail? (sorry Omnomnomnom...) – Thomas Jul 05 '14 at 00:48
  • To be more systematic, get a partner and a rope eash and walk out from the same starting point at right angles to one another, always in a straight line. If you're on a torus, both of you will return home, but one of you won't be able to contract the rope. – Jack M Jul 05 '14 at 09:14
  • Since a torus is done by crossing two circles if the radius of one is really close to the other when you send the dog around, the hole after the revolving is so small that the dog may step over it right? We have the same problem if we walk north to south verses east to west. On a sphere the two are the same, but on a torus they will be different assuming the genus isn't so small that we can just step over it. – Mr.Fry Jul 07 '14 at 07:45
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    This is based strongly on the assumption that the leash is not frictionless, which may not be true in a world of infinitely long leashes and two-dimensional dogs.. – Jason C Jul 07 '14 at 14:39
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    Are dogs and leashes a mathematically rigorous technique? – Robert Harvey Jan 16 '15 at 14:48

The Gaussian Curvature is an example of an intrinsic curvature, i.e. it is detectable to the "inhabitants" of the surface. The Gauss-Bonnet Theorem gives a connection between the Gaussian Curvature $K$ and the Euler Characteristic $\chi$. For a smooth manifold $M$ without boundary: $$\int_M K~\mathrm{d}\mu = 2\pi \chi(M)$$ The Euler Characteristic and the genus of the surface are connected by $\chi(M) = 2-2g$. A sphere has genus zero and so $\chi(S^2) = 2$, while a torus has genus one and so $\chi(T)=0$.

You could, as the ordinance survey people do, choose triangulation points on your surface, measure the Gaussian Curvature at those points and then use this to approximate the above integral.

Fly by Night
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Travel a lot and depict a map of the world. Then try give a colour to every state in your map, in order that neighbours have different colours. If you need more than four colours, you are on a torus.

This is just a reformulation of @Fly by Night's solution, since the chromatic number depends on the genus.

In a more deterministic way, on a torus you can embed a $K_5$, i.e. you can find $5$ points $A_1,\ldots A_5$ such that there exist $10$ non-intersecting paths from $A_i$ to $A_j$, on a sphere you cannot.

As an alternative, given two distinct points $A$ and $B$ on the surface, you can draw the locus of equidistant points (with respect to the geodetic distance) from $A$ and $B$. If such a locus has two connected components, you are on a torus.

Another possibility is to "comb" the surface. If you are able to, you are on a torus. And I bet that there is a plethora of opportunities given by the Borsuk-Ulam theorem, in general. For example, on a torus the wind (as a continuous vector field) can blow with a non null intensity in every point, on a sphere it cannot.

Or try to draw many concentric circles. If you are on a torus, sooner or later one of these circles must intersect itself.

And thanks to Giovanni Barbarino, on a toric surface there is always a point with a zero gravity, so there are some issues in building homes nearby.

Jack D'Aurizio
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    Needing more than four colours is sufficient, but not necessary for being on a torus. If one country successfully occupied a section of the torus (wrapped all the way around), then it would be indistinguishable from a country occupying the north pole of a sphere. – cloudfeet Jul 02 '14 at 18:45
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    @cloudfeet Not both north and south poles? The only way I can think of for a single region to guarantee that a torus can be colored with only four colors is for it to occupy a full loop in both directions (such that the torus-nature of the world could be proven by examining only that country's territory). – Brilliand Jul 02 '14 at 18:54
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    Yeah, you're right. :) It's still not *guaranteed* that a four-colour mapping will be impossible just because you're on a torus, though. – cloudfeet Jul 02 '14 at 19:14
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    +1 for the combing approach, though - that's really neat and elegant, and I think it should be listed first in this answer. :) – cloudfeet Jul 02 '14 at 19:15
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    +1 for the combining approach too, and I do like the circles idea (although no one would do this on Earth..) – Julien__ Jul 03 '14 at 04:43
  • in case a country lies on the entire handle, their administration of traffic would have realised this. Or we walk the leashed dog in every state, but this time don't let the dog leave that certain state. When you are sure about the non-existence of handle-conquering states, we can start to draw our world map. – Zhipu 'Wilson' Zhao Jul 09 '14 at 10:16
  • really very nice answer, sir +1 – jeanne clement Dec 06 '16 at 14:19

If the world you live in might have any shape homeomorphic to a sphere or a torus, then you cannot prove that it's not a torus without examining all of the surface. The reason for this that the surface might look almost exactly like a sphere, except for a small handle somewhere that, topologically, makes it a torus:

$\hspace{120px}$Sphere with handles
(Image from Wikimedia Commons; created and released into the public domain by Oleg Alexandrov.)

The illustration above shows a sphere with three (fairly large) handles; those handles could be shrunk to an arbitrarily small part of the surface without changing its topological genus.

So, what about if we assume that you've already examined every inch of the surface, and haven't found any small handles? How could you tell whether there might be any big handles that you didn't notice simply because you walked straight through them?

One solution would be to get an (infinitely) elastic loop of rope, initially all coiled into a single spot, and start expanding it outwards until it meets itself on the other side of the world. Then keep pushing the rope further away from the starting point and towards the initial contact point until it the moving rope has covered all of the surface.

If, by doing so, you can shrink the rope back to a single point without ever making it move back over a part of the surface it has already passed before, your surface is a sphere; if you're left with a loop of rope that you cannot get rid of, then you have a torus (or a surface of some higher genus).

Ilmari Karonen
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    I like the approach. There is the mildly disturbing consequence that [we do not live on a sphere](http://en.wikipedia.org/wiki/Arco_Naturale). – Jyrki Lahtonen Jul 02 '14 at 22:32
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    @JyrkiLahtonen also that our genus is [at least 2000](http://en.wikipedia.org/wiki/Arches_National_Park) – Chris Brooks Jul 03 '14 at 15:22
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    Well that's constantly changing... http://www.google.fr/imgres?imgurl=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2F9%2F9a%2FRattlesnake_Mountain_Tunnel.jpg&imgrefurl=http%3A%2F%2Fcommons.wikimedia.org%2Fwiki%2FFile%3ARattlesnake_Mountain_Tunnel.jpg&h=683&w=1024&tbnid=mU-IqcKXF77FzM%3A&zoom=1&docid=QEs298-VzHR0XM&ei=Mky9U6_rF7Da0QX8kIHgBg&tbm=isch&client=safari&iact=rc&uact=3&dur=330&page=1&start=0&ndsp=22&ved=0CCUQrQMwAQ – Julien__ Jul 09 '14 at 14:06
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    ¡Oh god! we are already living in a torus: I see a bridge yesterday. – Ernesto Iglesias Feb 28 '17 at 20:26
  • I thought the first comment was funny, but the combo with the second is deadly – Andrea Marino Mar 19 '21 at 23:29

Hopefully the land will be filled with (two dimensional) trees: take a very long rope and start connecting the trees in order that every tree is the vertex of a triangle. Do that till the whole world is covered with rope triangles. Now start counting the triangles formed with the rope, the rope pieces connecting the trees and the trees you used to form the triangles. Now you can compute $$\#\mbox{trees}-\#\mbox{pieces of rope}+\#\mbox{triangles}$$ If you get $2$ you have a sphere, if you get $0$ a torus.

If there are no trees you can use some poles :)

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    Cool approach! If there's a tiny "handle" hidden inside one of the triangles, then you'll miss it, but if all the topological features are larger than tree-scale then it's a really neat approach! – cloudfeet Jul 03 '14 at 10:43
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    If you want to cover all cases, you can very simply test each triangle for handles: lie a second rope right on top of the first (tracing the triangle again) and pull it tight. If it works, you're handle-free. If it *doesn't* pull tight, then you *may or may not be* on a torus, but you can continue this approach on a smaller scale (with toothpicks and string!). – cloudfeet Jul 03 '14 at 10:46
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    A triangle with a handle is not a triangle anymore: you should take more triangles using e.g. some tiny tread attached to something smaller than trees :) – Dario Jul 03 '14 at 11:44
  • Sorry, I might have mis-explained: say you're on the surface of a sphere with a really tiny handle (like Ilmari's picture, but much much smaller). If it's small enough, you could accidentally draw a "triangle" that contained the handle in the middle, and you wouldn't *know* you needed to take smaller triangles - unless you attempted to pull the edge in and it 'caught' on the anomaly. – cloudfeet Jul 03 '14 at 12:01
  • Yes, I get it. What i meant is that form my argument to hold you need that triangle are "real" triangles (i.e. homeomorphic to a disk). – Dario Jul 03 '14 at 12:05
  • Ah, OK - makes sense. :) However, if you just connect trees with ropes, you don't guarantee that. I hope I didn't put you on the defensive - I liked your approach, and wanted to add a clarifying comment that made it anal-retentively complete. :p – cloudfeet Jul 03 '14 at 17:04
  • I was just thinking about tiling the surface and using [Euler's Formula](http://en.wikipedia.org/wiki/Euler_characteristic), but you'd already done it. (+1) Of course, this answer relies on the fact that these mainifolds are compact. It's difficult to tile a non-compact manifold. – robjohn Jul 03 '14 at 17:30
  • @cloudfeet No problem :) – Dario Jul 03 '14 at 17:39
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    @cloudfeet: does building a bridge on the Earth change your perception of the shape of the Earth from a sphere to a torus? IMO, small handles don't change the perceived shape of the world. – robjohn Jul 03 '14 at 17:53
  • @robjohn: I wouldn't expect artificial structures to count as part of the planet, since otherwise buildings would count as "handles" unless all windows were sealed and each room could only be visited through one sequence of doors. On the other hand, tunnels would certainly alter the topology, whether they were truck-sized or mouse-sized. It's probably best to specify that a planet should be considered a "sphere" if it has no handles which are above a certain size, and a "torus" if it has handles which are above a certain (larger) size. If the largest handle is between the two sizes... – supercat Jul 05 '14 at 17:48
  • ...either classification would be deemed acceptable. Defining the smallest handle size of interest would make it possible to define things like the required size of triangles. – supercat Jul 05 '14 at 17:49
  • It's like asking what the Universe looks like if we go down in scale to "quantum foam" – Hagen von Eitzen Jul 06 '14 at 21:06

One way to determine a torus from a sphere would be to try to comb it. If all the wheat on the planet can tilted or brushed such that there is no cowlick then the planet is a torus.

Notice the cowlick, which tori do not have:

Not a torus

No cowlick:

Is a torus

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    Thanks for referring to the hairy ball theorem. Comb the planet! – Jacob Krall Jul 05 '14 at 03:49
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    Note, however, that this process is nondeterministic: Try combing. If singular, try again. The hazard is that there _are_ singular vector fields on the torus and one may just be horrendously unlucky in one's combing attempts. (Unless there is a theorem about local de-singularizations, which could augment the algorithm. Do you know of one?) – Eric Towers Jul 31 '15 at 19:27

Given that we live on a torus and most people think it's a sphere it is probably very hard to tell the difference.

enter image description here

Tom Collinge
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    I don't get, what is the point of your answer? – Anubhav Mukherjee Dec 10 '16 at 09:42
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    @Anubhav Well, if you live on a ball with a 6,000 km radius which somewhere has a comparatively tiny handle (making it equivalent to a torus) you would have to conduct an extremely exhaustive search to know that it isn't a sphere wouldn't you ? – Tom Collinge Dec 10 '16 at 14:53
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    may be it is hard, but what is the point of writing that "it is hard to tell the difference"? That is not an acceptable answer for me at least... – Anubhav Mukherjee Dec 10 '16 at 18:18
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    Given that there are multiple arches on the earth's surface, we actually live on a surface of high genus. – Cheerful Parsnip Apr 20 '17 at 01:18

I'm not a mathematician and in fact I dropped out of school, so please feel free to smack me down if this is wrong, but:

Can't you just start walking in a "straight line", while drawing your path on the ground as you walk. If you never get back to where you started, you are on a torus. If you get back to where you started, make a 90 degree turn, and walk again until you get back to where you started again. If your second path crosses your first line once, you are on a sphere. If it doesn't cross or it crosses more than once, you are on a torus. Otherwise you are on a sphere.

EDIT: This assumes that "torus" means "perfectly symmetrical donut", in the general case of torus = sphere+handles, it will not work.

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    I feel that to walk in a "straight line" and never get back where you started you'd need to walk along a line $x=at,y=bt$ where $a/b$ isn't rational. – Mark Fantini Jul 03 '14 at 05:33
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    Refer to Ilmari Karonen's answer. This helps for a torus with a "big" hole, but for a topological torus with a "small" handle that you walk either side of, you would conclude that you're on a sphere. This isn't intended as a smackdown, the question doesn't say you have to deal with handles. – Steve Jessop Jul 03 '14 at 09:09
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    @SteveJessop Thanks for that, yes you're right, by torus I assumed a perfectly symmetric"donut". – CaptainCodeman Jul 03 '14 at 09:17

"Mr. Sphere, is my world a torus or a sphere?"

Eric Towers
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The challenge here is that something can be topologically a sphere without having the exact geometry of a sphere. For example, if you are walking on the surface of an hourglass shape, it's still topologically a sphere, but if you thread a rope around the 'neck', it won't pull tight.

Instead, consider a torus as topologically identical to a big sphere with an extra loop on it that "warps" you to elsewhere on the sphere (a bit like a gym ball with a handle). We systematically search for loops which don't "pull tight", and test them to see if they're just protrusions or the 'warp' bit of a torus. This comes in two parts:

1: walk in an expanding spiral to search for loops: Fix one end of a rope to the ground, and walk in a circle, leaving a trail of chalk on the ground. Every time you complete a circle, give yourself a bit more rope and try again.

If at any point you cross your own rope, then you've found a loop you need to test. Alternatively, if you end up crossing your own chalk line, then you've found two loops to test (one for each direction on the chalk that you follow back to your home point).

2: test any candidate loops: For this, you basically attempt to spiral inwards, and see if you meet yourself or get 'warped' to another place on the sphere. With different-coloured chalk, trace the outside of the loop all the way around. Then repeatedly trace that edge again but slightly "inside" (that is, away from your home, into the area you haven't walked on yet).

If you end up spiralling yourself into a point, then you've shown that the area enclosed by your loop is topologically flat, and therefore consistent with being on a sphere. You continue searching for more loops to test, until you have covered the whole surface of your world.

Alternatively, if you encounter any lines in your original chalk colour, then that means the loop you found was wrapped around the 'warp' of a torus.

This method should find every possible loop in your world, so if you test them all and don't find a 'warp', then you must be on a sphere.

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I think that the higher rated answers answer the question except that they require a kind of total knowledge and impracticality.

What if we assume that light travels in the usual manner on the surface, in 'straight' lines? Then a program of sending out beams from a single, fixed point and measuring the possibly returned beam could yield sufficient information.

On a sphere the beam will always return and with the same attenuation from 180 degrees from the emitter regardless of the direction chosen. Whereas on a torus there will be a range of values with two minimums. The total flight time required could, I haven't checked, be made arbitrarily large depending on the angle chosen.

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    This is actually a terrific answer. It accounts for the 2-D beings, and it is (relatively) practical. This could be done with 20th century Human technology, which many other answers can not. – dotancohen Jul 07 '14 at 08:22
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    For most initial directions for the beam, it'll never return to the original point on a torus. – Ted Shifrin Jul 10 '14 at 16:02

Assume you are on a perfect sphere.

Then the circumference is the same at every point.

Therefore if you pick a point on the object, then walk in a straight line and measure the distance then pick a new point and direction, and measure that distance. On a perfect sphere those two distances will be equal.

Another way would be that as a flat lander you could see the back of yourself directly in front of you as you turn. The distance to your back should be constant.

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  • Wow, thanks for the nice (and scary) image of me seeing my own back all the time in front of me. – Yuriy S Apr 28 '16 at 08:56

$$\underline{\textbf{Dancing on a Surface}}$$

It's time to put on your dancing shoes!

Tie a rope to yourself and anchor the other end to the floor. Let another person do the same to themselves with their rope anchored to the floor at a different place.

Now dance, little flatlander. Dance!

Make sure that when your dance is finished you both end up at one of the anchor points (obviously you can't both end up at the same anchor point). Also make sure that you don't do a 'boring dance' by which I mean your ropes should be suitably intertwined that you can't tug on them so that the rope goes straight from you down to the floor.

I hope you're not tired because now you have to...

Dance again!

Do the same dance that you did before - this is important.

If you are dancing on a torus, then if a third person were to try to unravel the ropes so that they were going straight down from your waists to the anchor point, they would never be able to do it. If you are dancing on a sphere however, the third person will always be able to unravel the ropes and so you're done!

This is essentially an application of the fact that the braid group of two strings on the torus is torsion free, but the braid group of two strings on the sphere is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

Actually, this would work with more than just two people as well - the braid group of $n$ strings on the torus is torsion free for $n\geq 2$ and the braid group of $n$ strings on the sphere has torsion for all $n\geq 2$ (though the order of torsion elements will not always be $2$ and the braid groups of the sphere are not finite for any $n\geq 3$).

We can also use this to differentiate any other closed surface from the sphere.

The braid group $B_n(T_g)$ of $n$ strings on a closed surface of genus $g$ is torsion free if and only if $n\geq 2$ and $g\geq 1$.

Dan Rust
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i found: How to distinguish between surface of a sphere & surface of a torus

Take a flat circular ring of a diameter less than Min(diameter of sphere, inner diameter of torus) and freely place it once anywhere on the given surface of sphere or torus & check if the entire periphery of ring exactly touches the unknown surface then it is the surface of a sphere otherwise it is the surface of a torus (assuming perfect geometry of sphere & torus).

jeanne clement
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Since we're considering planetary ropes, sufficient resources to mark planetary surfaces, and combing entire planetary surfaces, how about something both "parallel" and with an analog in reality...

Put a satellite in orbit around the Flatlander's city (to get out of the acoustic pollution) and either measure the resonant acoustic spectrum of the world or generate a broadband source and measure the resultant ring-down spectra. Acoustic spectra of toroids and acoustic spectra of spheroids are distinguishable. For a real-world analog, see the discussion of data derived from COBE, WMAP, and Planck data. Torus example. General article. Experimental limitations:

  • If the universe is too big, greater detector sensitivity and/or more acoustic power will be needed.
  • If the local environment is too noisy, finding a better place for the detector will be required.
  • If the global metric is too nonuniform, the spectrum may be too complicated to understand. This affects intrinsic curvature methods in other answers. A particularly likely problem will be dispersion of natural world modes.
  • Some of the above can be partially relieved by putting up more satellites and engaging in interferometry.
  • If a torus, and the geometry is a large body with a miniscule handle, then very high frequencies may be required. Greater sensitivity and acoustic power may again be needed. This issue has also been raised in other answers. If it turns out that acoustic methods cannot resolve a compact potentially toroidal anomaly, then building particle accelerators at the site of the anomaly may be required to get fine enough resolution. Note that there's no a priori promise that processes inside the world are capable of resolving arbitrary complexities in the geometry of the world, so it may be impossible to resolve a potential anomaly.
  • If the geometry of the world is changing faster than the speed of sound, this method has problems. Of course all the rope/marking/combing solutions are even more impeded... (One may think to use light, but Flatland is filled with fog. This is a crucial component of Flatlanders identifying the rank of those they meet.)

Note: I'm not saying to put something up out of Flatland. For that method, see my other answer.

Eric Towers
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    There is no 'orbit' as the Flatlanders have no 'up' for the satellite to be in. – dotancohen Jul 07 '14 at 08:21
  • @dotancohen: "Note: I'm not saying to put something up out of Flatland. For that method, see my other answer." – Eric Towers Jul 07 '14 at 14:50
  • I don't understand what you meant here : "Put a satellite in orbit around the Flatlander's city" ? – Julien__ Jul 10 '14 at 21:59
  • @Julien__: Flatlanders live in cities which are separated by large empty expanses. A satellite in a city would be swamped by local noise, so a satellite would have to be put outside of the cities, in the empty spaces. (In real life, WMAP, COBE, and Planck benefit from this sort of quiet environment by being out in space instead of buried in a hot atmosphere.) If it were stationary, it would not adequately sample the background acoustic spectrum of the world, it could also be unfortunately located in a node caused by dispersion through/around cities, so it should be made to move around. – Eric Towers Jul 11 '14 at 03:33


Constant width spirals orthogonal to geodesics shown on a torus are unbounded. But on a sphere they are bounded between two parallels shown for a hemisphere. Imagine distance between adjacent teeth of a comb with same distance between tiny wheels on an axle. A hairy torus permits combing all along their length, but a hairy sphere combs only along restricted lengths up to the red circle Out of bounds polar caps.

Hairy Torus Hairy Ball Single wheel and axle

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If you succeed in finding a non-separating loop then you know you are not on a sphere.

Jessica B
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  • The question is how would a 2-D being find such a loop? – dotancohen Jul 07 '14 at 08:19
  • Trial and error with a rope (well, in this case a bicoloured ribbon would be better), much like many of the other solutions. This isn't a complete solution, as it only provides a way of certifying you are on a torus. But it seems to me to be an easier certificate than some of the others that require knowledge of the whole surface, whereas only two paths are needed here. This is also the method I was taught to define genus. An additional advantage, as I see it, is that this doesn't rely on a choice of geometry. – Jessica B Jul 07 '14 at 15:35

I would start by assuming a flat 2D surface. Then, survey a sufficiently large right angle. The relation between the three points should show whether the true surface is a torus or a sphere, partly in line with @Fly by Night 's answer.

Over time, one could also measure the tiny fluctuations in the distance and angle between the three points and gain a wealth of gravitational and seismological data about the world as well.

  • There is no gravity in a 2-D world perpendicular to the plane of existence. – dotancohen Jul 07 '14 at 08:18
  • @dotancohen , I would figure since gravitational waves are tracked using minute changes in distance/alignment in 3 dimensions, that the same could apply to our flatlanders on a torus/sphere. – vulpineblazeyt Jul 07 '14 at 18:14

A simple answer. (If you can measure the distance travelled.)

Take a few trips on both surfaces (just go straight till you reach the initial point). If it is a sphere you will be covering the same distance every time, but not on a torus. For this you need to fix a starting point and fix a direction, which the flatlanders are capable of. You need to change the direction of travel, of each trip, by an "odd" angle (not 90 degrees or its multiple). I guess 3 trips on each surface would do.


Topologically, two geodesics emanating from the same point on a sphere will intersect in an antipodal point, whereas this is not true on a torus. So with a friend (and enough time), you can do this.

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  • If one can draw a ring, and then draw another ring which crosses the first ring only once, then one is on a torus. Failure to do that, however, does not imply that one is on a sphere. It could be that one hasn't drawn the rings in the right place. – supercat Jul 05 '14 at 17:55
  • @supercat: see my "and enough time" comment.... – Guest Jul 05 '14 at 18:37
  • If one was on a sphere, by what means would you never know that circles one had constructed were geodesics? – supercat Jul 05 '14 at 18:40