I am learning the basic concepts of Topology, and playing now with the gluing diagrams (describing the fundamental domain of a topological space), this is an excerpt of a basic description I took from this page.

"In a gluing diagram, arrows or other markings are used show where a surface should be connected up with itself. A square without markings is just a square. It has boundaries in all directions. Now if we connect the left side to the right side, then a flatlander living in this space could go out the right side and come back in on the left. In fact, the flatlander could travel for ever in that direction without coming to a boundary. The top and bottom are still boundaries, though, so the flatlander couldn't travel far in either of those directions. "

These are samples of some basic configurations (left-right, up-down hope to be correct: cylinder, square, Möbius strip, torus, Klein bottle, and the real projective plane) and my question is below them:

enter image description here

The question I would like to ask is:

Is it possible to glue the surface with itself in the same point, like if it was an angle of reflection of a mirror? so if the flatlander goes through one of those points which are glued to themselves, the flatlander will appear gradually exactly in the same place but in the bouncing direction of the angle of reflection? If that is possible, how is the gluing diagram drawn?

Update (2015/08/25):

I have prepared an image of the question with a cylinder as example, "east" and "west" are glued, so the flatlander can walk along the surface and make a complete round, but the "north" and "south" are glued to themselves and the flatlander exiting will return exactly to the same point but "bouncing" in the point where is going out of the plane (like an angle of reflection in a mirror):

enter image description here

I have been reading some previous questions here at MSE, but I did not find hints about this point. I apologize because probably it is very basic. References or links to information about gluing diagrams covering this question are very appreciated, thank you!

P.S. for a very nice visual description of some of the above diagrams, there is a wonderful video named "The Shape of Space" (recommended!).

  • 1
    $\begingroup$ Are you familiar with the idea of an "orbifold"? I know essentially nothing about them, but I think they're basically what you're looking for, since they allow "mirrors" to exist, whereas manifolds do not. $\endgroup$ Aug 13, 2015 at 2:42
  • $\begingroup$ @MiloBrandt first time I hear about them, reading now Wikipedia, thank you very much! $\endgroup$
    – iadvd
    Aug 13, 2015 at 2:57

2 Answers 2


I am adding the answer I received by email from Doctor Tadashi Tokieda, he is the director of studies in mathematics at Trinity Hall, University of Cambridge, some bio here and here, I was following his Topology and Geometry open lectures at Youtube for the AIMS, so I dared to send him an email yesterday (I did not expect an answer, just tried) and received his answer!! (thank you very much Doctor Tokieda, it was a honor!)

He sent me a very nice easy to follow explanation of the concepts, so I am transcribing exclusively the explanation here as a complement (more adapted to layman terms) to the good answer of @AloizioMacedo. The drawings are mine, so I apologize if they are not very accurate. Here it is:

Let us take the example of a disk. Think of the disk as a billiard table and of its boundary circle as the cushion rails. If I understood aright, your picture is: shoot a particle, it slides along a straight line, but when it hits a cushion rail, it reflects, and slides along another straight line... Capture this picture by some sort of gluing diagram.

How about this? Take two copies of the disk, $A$ and $B$. Sew $A$ and $B$ together along their boundary circles, like a pita bread. Puff inside a little, so that the resulting closed surface $S$ looks like a very squashed sphere, with $A$ as the northern hemi-pita and $B$ as the southern hemi-pita. At every point along the equator of $S$, the surface is smooth and its tangent plane is vertical.

Imagine a particle sliding on $S$, and watch it from "above". While it is sliding on $A$, it appears to travel along more or less a straight line on a disk. As the particle approaches the equator, crosses the equator, and gets into $B$, you see its image from above approach the boundary circle of the disk, reflect off the boundary, and continue traveling on the disk more or less along a straight line. We had to say "more or less", because the curved $A$ and $B$ mean that the images of the trajectories seen from above are not exactly straight. But you agree that, as we squash $S$ more or more flat, the trajectories converge to straight lines. In summary, the picture you want on a domain (our example was a disk) can be realized by doubling the domain into a closed surface. The dynamical behavior you want on the domain is the projection of the dynamics on that surface.

enter image description here

There is a whole sub-field of mathematics called "billiards". There is also a well-studied topic of geodesics (trajectories of sliding particles) on Riemannian manifolds. Our discussion above shows that you can reduce the former to the latter, by taking the "squashing" limit.

See: Dynamical billiards at Wikipedia.


I am not sure if it is possible to glue it with the "opposite direction". Maybe this can be possible in some sense if you are working with orientable manifolds (since you are talking about "surfaces", I think you may be assuming the topological spaces to be $2$-manifolds) by reversing orientation after the process I will describe happens... but I'm not completely sure this will satisfy your intentions.

The process I will describe is general: it applies to any topological space $X$.

Take $X$ and take the disjoint union of $X$ with itself. This leaves a topological space $Y$ which is essentially two $X$'s that are disconnected (I'm not being rigorous enough, but the answer would be lengthy if I were.)

$Y$ is a topological space in its own right. Now, what you do is the following: you identify two points $x \in X$, one in each $X$, and let every other point be only identified with itself. By doing this, you create an equivalence relation $\sim$ on your $Y$. Now, if you consider the space $Z:=Y/ \sim$, this $Z$, with the natural topology that can be given to it (which can be characterized, for example, by the fact that it is the biggest topology that makes the projection $Y \rightarrow Y/ \sim$ continuous), should be the space you are searching for.

For example, if you take a closed ball $B$ and do this process, you should arrive at (something homeomorphic to) two touching balls as a result.

In fact, the things that you mention are all cases of this "quotient topology" I mentioned: note you are making equivalence classes in all of them, identifying points!

  • $\begingroup$ thank you for the explanation!an example of what I am trying to express is: in the gluing diagram (1) in the question,a cylinder,if the "flatlander" goes up to the "north" side (as we look at the graph,the top of the gluing diagram),in the moment it goes out,it will be back in opposite direction heading south,or in the moment he goes out in the south side he will be back in opposite direction heading north.If I understood correctly,your solution implies the existence of two X manifolds,while in my example,there is only one manifold,so I am not sure if it would be the same. $\endgroup$
    – iadvd
    Aug 10, 2015 at 4:19
  • $\begingroup$ Hmm... I'm not sure if I understood your question correctly then. Could you please elaborate more what you mean by "glue the surface with itself in the same point?". $\endgroup$
    – Aloizio Macedo
    Aug 10, 2015 at 5:00
  • $\begingroup$ the example above:when you exit, you return to the surface exactly in the same location where you left,and your direction is exactly the opposite direction of the direction you had when you went out of the surface through that side,so when you are going out you,are really getting again into the surface through exactly the same position.Probably the only way to do it is the way you explain,but then there is no gluing diagram for it, right? $\endgroup$
    – iadvd
    Aug 10, 2015 at 5:15
  • $\begingroup$ sorry to make you wait, and thank you for your answer. I have added the answer that I received yesterday from Doctor Tadashi Tokieda, I did not expected it and it was a great surprise, he gave me also an insight in layman terms similar to your explanation, and gave an insight about the "dynamical billiards" sub-field. I hope you enjoy it too! :) $\endgroup$
    – iadvd
    Aug 28, 2015 at 1:03

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