I would like to know what the following spaces $X$ and $Y$ look like. More precisely, I want to know if they are homeomorphic to some other known spaces. I define $X$ and $Y$ as a quotient of the square $[0,1]^2$ by gluing the edges (red with red, blue with blue):

$\qquad\qquad\qquad\quad$Unknown gluing

In other words, $X=[0,1]^2 / \sim$ and $Y=[0,1]^2 / \sim'$ where $\sim$ is the equivalence relation defined by $$(x,s) \sim (y,t) \iff (x,s)=(y,t) \;\text{ or }\; s=0, y=1, t = 1-x \;\text{ or }\; s=1, y=0, x = 1-t $$ and $\sim'$ is the equivalence relation defined by $$(x,s) \sim' (y,t) \iff (x,s)=(y,t) \;\text{ or }\; s=0, y=1, t = x \;\text{ or }\; s=1, y=0, x = 1-t $$

I tried to see with a handkerchief what I could obtain when gluing the edges as shown on the diagrams, but it was quite unsuccessful. I believe that I would obtain non-orientable spaces, maybe a wedge of projective space and something else. I failed to understand what $X$ and $Y$ are homeomorphic to.

However, I know what the following similar diagrams give :

$\qquad\quad$RP2 and S2 $\qquad\qquad\qquad\qquad$Klein+torus

Any suggestion will be appreciated, thanks in advance!


3 Answers 3


To see that the first space is homeomorphic to $\Bbb R \mathrm P^2$, cut it along the top-left to bottom-right diagonal and glue together the blue edges. You will end up with the standard representation of $\Bbb R\mathrm P^2$ in terms of a square with face identifications.

Your second space is a disk (2-cell) glued to a bouquet of two circles $a,b$ (after identifying the loops pairwise) along the loop $a^2b^2$. This is the Klein bottle.


The first diagram simplifies in that the two red edges cancel each other. The resulting gluing diagram is a standard diagram for the projective plane, namely a disk glued to a single circle by a double covering map of the boudary of the disc around the other circle.


Since you say you were trying to glue a handkerchief, I suppose you would like some geometric intuition on how these gluing diagrams are interpreted. The study of orbifolds (quotients of manifolds by symmetry groups satisfying certain constraints; any discrete group will do) provides a generalised, natural way to understand and notate these spaces. I will try to help you construct/visualise physical models of these spaces.

Here is my process of interpreting the gluing diagrams:

  1. Two matching arrows whose tips meet at a point signify that a fold should be performed whose crease passes through said point. This produces what we call a cone point. Why? Well, if you take a flat sheet of paper and fold it through a corner so that the edges meet, what you have is diffeomorphic to a cone. The existence of a cone point is designated by a positive integer $N$ whose magnitude indicates the steepness of the cone, although this doesn't matter up to homeomorphism.
  2. Two matching arrows that meet at a point, the head of one arrow and the tail of the other, signify a reflectional boundary. The reason for this is a bit more difficult to explain. We notate the existence of a reflectional boundary with the symbol $*$.

A reflectional boundary in a geodesic metric space is a geodesic, so we redraw these gluing diagrams with these vertices straightened out. i.e. the first one looks like this:

$\hskip2.1in$ enter image description here

The folding cannot be realised with a piece of paper, because this is a quotient space of the sphere (with positive Gaussian curvature):

$\hskip1.4in$ enter image description here

The first diagram has one cone point and one reflectional boundary. This makes it an $N*$ orbifold, as mentioned in the caption to the image above. I believe this folding pattern can be embedded isometrically in $\mathbb R^3$. After a quick calculation for an order $2$ cone point, it looks like this:

$\hskip2.15in$ enter image description here

This will be a lot more stretched and thin for the spherical tiling pattern shown above. The bounding disk at the bottom is to be interpreted as follows: when a geodesic passes through the boundary, it is reflected specularly. There's a hitch, however. You may have noticed that in the picture of the sphere above, more points are identified with one another than just those on the boundary. This is solved by allowing this shape we've created to have two sheets: enter image description here (Really, each of these sheets should be a cone half as thick, but I didn't think about that until now). The seam between these sheets is seen as a sudden change in colour where we should switch to the other sheet to keep the colour changing continuously as we follow a geodesic through the seam. Now if a geodesic passes through the disk boundary, it should be reflected specularly and emerge from the other sheet.

Here's the kicker: a really great way to envisage this with a physical model is to create a single copy of this shape and mark the seam. Now when you pass through the seam, or hit the disk boundary, allow yourself to keep drawing on the other side of the shape. By distinguishing between the two different sides of the $2$-dimensional surface, we allow ourselves to have two isometric sheets.

The second example is trickier. It has two reflectional boundaries which meet at two vertices. As a quotient space of the sphere:

$\hskip1.6in$ enter image description here

I'm not sure if there's a way to embed this isometrically in $\mathbb R^3$ without identified points. Leave a comment if you think of something.

The Spherical Images were taken from Conway's Symmetry of Things (ISBN: 1568812205)

I should note that the examples I've illustrated are spaces with uniform, positive curvature. You only asked about what these spaces are up to homeomorphism, but that's a bit boring ;)

  • $\begingroup$ Thank you for your nice answer! I didn't know the notion of orbifold. $\endgroup$
    – Watson
    Jul 24, 2016 at 17:53
  • 1
    $\begingroup$ All manifolds are orbifolds, but orbifolds allow for singluraties in the metric a.k.a. cone points whereas manifolds do not. If you're interested in learning more about them, I recommend Conway's Symmetry of Things as the first 5 chapters are wonderfully illustrated and very digestible even by the layman. $\endgroup$
    – Myridium
    Jul 24, 2016 at 18:04

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