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I am trying to identify the topological type of this fundamental polygon and I think it is the projective plane or the Klein bottle

fundamental polygon

If we treat the top green and red arrows a single arrow then we can call this $ab$

Doing a similar thing with the bottom two arrows we can combine them to get $a^{-1}b^{-1}$

Since they are inverses, can we swap the direction of $a^{-1}b^{-1}$ to get $ab$ in the opposite direction.... leading to the fundamental polygon of the projective plane?

I was thinking that other possibility is that it could be the Klein bottle since the new top and bottom arrows are in the same direction... but surely this would only work if $ab=a^{-1}b^{-1}$ ... i.e. the space is abelian... so maybe the Torus?!

I am really confused about this one so it would be good to gain some insight into what is happening, thanks

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    $\begingroup$ This is indeed a Klein bottle: consider the two edges $a,b$ as a unique edge. Glueing them you get a cylinder, whose two borders (the last edge: $c$) must be glued with opposite orientations: this gives rise to a Klein bottle. $\endgroup$
    – Crostul
    May 25, 2016 at 12:53
  • $\begingroup$ thanks, I think the thing that confused me was the $^{-1}$'s on the bottom red and green arrows $\endgroup$
    – amiz9
    May 25, 2016 at 13:21

2 Answers 2

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Closed surfaces are determined by the two properties:

  • Orientability
  • Euler characteristic

As you have already correctly determined, the considered surface is not orientable, because the edge $c$ is glued with the opposite orientation. Thus one has to determine the Euler characteristic $\chi$, which can be calculated by the formula: $$\chi = \#\{Vertices\} - \#\{Edges\} + \#\{2-cells\}$$

In your example, the number of edges (=3) and number of $2$-cells (=1) are easily determined. To determine the number of vertices, one has to understand which of the $6$ vertices in the picture are identified. It turns out that there are 2 distinct vertices in the quotient.

Hence, one has $\chi = 2 - 3 + 1 = 0$, thus the quotient surface is a Klein bottle (The projective space has Euler characteristic $\chi= 1 - 1 + 1 = 1$).

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  • $\begingroup$ Thank you. I understand the argument, except I am a little confused about the number of edges and vertices. I would be inclined to think of $a$ and $a^{-1}$ as different edges. But they are the same just with different orientability, right? For the vertices, I see that the 4 vertices either side of the two $c$ edges correspond to 2 identified vertices.... are the remaining two vertices in the diagram neglected since we can combine $a$ and $b$ into one arrow? $\endgroup$
    – amiz9
    May 25, 2016 at 13:08
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    $\begingroup$ Because gluing the surface is a process, there are two objects one has to consider: The hexagon (before the gluing), and the glued surface. Before gluing, the hexagon has $6$ edges ($2$ of each color) and $6$ vertices. In the gluing process, the edges of the same color are identified, so in the end there are only $3$ edges. For the vertices, there is in fact only one vertex adjacent to the edges named $c$: The a priori different end point are identified when the two green (or red) edges are glued together. $\endgroup$
    – Oles Wohnzimmer
    May 25, 2016 at 13:16
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The notation in your picture is inconsistently overdetermined, and this seems to be causing you confusion.

The arrows on the sides are already enough to indicate orientation, so you should not also put a superscript $-1$ on any arrow labels. Thus you should replace the $a^{-1}$ by an $a$ and the $b^{-1}$ by a $b$.

It is then clearer that you can combine the upper and lower green-red-arrow-pairs into a single arrow pair, and then it is just the standard gluing diagram for the Klein bottle, as you say.

The appropriate place for "$-1$"s to appear is when you go around the gluing diagram and write down the corresponding word in $a,b,c$ and their inverses. Thus, starting from the upper right $a$ and going counterclockwise, the word is $abcb^{-1}a^{-1}c$, where you write a $-1$ superscript when going backwards against the direction of the arrow.

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  • $\begingroup$ So in general if two consecutive arrows are in the same direction then we can treat them as one arrow in that direction? I still do not understand why we can replace $a^{-1}$ by $a$ since they are different generators- I though $a^{-1}$ works in the opposite direction to $a$. Could you help me understand this? ,,, or is $a=a^{-1}$ since we are already given that they act in the same direction? $\endgroup$
    – amiz9
    May 25, 2016 at 12:59
  • $\begingroup$ I added a paragraph to address your questions about the $-1$ superscript. $\endgroup$
    – Lee Mosher
    May 25, 2016 at 12:59
  • $\begingroup$ So if we are given $a=a^{-1}$? $\endgroup$
    – amiz9
    May 25, 2016 at 13:00
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    $\begingroup$ We are never given $a=a^{-1}$. The letter $a$ is a label which we assign to a pair of oriented sides which are to be identified in the quotient space. The symbol $a^{-1}$ is a label that applies to the same pair of sides but with the opposite orientations. $\endgroup$
    – Lee Mosher
    May 25, 2016 at 13:02
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    $\begingroup$ It is correct if it is said more carefully. If you have two consecutive arrows in one segment of the diagram, and two consecutive arrows in another nonoverlapping segment of the diagram, and if in both of those segments the arrows have identical labels and orientations, then yes, each of those segments may be replaced by a single arrow. $\endgroup$
    – Lee Mosher
    May 25, 2016 at 13:13

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