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I am aware that the 2-hole torus can be identified with the octagon with the equivalence relation as given in this picture:

https://www.quantamagazine.org/wp-content/uploads/2012/09/Haken_sized_Figure02.jpg

However, today in a topology revision lecture, the lecturer said the 2-hole torus can be represented by the octagon with opposite sides identified. This is different to the picture, so was our lecturer wrong? If so what surface is given by the octagon with opposite sides identified?

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  • $\begingroup$ I think he was right. On the picture you gave, there is no evidence how you identify the sides. However, you can see on the double torus that all the lines meet in one point, which basically gives you the hint that the opposite sides should be identified. $\endgroup$ Apr 17 '15 at 14:38
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    $\begingroup$ @LaurentHayez Clearly in the picture any two sides with the same letter are identified? $\endgroup$ Apr 17 '15 at 14:41
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The space resulting from gluing opposite edges of an octagon naturally depends on the orientation of the gluing, but assuming "antipodal edges" are glued with the same orientation, as when building a torus from a square, the result is indeed a two-holed torus.

Gluing opposite edges of an octagon

It's easy to see that

  • Two vertices that are endpoints of a dashed segment are identified, so all eight vertices are identified. Consequently, the interior angles of the octagon are joined cyclically; after identification, the vertex has a disk neighborhood.

Vertex identifications of an octagon

  • Each edge is identified with the opposite edge, and no others, in a manner compatible with the orientation of the octagon.

It follows that the identification space is an orientable surface having a cellular decomposition with one vertex, four edges, and one $2$-cell. The Euler characteristic is $\chi = 1 - 4 + 1 = -2$. By the classification of surfaces, the glued octagon is a two-holed torus.

(Presumably the same conclusion can be reached by surgery, modifying the octagon to the "conventional" gluing, though I haven't checked carefully.)

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  • $\begingroup$ Do our solutions come to the same result? Because I chose to glue the opposite sides of the octagon with a different orientation. In fact my solution comes from my current differential geometry class, and from this book books.google.ch/…. But I'm still a undergrad student, so there's no guaranty that my solution is correct. $\endgroup$ Apr 17 '15 at 18:12
  • $\begingroup$ @LaurentHayez: Modulo mistakes on my part, yes, the two gluings must be topologically equivalent. :) (The gluing in your answer is the "standard" presentation of the two-holed torus, and certainly correct.) The way I read the OP's question, their instructor stated that the "antipodal" gluing gave a two-holed torus, and they wanted to confirm this assertion. $\endgroup$ Apr 17 '15 at 18:37
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Edit/disclaimer:

Although the solution below does indeed identify the two-hole torus with an octagon, it does not answer OP's question which was the two-hole torus can be represented by the octagon with opposite sides identified. I do not explain how to identify the opposites sides of an octagon, but rather how to glue two representations of a torus to get a two-hole torus.


Original answer:

Basically, what you want to do is to glue two tori.

Both tori can be identified with the following representation: enter image description here

Now to glue them together, you want to cut a small piece of each torus, and glue them in a certain way on the area you cut. We are going to cut a small triangle at points $p$ and $q$ as following way:

Note that $p$ and $q$ are note at the same place in both tori!

Now we glue them together, $p$ goes on $p$ and $q$ goes on $q$ which gives:

enter image description here

Redrawing it as an octagon, it gives:

enter image description here

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  • $\begingroup$ very useful thank you. $\endgroup$
    – user111750
    May 9 '16 at 18:48
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    $\begingroup$ Wasn't the question about opposite sides glued together? $\endgroup$ Nov 16 '17 at 8:28
  • $\begingroup$ @EricDuminil Yes you are right. I misread the question before answering, but as my answer was upvoted, I did not delete it. $\endgroup$ Nov 21 '17 at 7:39

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