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The answer to this question shows two different fundamental polygons to represent a double torus, i.e.

enter image description here enter image description here

Let's call $I=A,II=B,III=C$ and $IV=D$, start from left and go counter-clockwise, then their identifications of their sides are:

  • left: $\,\,\,A\phantom{^{-1}}D^{-1}C^{-1}B^{-1}A^{-1}D\phantom{^{-1}}C\phantom{^{-1}}B\phantom{^{-1}}$
  • right: $A^{-1}B\phantom{^{-1}}A\phantom{^{-1}}B^{-1}C^{-1}D\phantom{^{-1}}C\phantom{^{-1}}D^{-1}$

What's difference between them? Are there more $8$-sided fundamental polygons for the double torus?

UPDATE: I just found the two above listed here:

enter image description here enter image description here

flipped and relabelled versions of the above...

draks ...
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2 Answers2

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There are four octagon gluing patterns (up to rotation and relabelling) which give a double torus. Starting with the two that you have depicted, here they all are:

  • a b c d A B C D
  • a b A B c d C D
  • a b A c B d C D
  • a b c A B d C D

To convert these into pictures, start from one corner of the octagon and walk around it in the counterclockwise direction. When you see lower case, write that letter and a counterclockwise arrow. When you see upper case, write the lower case letter and a clockwise arrow.

What distinguishes these patterns is that all 8 corners are identified to give $V=1$ point. The 8 sides are identified in pairs to give $E=4$ edges, and of course there is $F=1$ face. The Euler characteristic formula gives $$\chi = V-E=F = 1 - 4 + 1 = -2 $$ For an oriented surface of genus $g$ --- the double torus has $g=2$ --- one has $\chi = 2-2g$. So all together we get $-2=2-2g \implies g=2$, as expected.

Other (oriented surface) octagon gluing patterns will identify the 8 corners into either 3 or 5 points, as it turns out, with (respective) Euler characteristics $\chi = 0$ or $2$, giving the surface of genus $1$ which is the torus), or of genus $0$ which is the sphere).

Rigorous proofs of all these relations between Euler characteristic and genus can be found in any treatment of the Classification of Surfaces. The other statements here, regarding octagon gluing patterns, are proved by simply enumerating combinatorial possibilities.

Lee Mosher
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  • Interesting. Do you think it's possible to use all kinds of possible 8-letter words of the octagon to get a consistent labelling of octagons in the hyperbolic plane? This would answer this question as well: https://math.stackexchange.com/q/3341839/19341 – draks ... Sep 08 '19 at 15:29
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The right-hand one emphasizes the double torus (I prefer "genus two surface") as a Connected Sum of two (one-holed) tori. The left hand one emphasizes the genus two surface as a Translation surface. From the perspective of topology, the surfaces are identical, but their construction differs. Not all ways of labeling the sides of the octagon will give you a genus-two surface. $A^2B^2C^2D^2$ will give you something nonorientable. I would conjecture that any way of writing an 8-letter word in $A,B,C,D$ and their inverses such that each letter appears once positive and once negative and no label is next to its inverse will work, but I don't have a proof.

However, the theory of translation surfaces suggests that there are infinitely many ways to glue the sides of a polygon to produce a genus-two surface.

Rylee Lyman
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  • Maybe some symmetries reduce the number of possible 8-letter words down to 4, as in Lee's answer. Any idea how to give the other two a geometrical meaning? – draks ... Sep 08 '19 at 15:26
  • Thanks for pointing out the translation surface aspect. I think I found a representation of such a hyperbolic surface here: https://math.stackexchange.com/q/3341839/19341

    What do you think?

    – draks ... Jan 08 '22 at 11:03