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I've faced some difficulty in how to equip an orientable compact connected surface of genus n with a CW-structure using the 4g-gon model.

I understand how a torus is constructed:

-Start with 1 point (a 0-cell)

-Add two lines which their start and end point are mapped to that point (2 1-cells)

-Add a 2-surface that has two opposite "edges" mapped to one of the lines and the other two mapped to the other line. (a 2-cell)

This gives us a Euler characteristic $1 - 2 + 1 = 0$.

However when I try to extend what I did to a 2-torus I do:

-Start with 1 point (a 0-cell)

-Add three lines which their start and end point are mapped two the point (3 1-cells)

-Add two 2-surface, attaching the two opposite "edges" of each to one of the lines and the other two mapped to the third line which they have in common.

This gives an Euler characteristic $1 - 3 + 2 = 0$. However the double torus should have an Euler characteristic $-2$.

If I extend this process to form an n-torus, it seems that I seem always will get $0$, and it's when that it should be $2 - 2n$.

I am probably misunderstanding how an n-torus is constructed and what I am constructing is not homotopically equivalent to an n-torus except perhaps in the case $n = 1$.

I was wondering if somebody could elucidate this issue?

Farrokh
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Julian Sutherland
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    Do you know the "standard $4n$-gon picture"? In a web search you'll have more success by calling the genus "$g$". Here's the case $n = 2$. :) – Andrew D. Hwang May 27 '14 at 17:56
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    Yes I was, but I had difficulty with visualising the folding beyond the torus, so that was extremely helpful! I now see how what I constructed was not a 2-torus. So in general it seems to be one point ($0$-cell) with $2n$ lines attached to it and one $2$-cell then covering the whole thing, which translates to an Euler characteristic of $2 - 2n$, which is correct. My difficulty in constructing an n-torus as a CW-complex then is in general how does one attach the $2$-cell? – Julian Sutherland May 27 '14 at 18:17
  • To complement the comment of @user86418, another standard cell-complex construction of surfaces with genus $g\geq 2$ is via the 'pair of pants' decomposition where the pair of pants themselves have a standard cell-structure with two $2$-cells, nine $1$-cells and six $0$-cells (the obvious way that you would construct a closed disk with two smaller open disks removed). – Dan Rust May 28 '14 at 16:22

1 Answers1

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Just so this question has an answer, here's a more detailed sketch of the "standard $4n$-gon" gluing.

Label the (oriented) $1$-cells $a_{i}$ and $b_{i}$, with $i = 1, \dots, n$. Divide the boundary of the $2$-cell into a $4n$-gon. Pick an arbitrary edge, and proceed counterclockwise around the boundary, labeling successive edges $a_{1}$, $b_{1}$, $a_{1}^{-1}$, $b_{1}^{-1}$, $\dots$, $a_{n}$, $b_{n}$, $a_{n}^{-1}$, $b_{n}^{-1}$. (Some of these choices are not topologically important, just made for definiteness.) Attaching the boundary of the $2$-cell to the $1$-skeleton as indicated accomplishes the desired gluing.

Remarks: Geometry and the Imagination by Hilbert and Cohn-Vossen contains a picture of the case $n = 3$. If memory serves, Algebraic Topology by Massey also has helpful diagrams.

It may help to draw a genus-$n$ surface and the $2n$ curves analogous to the upper left diagram in the linked post. (In that diagram, $a \mapsto a_{1}$, $b \mapsto b_{1}$, $c \mapsto a_{2}$, and $d \mapsto b_{2}$.) Then convince yourself that the product of the commutators, $\prod_{i} a_{i} b_{i} a_{i}^{-1} b_{i}^{-1}$, is homologous to a small loop around the $0$-skeleton, i.e., the vertex of the $4n$-gon. (The $2$-cell is attached to the "thickened $1$-skeleton" by filling in this loop with a disk.)

The bottommost picture in Neal's answer indicates how to make the construction inductive, i.e., how to cut a surface of genus $(n + 1)$ into surfaces of genus $n$ and genus $1$.

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Andrew D. Hwang
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