A few questions on pair of pants, surfaces. A pair of pants is another word for a disk with two holes.


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*How do I see that a surface of genus $g$ can be decomposed into $2g - 2$ pairs of pants?

*How do I see that the number of such decompositions, up to combinatorial equivalence, is equal to the number of graphs with $2g - 2$ vertices, and $3$ edges at every vertex?

*What is the number of such graphs for $g \le 5$?

 A: *

*Any surface of genus g can be continuously deformed into a linear chain of g holes glued together. Call the direction along the chain the $x$-direction, and the perpendicular direction the $y$-direction. Cut the chain in half along the $x$-direction, and cut each hole along the $y$-direction on both sides of the first cut, except for the two holes at the ends. This partitions the surface into pairs of pants, giving two for each hole (associate each pair of pairs of pants to their adjacent hole in the $+x$-direction), except for the holes at the ends, hence $2g - 2$ pairs of pants.

*For each decomposition, contract each pair of pants down to three edges joined at a vertex. This associates each decomposition to a graph of $2g - 2$ vertices (one for each pair of pants), with three edges at every vertex. Since every such graph can be obtained this way (simply uncontract the graph to find the decomposition it came from) and exactly one graph is associated to every decomposition, this association defines a bijection. The number of decompositions is therefore equal to the number of graphs of $2g - 2$ vertices with three edges at every vertex.

*This is the number of simple cubic graphs, the number of which on $4$, $6$, and $8$ vertices ($g \in \{3, 4, 5\}$) is $1$, $2$, and $5$, respectively (there are zero such graphs on $2$ vertices). Therefore, the number of these graphs for $g \leq 5$ is 8.
