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Sorry, title probably not great, but I can't think of a better one. Here's my question. Suppose I have a $2n$ gon, with the edges identified in pairs, and no neighboring edges identified. Suppose also that I have have already done the standard thing and cut and pasted to eliminate "unnecessary" edge pairs, in other words I have the fundamental domain for surface of genus $n/2$. Now of course if I allow myself to cut and paste between any vertices I can rearrange this into the standard fundamental polygon for the surface. But suppose instead that before I allow myself to cut and paste to do this, I triangulate the polygon by connecting vertices, and then only allow myself to cut and paste respecting this triangulation. I can no longer necessarily get to the standard polygon, but has anyone seen anything about how close I can get in this case? In particular, I guess the question is how low can I make the total interleaving, say can I guarantee that no two identified sides will have more than N edges between them? I played with Mathematica about this a while ago, with $n=4$ and $n=6$, and it seemed that I could make sure that no paired edges ever had more than 2 others between them, but doubt that really means anything.

Thanks.

Here is an attempt to make clearer what I'm interested in and what I mean I agree with Lee that the question is probably confusing, I found it very difficult to explain without being able to draw a good picture and move things around. Let me try again. I am referring to the standard polygonal proof, and yes of course you might have to do a series of such cuts to get the final "standard" form. As far as what "no two sides will have more than N edges between them" means, think of starting at one edge and walking around the polygon, counting edges until you hit the edge identified with the edge you started on. What I am interested in is trying to minimize the maximum such count, in other words do this for each edge pair and for every possible rearrangement of the polygon, and then find the minimum of the maximum value. What this will depend on is the original triangulation. If you have no triangulation and allow for any cuts to rearrange, then of course the min will be $N=1$ since you will end up with $ABA^{-1}B^{-1}...$, but what I am interested in is if in an $n$ ton you start by triangulating with $n-3$ cuts connecting pairs of vertices and you are only allowed to use those to rearrange. Hope this makes it clearer.

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    $\begingroup$ I am intrigued by your question but it needs a thorough rewrite. It is unclear and confusing. When you write "if I allow myself to cut and paste between any vertices I can rearrange this into the standard fundamental polygon for the surface", are you referring to the standard polygonal cut-and-paste proof, where you cut the polygon along some diagonal and reglue along some edge pair? If so, the "rearrangement" that you refer to can be a long sequence of such cut-and-reglue operations, which you wording does not seem to acknowledge. $\endgroup$
    – Lee Mosher
    Nov 29, 2012 at 16:48
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    $\begingroup$ What does "how low can I make the total interleaving" mean? What does it mean to "guarantee that no two identified sides will have more than $N$ edges between them"? For example, what unexpressed quantities do you expect that $N$ will depend on? When you write "it seemed that I could make sure that no paired edges ever had more than 2 others between them", I cannot tell precisely what cut-and-reglue operations you are referring to here, so I do not know enough to be able to reproduce what you are saying. $\endgroup$
    – Lee Mosher
    Nov 29, 2012 at 16:51

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Here's an example. Start with a $4g$-gon $P$ with opposite sides glued, which yields a closed surface of genus $g$. Between any side-pair which is glued, there are $2g-1$ sides, which is the maximum possible. Now pick a single diagonal $L$ of the $4g$-gon, connecting any non-adjacent pair of vertices. Cut along $L$, dividing $P$ into two polygons $P_1,P_2$. It may be that $L$ connected an opposite pair in which case $P_1,P_2$ have the same number of sides, otherwise one of them, denoted $P_1$, has the smaller number of sides; with this notation, every side of $P_1$ is glued to a side of $P_2$. Let $L_i$ be the side of $P_i$ obtained by cutting $L$. Pick a side $A_1$ of $P_1$ not equal to $L_1$, and glue it to the corresponding side $A_2$ of $P_2$. Let $P'$ be the new polygon, in which $L_1,L_2$ now appear as a pair of glued sides. No matter how $L,A_1,A_2$ were picked, $L_1,L_2$ will be an opposite pair in $P'$, with $2g-1$ sides between them.

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  • $\begingroup$ Yes, but I'm interested in what happens if you do $4g-3$ cuts initially, thus cutting the original polygon into triangles which can be cut and pasted. Unless I'm misunderstanding you are only doing one initial cut, which severely limits what pasting you can do. $\endgroup$
    – Jeff McGowan
    Dec 3, 2012 at 17:44
  • $\begingroup$ OK, that helps clarify the question further, I was unsure whether that was what you meant. Anyway, I don't have any actual answer to offer that is any better than this one, I would just play with more examples. The opposite side gluing seems a good example to pursue further. $\endgroup$
    – Lee Mosher
    Dec 3, 2012 at 18:00

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