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The topology of a closed surface can be constructed by identifying edges of a fundamental polygon of an even number $2n$ of edges. Labeling the edges and using $\pm 1$ exponents to indicate direction, the construction can be specified by a string of $2n$ symbols: $a b a^{-1} b^{-1}$ for the torus, $a a b b$ for the Klein bottle, etc. (Reflecting a comment by Benjamin Steinberg:) Each letter in a fundamental polygon string appears exactly twice.

Let's say two symbol strings are equivalent if they are related by a combination of (a) circular permutation, (b) reflection/reversal, or (c) symbol permutation/relabeling. For $n=1$, there are two distinct strings, $aa$ and $aa^{-1}$. For $n=2$, I think (not certain) these are the combinatorially distinct strings: $$ aabb,\; aa^{-1}bb,\; a^{-1}abb,\; aa^{-1}bb^{-1},\; a^{-1}abb^{-1},\; $$ $$ abab \;, aba^{-1}b,\; aba^{-1}b^{-1},\; a^{-1}ba^{-1}b $$


FundPoly
Two questions:

Q1. Does every possible such string correspond to some surface?

Q2. Might two combinatorially distinct strings correspond to the same surface?

Perhaps $aa^{-1}bb^{-1}$ and $a^{-1}abb^{-1}$ describe the same surface, as they only differ in $aa^{-1}$ vs. $a^{-1}a$?

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Doesn't permutation of $a$ and $a^{-1}$ leave the surface invariant? Or am I saying something really stupid? –  Per Alexandersson Jun 27 at 12:16
    
More likely abab and a'ba'b are the same surface. (Not feeling up to superscripts right now.) –  The Masked Avenger Jun 27 at 12:22
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I think that any identification of this form corresponds to a triangulated surface. E.g. in the case of the square, take the 3x3 grid and subdivide each square in two triangles. Every side is adjacent to two triangles, and every vertex has link homeomorphic to $S^1$. Now it should be a matter of counting: there are only so many surfaces you can represent with a $2g$-gon (since $\chi$ is bounded by a linear function of $g$). –  Marco Golla Jun 27 at 12:26
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aabb is nonorientable; also, it has one vertex, two edges, and one face, so Euler characteristic 0; the same is true for aba'b, so they are homeomorphic (in fact, each is a Klein bottle). This sort of analysis can be used to determine all of the diagrams. –  Gerry Myerson Jun 27 at 12:44

1 Answer 1

up vote 14 down vote accepted
  1. Yes. 2. Yes. (I suppose that the surfaces are "the same" if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic, and the 2-valued invariant, orientability. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$. And orientability has two values. So you have at most $2n$ topologically different surfaces from strings of length $2n$.

And you see that for given length there are much more classes of words than $2n$.

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Thanks for the clear answer! –  Joseph O'Rourke Jun 27 at 12:47
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Isn't this a proof that the answer to 2. is Yes, i.e. that combinatorially distinct strings can correspond to the same surface ? –  Arnaud Jun 27 at 12:58
    
Arnaud, thanks for your correction. –  Alexandre Eremenko Jun 27 at 13:09
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I am confused. There are one-relator groups with even length relations that are not surface groups. The string $ab^6a^{-1}b^{-8}$ defines a non-Hopfian Baumslag-Solitar group and hence is not a surface group. –  Benjamin Steinberg Jun 27 at 14:20
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To get a surface each letter should appear exactly twice. Maybe this was implicit in the question. –  Benjamin Steinberg Jun 27 at 17:31

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