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Given an edge word $X$, let $\sigma(X)$ denote the corresponding surface (or equivalence class of surfaces defined up to homeomorphism).

I've been reading some notes on the classification of surfaces via edge words (found in this pdf). The authors mention that the connected sum of surfaces corresponds to the concatenation of the corresponding edge words, i.e. $\sigma(W_1 W_2)=\sigma(W_1)\,\sharp\, \sigma(W_2)$, where $W_i$ are arbitrary edge words, and $\sharp$ denotes the connected sum.

Later on, the authors also mention that $\sigma(X)\simeq \sigma(X^{-1})$ for any edge word $X$, which also seems very natural: reading the boundary in the opposite direction shouldn't change the corresponding surface.

However, I'm failing to see how these two conditions are consistent with each other. Don't they also imply that, for any edge word such as $ab$, we would have $\sigma(ab)\simeq \sigma(ab^{-1})$? We'd have $$\sigma(ab)=\sigma(a)\,\sharp\,\sigma(b) \simeq \sigma(a)\,\sharp\,\sigma(b^{-1}) \simeq \sigma(ab^{-1}).$$ A possible fallacy I see in the above argument is that if two surfaces are homeomorphic, $S_1\simeq S_2$, that does not mean that we also have $S\,\sharp\, S_1\simeq S\,\sharp \,S_2$. But if this was the case, then I don't understand how the classification result works: the trick is to manipulate edge words in order to progressively "extract" copies of torii and projective planes. But if we cannot apply equivalences of edge words within a connect sum, wouldn't we have to stop the argument after the first torus/projective plane has been extracted?

glS
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1 Answers1

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There is no contradiction. $\sigma(ab)$ and $\sigma(ab^{-1})$ are both a disk, since no edges are glued together at all.

I'm not sure if the article even deals with surfaces with boundary, but at first glance I think the results you mention still hold.

EDIT: I see the problem. I think the result just only work for closed surfaces, since e.g. $$ \sigma(a)\mathop\# \sigma(a^{-1}) \simeq disk\mathop \#disk \simeq tube \not\simeq sphere \simeq\sigma(aa^{-1}) . $$ It fails because we connect the surfaces at interior points, which makes the proof fail for surfaces with boundary. The article doesn't really mention it, but I'm pretty sure it's assumed that an edge-word contains each letter exactly twice, so the edges are glued in pairs. Also, the two edge words should be disjoint, i.e. not contain any common letters, since they need to represent two surfaces in their own right, i.e. they don't "know" each other. So something like $$ \sigma(aa^{-1}.bcb^{-1}c^{-1}) = \sigma(aa^{-1})\mathop\#\sigma(bcb^{-1}c^{-1}) = \sigma(aa^{-1})\mathop\#\sigma(cbc^{-1}b^{-1}) = \sigma(aa^{-1}cbc^{-1}b^{-1}) $$ is allowed.

Milten
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  • for this simple case, sure, but with the same reasoning I could invert any member of an edge word. For example, wouldn't I be able to say something like $\sigma(aa^{-1})=\sigma(a),\sharp, \sigma(a^{-1})=\sigma(a),\sharp,\sigma(a)=\sigma(a^2)$, which is clearly false? – glS Jul 09 '21 at 10:09
  • @glS Right, I see. Take a look at my edit. – Milten Jul 09 '21 at 10:45
  • @glS Actually, the most important part is that the words are disjoint. See this question about the case of surfaces with boundary. Though care must be taken in this case that "lonely letters" (i.e. those corresponding to a boundary) are not next to each other. – Milten Jul 09 '21 at 10:58
  • ah, that makes sense, thanks! – glS Jul 09 '21 at 11:20