Presentation of the monoid of surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:54:28Z http://mathoverflow.net/feeds/question/20438 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20438/presentation-of-the-monoid-of-surfaces Presentation of the monoid of surfaces Martin Brandenburg 2010-04-05T23:28:38Z 2010-04-06T03:23:47Z <p>In the following every surface is assumed to be connected. I've read that the commutative monoid of homeomorphism classes of closed surfaces is generated by $P$ (projective plane) and $T$ (torus) subject to the only(!) relation $P^3=PT$. Here the product is given by the connected sum. Now what about the commutative monoid of homeomorphism classes of compact surfaces (with boundary)? Does it also have a nice presentation? I think it is generated by the $P[k]$ and the $T[k]$, where $[k]$ means that $k$ holes have been inserted, and $k$ runs through the natural numbers. What relations do we need? And how do you prove that no others are needed? </p> <p>Another question, which is rather informal: Do you think that it's worth to read the proofs of these classical classifications? I know the importance of the results, but I suspect that the proofs are just technical.</p> http://mathoverflow.net/questions/20438/presentation-of-the-monoid-of-surfaces/20443#20443 Answer by Tom Church for Presentation of the monoid of surfaces Tom Church 2010-04-05T23:55:54Z 2010-04-05T23:55:54Z <p>I assume you want your surfaces-with-boundary to be compact? Anyway, this cannot be generated by the $P[k]$ and $T[k]$, since you are leaving out the genus 0 surfaces (spheres with holes). Since connect-sum-with-a-disk is the same as removing an open disk, I would work instead with the generators $P$, $T$, and the disk $D$; the $D$ won't interact with the other generators. Since every surface-with-boundary is a surface-minus-some-disks, it seems a presentation is $\langle P,T,D\vert P^3=PT\rangle$.</p> <p>In answer to your second question: it can be very worthwhile to know the proofs of these classical facts, but that doesn't mean you need to learn the classical proofs. For the classification of surfaces, I have seen Benson Farb give a very nice proof (fitting with the "modern" perspective on mapping class groups etc.) hinging upon the fact that the sphere has the maximal Euler characteristic among surfaces. If I can find any notes of that lecture or a written version, I'll update with a link.</p> http://mathoverflow.net/questions/20438/presentation-of-the-monoid-of-surfaces/20468#20468 Answer by Andy Putman for Presentation of the monoid of surfaces Andy Putman 2010-04-06T03:23:47Z 2010-04-06T03:23:47Z <p>I agree with Tom and Ryan that it is worthwhile to learn proofs of the classification of surfaces. I think that the result get a bit of a bad rap since the "standard" combinatorial proof that everyone used to learn (which appears in Seifert-Threlfell's book and Massey's book) is complicated and unenlightening. However, there are now a number of nicer proofs available. Here are a few of my favorites.</p> <p>1) If you like Morse theory, there is a nice proof in Hirsch's book on differential topology.</p> <p>2) There is a slick combinatorial proof in Armstrong's book "Basic Topology". I believe that this is the source for the proof mentioned above by Tom Church that Benson Farb likes to give.</p> <p>3) In Fomenko-Matveev's book "Algorithmic and Computer Methods for Three-Manifolds", there is a nice proof using handle decompositions.</p> <p>4) There is finally John Conway's "ZIP proof", which was written up by Francis and Weeks in their paper "Conway's ZIP Proof".</p> <p>All of these proofs assume that the surface has been equipped with either a triangulation (for numbers 2-4) or a smooth structure (for 1). For nice approaches to this, see the answers to my question <a href="http://mathoverflow.net/questions/17578/triangulating-surfaces" rel="nofollow">here</a>. However, when you are first approaching these types of results, I would recommend just assuming that the surfaces can be triangulated or smoothed.</p>