# The fundamental group of a closed surface without classification of surfaces?

The fundamental group of a closed oriented surface of genus $g$ has the well-known presentation

$$\langle x_1,\ldots, x_g,y_1,\ldots ,y_g\vert \prod_{i=1}^{g} [x_i,y_i]\rangle.$$

The proof I know is in two steps: 1. draw your favorite presentation of the surface onto a sheet of paper and compute the fundamental group using Seifert-van Kampen. 2. Appeal to the classification of surfaces to prove that any surface is diffeomorphic to what you drew. Here is my question:

''Can one avoid using the classification of surfaces? More specifically, can one prove that a discrete subgroup of $PSL_2 (R)$ that acts freely and cocompactly on the upper half plane must be of the above form - using a group-theoretic argument and without refering to the classification of surfaces or something that comes close to it?''

By ''something that comes close to it'', I mean an argument using Morse theory or another device that decomposes a surface into simpler parts.

Background: while contemplating again about the well-known paper by Earle and Eells ('A fibre bundle description of Teichmueller theory'), I realized that their main arguments can be upgraded slightly to give at once

1. closed surfaces are determined up to diffeomorphism by their fundamental groups
2. each isomorphism of fundamental groups is realized by a diffeomorphism which is unique up to isotopy (Dehn-Nielsen-Baer-Epstein)
3. the group of diffeomorphisms homotopic to the identity is contractible (the original Earle-Eells result)

and I would like to know whether this also gives the classification cheaply. According to the above two step argument, I would be happy with an argument that proves that two surfaces with the same genus (defined by the relation $\chi = 2 -2g$) must have isomorphic fundamental groups.

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I remember hearing that it is known that the only spaces satisfying $2$-dimensional Poincare duality (for arbitrary coefficient systems) are those which are weakly homotopy equivalent to the usual surfaces. I don't have a reference, and I don't know how the proof goes, but such a result would include the result that every surface is homotopy equivalent to one of the usual ones and therefore has the right sort of fundamental group. –  Tom Goodwillie Jul 10 '13 at 21:47
@TomGoodwillie : That theorem is due to Eckmann; a suitable reference is B. Eckmann, Poincare duality groups of dimension two are surface groups, Combinatorial Group Theory and Topology (eds. S.M. Gersten and J. Stallings), Annals of Math. Studies 111 (1987), 35–51, Princeton Univ. Press, Princeton. –  Andy Putman Jul 10 '13 at 23:22
By the way, it is conjectured that every finitely presentable Poincare duality group is the fundamental group of a compact aspherical orientable manifold (generalizing the above statement for surfaces). I believe that this is open even for $3$-dimensional Poincare duality groups. The condition "finitely presented" is necessary; Mike Davis has constructed non-finitely-presentable Poincare duality groups in M. Davis, The cohomology of a Coxeter group with group ring coefficients, Duke Math. J. 91 (1998), 297-313. –  Andy Putman Jul 10 '13 at 23:28
@Andy: the proof by Eckmann uses deep things such as a version of Stallings' splitting theorem, so it is quite complicated. Nevertheless it was a pleasant read, thanks for the reference. –  Johannes Ebert Jul 11 '13 at 17:36
@JohannesEbert : It's a deep theorem -- I doubt that there is a simple proof. –  Andy Putman Jul 11 '13 at 17:41

Yes, the magic words are "The Poincare polygon theorem". For (considerably) more detail, see Fine and Rosenberger "Algebraic generalizations of discrete groups: A path to combinatorial group theory through One-relator products" (section 4, I believe).

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I will answer the question of whether this also gives the classification cheaply.

No.

It gives the classification at the expense of proving that every surface group has a free cocompact action on the upper half plane (or on the euclidean plane, or on the 2-sphere). This in turn depends upon proving:

• Every surface has a triangulation (Rado's Theorem, and now you've done 90% of the work of the classification theorem)...

• Every triangulated surface has a compatible smooth structure...

• Every smooth surface has a compatible conformal structure...

and, finally

• Every conformal structure has a compatible hyperbolic metric, euclidean metric, or spherical metric (the uniformization theorem).
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Steps 3 and 4 are involved anyway in the argument and I would consider using them as "cheap" in this context. –  Johannes Ebert Jul 11 '13 at 6:57