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Let $X$ be a connected compact smooth manifold. If $X$ is boundaryless, we can choose a Riemannian metric for $X$ so that $\pi_1(X)$ acts geometrically (ie. properly, cocompactly, isometries) on the universal cover $\tilde{X}$. Because it is know that a group acting geometrically on a simply connected geodesic space is finitely presented (see Bridson and Haefliger's book, Metric spaces of non-postive curvature), we deduce that $\pi_1(X)$ is itself finitely presented.

What happens when $X$ has a boundary?

Seirios
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2 Answers2

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One can even do a bit better:

Every compact topological manifold (possibly with boundary) is homotopy-equivalent to a finite CW-complex.

That every ANR (absolute neighborhood retract, and every topological manifold with or without boundary is ANR) is homotopy-equivalent to a CW complex, was known as Borsuk conjecture. This conjectire was proven by West in 1974 (here) and, later on, a simpler proof was given by Chapman:

Chapman, T. A., Invariance of torsion and the Borsuk conjecture, Can. J. Math. 32, 1333-1341 (1980). ZBL0539.57009.

In the case of topological manifolds it is also a theorem in the (unreadable) book by Kirby and Siebenmann that every compact manifold is homotopy-equivalent to a finite CW-complex.

Moishe Kohan
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  • by the way: both articles are written by Chapman, perhaps its the wrong link? this seems to be correct article: http://www.jstor.org/discover/10.2307/1971155?uid=3737760&uid=2&uid=4&sid=21104002130717 ? – user135041 May 11 '14 at 14:47
  • @Herbert: Yes, wrong link, I will correct it. In any case, I find Chapman's paper to be more readable (still, it uses a lot of heavy machinery). – Moishe Kohan May 11 '14 at 14:53
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Differentiable manifolds can always be given the structure of PL manifolds, which can be triangulated into simplicial complexes. By shrinking a spanning tree of the 1-skeleton of this simplicial complex, we can obtain a CW complex $X$ with a single $0$-cell. This complex is no longer a manifold, but has the same fundamental group as the original manifold, since quotienting out by a contractible subspace is a homotopy equivalence.

If the manifold is compact, it has a simplicial decomposition with a finite number of cells. This carries over to $X$. But the fundamental group of a $CW$ complex with a single $0$-cell has a presentation with a generator for each $1$-cell and a relation for each $2$-cell. Thus $X$, and therefore the original manifold, has a finitely presented fundamental group.

MartianInvader
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