Is the fundamental group of a compact manifold finitely presented? 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?

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.
