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I am reading a book (Mapping Class Group by Farb and Margalit) and it says (in a proof of one theorem):

If $S$ admits a hyperbolic metric (they define such a surface to be of finite area and complete) and we had $\pi_1(S)\cong \mathbb{Z}\ $ then the surface would have an infinite volume which is a contradiciton. Hence $\pi_1(S)$ is NOT isomorphic to $\mathbb{Z}$.

Questions

  1. If we have a Riemannian manifold of finite area, does it have a finite volume also? I am interested in the case of hyperbolic surfaces.
  2. Why $\pi_1(S)\cong \mathbb{Z}\ $ implies the volume of $S\ $ is infinite?

Can someone help me, please?

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    @Marek I do not have a definition in the book (Mapping Class Group-Farb & Margalit) I am using! But I am intrested in the case of surfaces only. If volume=area for surfaces, then Q1 is done.2012-09-10

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$\pi_1(S)= \mathbb{Z}$ means that $S$ is topologically an open annulus. Actually, there is a classification of annuli (Look at John Hubbard's Teichmuller Theory , volume 1, section 3.2, page 63) by using the uiformization theorem , which says the annuli (non copact Riemann surface with fundamental group $\mathbb{Z}$ ) are isomorphic to $C- {0}$ = $C / \mathbb{Z}$ , $D-{0} = \mathbb{H}/ z\to z+a, a\in \mathbb{R}-0$ , or a round annulus with inner radius 1 and outer radius r = $H/ z\to \lambda.z$ for some suitable $\lambda$ depending only on r. Now if you look at the fundamental domains F of each of these $\mathbb{Z}$-action you will get infinite hyperbolic area (for example, F for $H/ z\to \lambda.z$ is the open semi-annulus in $\mathbb{H}$ with vertices $1,\lambda, -1, -\lambda$, which has infinite hyperbolic area (use the hyperbolic metric on $\mathbb{H}$).Similarly, you can try to find (it is easy) the F for the other actions and show they have infinite area. So, ALL the annuli have infinite area.