3
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ Could you clarify in how many dimensions you are working and what are your definitions of area and volume? Since it seems to me you are only talking about surfaces and for those volume = area, at least in my book... $\endgroup$
    – Marek
    Sep 10, 2012 at 10:02
  • $\begingroup$ @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. $\endgroup$
    – JimWang
    Sep 10, 2012 at 18:00

1 Answer 1

2
$\begingroup$

$\pi_1(S)= \mathbb{Z}$ 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.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .