1
$\begingroup$

This is a question about the proof of Proposition 1.4 in Farb and Margalit's "Primer on Mapping Class Groups" (in v. 5.0, it is on page 37 in the PDF, which you can download here). The proposition states

Let $\alpha$ be a non-nullhomotopic simple closed curve on the (hyperbolic) surface $S$; then $[\alpha]\in\pi_1(S)$ is primitive.

Most of the proof I'm OK with, except right in the beginning, when they write

...let $\phi\in\text{Isom}^+(\mathbb{H}^2)$ be the hyperbolic isometry corresponding to some element of the conjugacy class of $\alpha$.

Two questions:

1) What do they mean by the hyperbolic isometry $\phi$? Don't different elements of $\pi_1(S)$ correspond to different elements of $\text{Isom}^+(\mathbb{H}^2)$? (Here $\pi_1(S)$ is acting as deck transformations on $\mathbb{H}^2$.)

2) Why is there a hyperbolic isometry corresponding to $\alpha$? For example, if $\alpha$ is a simple loop around a puncture point, then shouldn't any such $\phi$ be parabolic?

  • 0
    They do seem to be using the fact that the isometry has an axis, so maybe it was an oversight on their part. Indeed, if you can do the parabolic case separately then you're done.2011-02-24

1 Answers 1

2

For (1): They mean pick any element of the conjugacy class, and look at the corresponding $\phi$. It doesn't matter which one you look at because being primitive is a conjugacy invariant.

  • 0
    Yes, you are probably right. A combination of ambiguous wording and my own non-understanding :)2011-02-24