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?

  • $\begingroup$ Dear Steve, Could it be that you are interpreting hyperbolic in the sense of "hyperpoblic, parabolic, or elliptic"? If so, that is part of the confusion, because (I think) all they mean when they write hyperbolic isometry is that $\phi$ is an isometry of the hyperbolic plane. Regards, $\endgroup$ – Matt E Feb 24 '11 at 4:40
  • $\begingroup$ @Matt E: I don't think so, because in the next paragraph of the proof they use the axis of $\phi$. I mean, it is actually pretty easy to do the parabolic case separately (only arises when there's a puncture, so $\pi_1(S)$ is free), but they seem to be ignoring that case. $\endgroup$ – user641 Feb 24 '11 at 6:28
  • $\begingroup$ OK, I retract my suggestion! Best wishes, $\endgroup$ – Matt E Feb 24 '11 at 6:44
  • $\begingroup$ 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. $\endgroup$ – aaron Feb 24 '11 at 17:35

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.

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  • $\begingroup$ Yes, you are probably right. A combination of ambiguous wording and my own non-understanding :) $\endgroup$ – user641 Feb 24 '11 at 6:30

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