Proof of the bigon criterion

Question

In Farb and Margalit's "A Primer on Mapping Class Groups" we have the following

Proposition 1.7: Two transverse simple closed curves in a surface $S$ are in minimal position iff they do not form a bigon.

The proof for the reverse direction for the case $\chi(S) < 0$ starts as follows:

Assume simple closed curves $\alpha, \beta$ form no bigon. Let $\tilde{\alpha}, \tilde{\beta}$ be nondisjoint lifts of $\alpha, \beta$. By a previous Lemma, $\tilde{\alpha}$ intersects $\tilde{\beta}$ in exactly one point $x$. It cannot be that the axes of the hyperbolic isometries corresponding to $\tilde{\alpha}$ and $\tilde{\beta}$ share exactly one endpoint at $\partial \mathbb{H}^2$, because this would violate discreteness of the action of $\pi_1(S)$ on $\mathbb{H}^2$; indeed, in this case the commutator of these isometries is parabolic and the conjugates of this parabolic isometry by either of the original hyperbolic isometries have arbitrarily small translation length.

Question 1: How do we know the isometries corresponding to the lifts are hyperbolic and not parabolic? Earlier in the text, it is mentioned that if a closed curve is homotopic into a neighbourhood of a puncture, then the corresponding isometry is parabolic.

Question 2: What precisely is meant by translation length here? It seems to me like the conjugate of a parabolic isometry by some hyperbolic isometry (sharing an endpoint) will again be parabolic, so $\inf d(x, f(x))$ would be $0$. Is a different translation length meant or is there a reason why the conjugate ends up being hyperbolic rather than parabolic?

Question 3: Why does the conjugate (regardless of being parabolic or hyperbolic) end up being of arbitrarily small translation length?

Answer 1

For question 1, suppose that the corresponding lift is a parabolic isometry $\phi$. So the curve, which I'll call $c$, lifts to a $\phi$-equivariant map $\tilde c : \mathbb{R} \to \mathbb{H}^2$. Letting $P$ be the point on the circle at infinity which is fixed by this parabolic, it follows that the two ends of $\tilde c$ converge to $P$, and further more $\tilde c$ is $\phi$-equivariantly homotopic to any horocyle. Since there is a horocycle arbitrarily close to $P$, it follows that $c$ is homotopic to a curve that is arbitrarily far out the cusp.

For Question 2, the translation length of an isometry $\phi : \mathbb{H}^2 \to \mathbb{H}^2$ is the infimum of $d(x,\phi(x))$. If $\phi$ is finite order (which does not happen in the deck action of $\pi_1(S)$) then the infimum equals zero and is realized at some point in $\mathbb{H}^2$. If $\phi$ is parabolic then the infimum equals $0$ but is not realized at any point of $\mathbb{H}^2$. If $\phi$ is hyperbolic then the infimum is positive and it is realized by each point on a unique line called the axis of $\phi$, that line being the unique geodesic which connects the two points of the circle at infinity that are fixed by $\phi$.

In Question 3 and the second part of Question 2, it looks like you are confusing "conjugate" with "commutator". The commutator of two isometries $\phi,\psi$ is by definition $[\phi,\psi] = \phi \psi \phi^{-1} \psi^{-1}$. If $\phi,\psi$ are hyperbolic isometries sharing a fixed point $P$ on the circle at infinity, then you can easily check that $[\phi,\psi]$ is either the identity or a parabolic isometry fixing $P$. The text you quoted then goes on the refer to the "conjugates of the commutator by either of the two isometries", which refers to either the conjugate of $[\phi,\psi]$ by $\phi$ which would be $\phi^{-1} [\phi,\psi] \phi = \psi \phi^{-1} \psi^{-1} \phi$, or $\psi^{-1} [\phi,\psi] \psi = \psi^{-1} \phi\psi\phi^{-1}$. So if $\phi,\psi$ started out as two hyperbolic isometries with the same fixed point, their commutator $[\phi,\psi]$ is parabolic (as said above) and hence has translation number zero, and therefore any conjugate of the commutator, such as $\phi^{-1} [\phi,\psi] \phi$ is also parabolic and hence has translation number zero.