Geodesic cuffs of pairs of pants in a hyperbolic manifold- why are they disjoint?

I'm trying to understand Kahn-Markovic's celebrated Immersing almost geodesic surfaces in a closed hyperbolic three manifold. There is something probably quite basic which I can't figure out.

We have $M^3= \mathbb{H}^3/\mathcal{G}$ a closed hyperbolic $3$-manifold, where $\mathcal{G}$ is a Kleinian group. Let $\Pi^0$ be a topological pair of pants with cuffs $C_0$, $C_1$, and $C_2$, and let $\rho\colon\thinspace \pi_1(\Pi^0)\to \mathcal{G}\subset PSL(2,\mathbb{C})$ be a faithful representation. Let $\gamma_i$ denote the geodesic in $M$ that represents the conjugacy class of $\rho(C_i)$ in $\mathcal{G}$ for $i=0,1,2$ (these become cuffs for pairs of pants which Kahn-Markovic construct inside $M$ and glue together to form an immersed almost geodesic surface).

Question: Why are $\gamma_0$, $\gamma_1$, and $\gamma_2$ disjoint?

It is essential to the construction that the cuffs indeed be disjoint, because if they intersect, reduced complex Fenchel-Nielsen coordinates don't exist (there is no "foot"). In fact, Kahn and Markovic need many pairs of pants inside $M$ to coexist, so not understanding why the cuffs are non-intersecting even for a single pair of pants is particularly frustrating.

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Dan, in general these geodesics intersect, just think about free subgroups in $PSL(2,R)$ uniformizing one-holed torus. – Misha Dec 19 '12 at 17:51
+1 for the title itself! – Suvrit Dec 19 '12 at 21:02

If you'd like to visualize embedded geodesics, you can imagine the lifts to the unit tangent bundle, or for a given pair of pants, it will lift to an embedded pants in some covering space corresponding to the image of the fundamental group $\rho(\pi_1(\Pi^0))$. To understand the feet of $\gamma_i$, then you can work in this covering space. The feet will be at the points of the shortest geodesics connecting the three boundary components in pairs (seams). There's a canonical involution sending $\gamma_i$ to its inverse, fixing the seams, and thus the feet lie equally spaced about each geodesic. Then project the whole picture back down into $M$ to get the feet.
One might also want to go all the way up to $\mathbb{H}^3$ and visualize the right-angled hexagons associated to $\rho(\pi_1(\Pi^0))$, although even in $\mathbb{H}^3$ there are cases where the hexagons degenerate to being non-embedded. – Lee Mosher Dec 19 '12 at 18:32
Where do Kahn and Markovic guarantee no intersections between cuffs? I'm looking at the construction of $\delta$ in Section 4.5. Thank you for the explanation of "feet" when there are intersections! – Daniel Moskovich Dec 20 '12 at 7:48