Shearing in hyperbolic 3-manifolds I'm new to 3-manifolds, and while reading an article (arXiv link) by Hongbin Sun about virtual domination of hyperbolic manifolds, I got a little bit confused, he says about $1+\pi i$-shearing (page 16, for example, Step III). But I still can't understand what it means. Well, I have complex length, and geodesics, I read about good pants construction (only main ideas), and still don't really understand that. Maybe I have some problems with understanding shearing parameter $s(C)$.
Thank you for help
 A: Suppose that $\gamma$ is an oriented geodesic in a hyperbolic three-manifold $M$.  Then $\gamma$ has a length, which we denote $\lambda_\gamma$.  Pick a normal vector $v$ at a point $x \in \gamma$.  We perform parallel transport to $v$, to take it once (forward) around $\gamma$.  This rotates $v$ through some angle, say $\theta_\gamma$.  We call $\lambda_\gamma + i \theta_\gamma$ the complex length of $\gamma$.  
Now suppose that $P$ is a totally geodesic pair of pants in a hyperbolic manifold, with all boundary components of length $R$.  (This is a basic unit in later efforts to build an almost geodesic surface in $M$.)  I'll call the boundary components cuffs.  We can draw the seams in $P$ - these are the three geodesic arcs in $P$ that connect the cuffs in pairs. When $R$ is very large, the seams are very short.  The seams meet the cuffs at right angles, and so provide a pair of normal vectors that are exactly distance $R/2$ apart and which are taken to each other by parallel translation.  We (that is, Kahn and Markovic) call these vectors the feet of the seams on the cuffs.   Suppose that $\gamma$ is one of the cuffs of $P$.  So the complex length of $\gamma$ is $R$ - the rotational part vanishes. 
Suppose that $Q$ is another pair of pants, also totally geodesic and having all cuffs of real and complex length $R$.  Suppose that $Q$ also has $\gamma$ as a cuff.  Then the feet of $P$ on $\gamma$ and the feet of $Q$ on $\gamma$ differ by some complex length, called the shear.  One nice situation is when shear is $i\pi$.  Then the feet are at the same point, and the seams glue to give a "dual" geodesic.  However, as $R$ becomes large, this "dual geodesic" pinches, and our surface will not have good properties.  So, instead, we take the shear to be $1 + i\pi$. That is, the feet of $P$ and $Q$ on $\gamma$ are separated by a real distance (one) and we must rotate the foot of $P$ through an angle of $\pi$ to match the foot of $Q$ (under parallel transport).  
I'll also point out that the sign of the real part is important (the sign needs to be everywhere the same - always plus one or always minus one, but not a combination of the two!).  But perhaps this is enough for now.
