Meaning of " Open Book cannot be Stabilized Further"? I'm going over some old notes on Giroux's theorem on the equivalence ( bijection, actually) between open books ( up to positive stabilization)  for 3-manifolds and contact structures ( up to isotopy.) 
I'm curious about a statement referring to:  open books that cannot be further (de)stabilized. I'm having trouble finding a reasonable interpretation for this phrase. Moreover, I only know vaguely that the process of destabilization/deplumbing is the inverse process of that of stabilization (meaning that destabilization undoes the effects of stabilization.), but I don't have a clear, specific-enough description of the process of destabilization. 
Let's see stabilization first:
   So, if we have an open book decomposition $(B, \pi)$ , where $B$ is the binding of the open book, i.e., $B$ is a fibered knot, and $\pi$ is the projection map so that $(M-B) \rightarrow S^1$ is a locally-trivial fiber bundle with fiber a surface $\Sigma$, we then "stabilize" by attaching a 2-D 1-handle and we choose a curve $C$ going once through the cocore of the handle. Then we do a positive Dehn twist about $C$ .The manifold $M':=M \cup$ 1-handle that results from this process of stabilization is homeomorphic to $M$, and the open book whose monodromy is composed with the Dehn twist still supports the original contact structure. 
So, naively, it would seem we can destabilize and undo the effects of stabilization as in the last paragraph, by removing a 2-D 1-handle  , removing a curve C going through the
cocore once, and composing with a  +/-  Dehn twist about C  ( the sign of the twist would depend on whether we are doing a +/- destabilization). But this seems too easy.
Equivalently, the process of stabilizing is equivalent to plumbing together the bindings (let's work on 3-manifolds for now ), and ending up with surfaces/fibers with transverse self-intersections, and an associated plumbing diagram, e.g: http://en.wikipedia.org/wiki/E8_manifold.
 But I don't see how to use this to define destabilization ( deplumbing?) , nor how to find associated results to destabilization, , e.g., obstructions to the process -- there must be some restrictions, since otherwise, we could always have genus-zero pages in our books.
Can anyone please explain this process and/or give a ref. for results associated with it? It would be great too, if the sources included plane-field invariants like d_3 , and how these invariants are affected by (de) stabilization.
Thanks.
 A: I would think that the best reference for these topics are still John Etnyre's Lectures on open book decompositions and contact structures. I will try to address some of your questions.
When dealing with open books, the distinction between open book and abstract open book should be kept in mind.
An open book is a triple $(M,B,\pi)$, where $M$ is a 3-manifold (let's stick to dimension 3), $B\subset M$ is a fibred link and $\pi$ is a (nice) fibration of the complement. A stabilisation of an open book is just a connected sum with the (positive or negative) Hopf link/fibration. This clearly tells you that, in some cases, destabilisation is obstructed by knot-theoretic conditions (e.g. you need to have at least one unknotted component in the binding).
An abstract open book is a pair $(F,h)$ where $F$ is a compact surface with nonempty boundary and $h$ is a self-diffeomorphism of $F$ preserving a neighbourhood of $\partial F$ (the monodromy). A stabilisation of an abstract open book is a Murasugi sum with a (positive or negative) Hopf band $(A,h_\pm)$.
In order to be able to do a destabilisation, in particular, you need to find a 1-handle $H$ of $F$ and a curve $c$, running through that hand exactly once, such that composing the monodromy with a (negative or positive) Dehn twist along $c$ yields a map $h'$ on $F$ that's isotopic to a map $h''$ which is the identity on $H$. Then, and only then you can destabilise.
I think that an example helps: take the annulus $A$ and the monodromy $h$ that's the square of the positive Dehn twist along the core of $A$. There's only one way to split a 1-handle off of $A$, and that is by looking at $A$ as a ball with a handle. There's only one curve on the surface that runs through the handle once, up to isotopy, call it $c$. But composing $h$ with a single Dehn twist along $c$ doesn't produce a monodromy that's trivial on the handle, not even up to isotopy, so you can't destabilise.
This can be seen at the level of (non-abstract) open books as well, since either component of the binding (i.e. of the boundary of the surface) is not nullhomologous in $\mathbb{RP}^3$ (which is the three-manifold you obtain).

Finally, as for the $d_3$-invariant, this is easier: positive (de)stabilisations preserve it, and negative (de)stabilisations decrease (increase) it by $1$. There's a nice post on the Electric Handle Slide blog where the issue is brought up.
