How does einselection fit into quantum decoherence? My understanding of decoherence is the following:
There is a system state vector $|\Psi\rangle_s$ and the environment state vector $|\Psi\rangle_\epsilon$ which are combined to form $|\Psi\rangle_s \otimes |\Psi\rangle_\epsilon = |\Psi \rangle_{s \epsilon}$.
If the interaction Hamiltonian $\hat{H}_{s \epsilon} $is time-independent the combined state evolves such that
$$|\Psi(t) \rangle_{s \epsilon} = e^{- i \hat{H}_{s \epsilon} t / \hbar} |\Psi \rangle_{s \epsilon} \, .$$
Although the system state vector doesn't act unusually, its density matrix $\text{Tr}_{\epsilon}(\rho_{s \epsilon})$ has an unusual property, its off diagonals approach zero extremely quickly. As a byproduct, the entropy of this matrix increases. 
I'm trying to understand how einselection/superselection fits into this picture. Is it merely that only in a particular basis the off diagonals get reduced? Somehow this fits with observables, but I'm not sure. 
 A: Your guess is correct: "einselection" refers to the fact the density matrix becomes diagonalized only with respect to a special basis (more generally, a special collection of mutually orthogonal subspaces), and these are the basis vectors that represent to the possible outcomes of the "measurement." We can then say that any observable which is diagonal in that same basis has been "measured," in the sense that we might as well apply Born's rule with respect to that observable.
I'm using the "measurement" language for convenience, but of course there are different degrees of decoherence. The reduced density matrix may remain approximately diagonalized without asymptotically approching a perfectly diagonal form. This is the case, for example, in a typical "position measurement" of a particle, because real position measurements don't have infinite resolution. Decoherence doesn't solve the measurement problem, but it is still a good indicator of which observable(s) have been measured.
By the way, decoherence is sometimes criticized because it refers to a particular factorization of the Hilbert space, which is artifical. That's a valid criticism of that way of formulating decoherence, but it can also be formulated in a more general way that doesn't require any factorization of the Hilbert space at all. (I'm bringing this up because it may help answer the question more fully.) Instead of choosing a factorization of the Hilbert space, we only need to recognize (1) which observables are associated with the system of interest — which we have to do anyway in order to use quantum theory at all — and (2) which observables could feasibly be measured using the limited resources that are actually available in the real universe. If $A$ is some observable associated with the system of interest, then any state-vector can always be written as a linear combination of eigenstates of $A$. We say that decoherence (specifically "einselection" in $A$'s eigenbasis) has occurred whenever no future feasibly-measurable observable can mix those eigenspaces with each other. In that case, if the non-mixing property is close enough to being exact, we can safely apply Born's rule with respect to $A$. 
The more familiar formulation of decoherence is a special case of this, namely the case in which the observables associated with the system of interest are just those observables that act like the identity operator on one of the two factors (the "environment" factor). 
