Surely this is obvious? Given a unitary automorphism of $X^\perp$, just extend it to an automorphism of $V$ by defining it to be trivial on $X$. (Is it the idea of embedding a non-quasi-split space inside a quasi-split one which troubles you? If so, play with some explicit matrices until it stops troubling you.)

@Satan'sMinion You are quite right: I mixed up which of the morphisms has to be proper in proper base change – Spec(L) \to Spec(K) is proper, but that isn't the hypothesis. I should have appealed to smooth base change, as Corollary 1.3.1 of these notes by Landesman virtualmath1.stanford.edu/~conrad/Weil2seminar/Notes/L7-8.pd‌​f

There exists no $\mathbf{Z}_p$ extension (cyclotomic or anticyclotomic or whatever) contained in a $PGL_2$ extension, just by elementary group theory: $\mathbf{Z}_p$ is not a quotient of $PGL_2(\mathbf{Z}_p)$. This is precisely why the PGL2 extensions are studied: because they are the first obvious example of p-adic Lie extensions which are "essentially" nonabelian, having no nontrivial abelian extensions inside them.

"In a few instances one can give an algebraic characterisation of the generator of the characteristic ideal" — which cases do you mean, and how?

This question is now so strewn with wreckage from previous iterations as to be incomprehensible. Please delete it, and open a new one if you still have any actual questions to ask.

In your example, I would argue that cyclotomic units aren't a purely algebraic concept: they rely on the fact that we can generate the abelian extensions of $\mathbb{Q}$ using values of a transcendental function (the exponential map).

I would disagree vigorously with your implicit suggestion that Euler systems are "purely algebraic". In all cases where we have a reasonably canonical construction of an Euler system (not just a proof that some module of Euler systems is nonempty), this construction involves automorphic methods.

Please, just stop commenting and start actually thinking! We've already established that (with your somewhat weird definitions) all the rings $\mathbb{Z}_p\langle \tfrac{x}{r}\rangle$ for $0 \le r < 1$ are all actually the same ring. So please don't waste your time and ours trying to microscopically analyse the maps in between them.

@Wojowu I've never found this viewpoint terribly convincing. Yes $Spec O_k$ behaves like a curve with some points removed; but the machinery that purports to "put them back in" is itself treating the Archimedean places in a radically different way from the non-Archimedean ones, so the claim that the resulting object restores the symmetry between arch and non-arch places seems highly suspect to me.

But you must have ignored a great deal of "too many comments" warnings to get this far.

Your maps $f_r$ and $g_r$ are never both well-defined (one is only defined if $r < 1$, the other if $r >1$, and I'm too lazy to work out which).

That's an important principle but it does have limitations – one sees this much earlier on in the theory, with Dirichlet's unit theorem, where the Archimedean signature determines the rank of the unit group in an essential way.