Perhaps I made a mistake, but I get a different value of $p_0(10^4)$ (0.4912 instead of 0.4700). For larger $m$ the value seems even closer to 1/2. As a control the computation for $m=10^4$ has a maximum of $f^{\circ k}(73)$ equal to $1341801280048839911857201274496$ (for $k=2584$), so roundoff errors when computing $n\cdot \sqrt{2}^{\pm 1}$ could have occured in your computation. Could some confirm the above?

Quite from the manual: "Warning. Make sure you understand the above! By default, most of the bnf routines depend on the correctness of the GRH. In particular, any of the class number, class group structure, class group generators, regulator and fundamental units may be wrong, independently of each other. Any result computed from such a bnf may be wrong. The only guarantee is that the units given generate a subgroup of finite index in the full unit group. You must use bnfcertify to certify the computations unconditionally."

Note also that bnfisintnorm (as a number of other gp/pari functions) is only guarenteed to work correctly if one assumes GRH!

A reference is Theorem 5.4A(iii) and exercise 5.4.1 in Dixon and Mortimers "Permutation Groups" I think.

You are right. Let me reformulate what JMO and Osse shows in more algebraic terms: First of all everything behaves well with respect to direct products so we can assume that the root datum is indecomposable (note I'm not saying simple here). Then we have two disjoint possibilities: 1) The root datum is simple and adjoint of type $B_n$ ($n=1$ is also allowed) OR 2) we have R=D in the notation above.

Oops, I see now that I made a typo in the JMO reference, it should have been Proposition 3.2(vi) and not (iv).

Quick answer: No the Matthey paper is unfortunately unpublished (but he refers to Osse). Both JMO and Osse answers the question: To what extent can one reconstruct the root datum from the Weyl group action? The short answer is that the only indeterminacy comes from direct factors of $G$ isomorphic to $\text{Sp}_n$ or $\text{SO}_{2n+1}$. From this one gets the center formula after some work.

I'm unfortunately somewhat ignorant when it comes to algebraic groups. At least the statement is true for compact Lie groups and the proofs just use root system yoga. So is the following statement correct: $T^W = Z(G) \times (\mathbb{Z}/2)^r$ where $r$ is defined as before (ignoring the restriction on $\text{char}(k)$)?

Unfortunately the inequality $\lvert H_2(\Gamma,\tilde{\mathbb{Z}})\rvert \leq 2\cdot \lvert H_2(\Gamma)\rvert$ doesnt hold: Let $\Gamma = \text{SD}_{16} = \left<x,y | x^8, y^2, x^y=x^3 \right>$ and $G=\left< x^2, y\right> \cong D_8$ (an index $2$ subgroup). Then $H_1(\Gamma) \cong H_1(G) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2$ and the kernel $H_1(\Gamma) \xrightarrow{\text{tr}} H_1(G)$ has order $2$. The Schur multipliers equals $H_2(\Gamma)=0$ and $H_2(G)=\mathbb{Z}/2$. The long exact sequence then shows that $H_2(G,\tilde{\mathbb{Z}})$ has order $4$.