geometrically connected and simply connected projective surface that is a counterexample to the Hasse principle.

Let us prove (1). It is well known that any group acting on a finite connected tree fixes a vertex or an edge. (This is easily proved by induction on the diameter of a tree, that is, on the length of a longest path contained in it.) We apply this to the action of the Galois group $\Gamma_k$ on $X(C)$. If $\Gamma_k$ fixes a point of $\ovl\Pi$ or $\ovl\Psi$, then $C(k)\not=\emptyset$. If $\Gamma_k$ fixes a point of $\ovl\Lambda$, then $C$ has an irreducible component $C_0$ which is a geometrically integral geometrically rational curve. Let $\wtl C_0$ be the normalisation of $C_0$. Since $X(C)$ is a tree, the morphism $\wtl C_0\to C_0$ is a bijection on points. Thus if we can prove that $C_0({\mathbb{A}}_k)\not=\emptyset$, then $\wtl C_0({\mathbb{A}}_k)\not=\emptyset$, and by the Hasse–Minkowski theorem $\wtl C_0(k)\not=\emptyset$, so that finally $C_0(k)\not=\emptyset$. Since $X(C)$ is a connected tree, each connected component of $\ovl C\setminus \ovl C_0$ meets $\ovl C_0$ in exactly one point. Let $k_v$ be a completion of $k$. Suppose that $C_0(k_v)=\emptyset$. Since $C(k_v)\not=\emptyset$, at least one of the connected components of $\ovl C\setminus \ovl C_0$ is fixed by 