A note on stamping

Abstract

The stamping deformation was defined by Apanasov as the first example of a deformation of the flat conformal structure on a hyperbolic 3-orbifold distinct from bending. We show that in fact the stamping cocycle is equal to the sum of three bending cocycles. We also obtain a more general result, showing that derivatives of geodesic lengths vanish at the base representation under deformations of the flat conformal structure of a finite-volume hyperbolic 3-orbifold.

Rigidity of representations in SO(4,1) for Dehn fillings on 2-bridge knots

“…For γ ∈ Γ, let tr γ : R(Γ, SO 0 (4, 1)) → R denote the trace function. Since tr γ is constant on orbits by conjugation, d tr γ : Z 1 (Γ, so(4, 1)) → R vanishes on B 1 (Γ, so(4, 1)), and it induces a linear map, d tr γ : H 1 (Γ, so(4, 1)) → R. (See also [1] for properties of the trace function) Lemma 3.9. For a representation ρ : Γ → SO 0 (3, 1) ⊂ SO 0 (4, 1), it holds…”

“…In particular, its holonomy representation is the only discrete and faithful representation of π 1 (M p/q ) in SO 0 (3, 1) up to conjugation. Here SO 0 (3,1) denotes the identity component of SO (3,1) and is isomorphic to Isom + (H 3 ), the orientation preserving isometry group of hyperbolic three space.…”

“…For the unfilled manifold M , M. Kapovich proved global rigidity [7]. That is to say, every discrete and faithful representation of π 1 (M ) in SO 0 (4, 1) is conjugated to a representation in SO 0 (3,1), and therefore to the holonomy of the hyperbolic metric. Moreover M. Kapovich [7] proved infinitesimal rigidity for infinitely many Dehn fillings M p/q , which are therefore locally rigid: there is no continuous non-trivial deformation in SO 0 (4, 1) of the holonomy of M p/q .…”

“…Let M be the exterior of a hyperbolic two-bridge knot. Then, for infinitely many p/q ∈ Q ∪ {∞}, the Dehn filled manifold M p/q has no discrete and faithful representation in SO 0 (4, 1) other than its holonomy in SO 0 (3,1).…”

“…Since ρ 0 is contained in SO(3, 1), the Lie algebra splits as π 1 (M )-module as follows so(4, 1) = so(3, 1) ⊕ R 3,1 , where R 3,1 denotes the Minkowski space equipped with the linear action of SO (3,1). That splitting induces a direct sum of cohomology groups…”