@MISC{Kjos-hanssen04arigid, author = {Bjørn Kjos-hanssen}, title = {A rigid cone in the truth-table degrees with jump}, year = {2004} }

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Abstract

Each automorphism of the truth-table degrees with order and jump is equal to the identity on a cone with base 0 (4). A degree structure is said to be rigid on a cone if each automorphism of the structure is equal to the identity on the set of degrees above a fixed degree. It is known [4] that the structure of the Turing degrees with jump is rigid on a cone. This is shown by applying a jump inversion theorem and results on initial segments. In this paper it is shown using a weaker jump inversion theorem that also the structure of truth-table degrees with jump is rigid on a cone. For definitions relating to initial segments we refer to [13]. 1 Initial segment construction Lemma 1.1. Suppose for each e, g lies on a tree Te which is e-splitting for some c for some tables with the properties of Proposition 4.9, in the sense of [5]. Then g is hyperimmune-free. Proof. For each e ∈ ω there exists e ∗ ∈ ω such that for all stages s and all oracles g, if {e ∗ } g s(x) ↓ then {e ∗ } g (x) = {e} g (x) and {e} g s(y) ↓ for all y ≤ x. If g lies on Te ∗ then it follows that {e}g is total and {e ∗ } T (σ) (x) ↓ for each σ of length x + 1. Hence {e} g = {e ∗ } g is dominated by the recursive function f(x) = max{{e} T (σ) (x) : |σ | = x + 1}. Proposition 1.2. Let L be a Σ 0 4(y)-presentable upper semilattice with least and greatest element. Then there exist t, i, g such that 1. t: ω → 2 is 0 ′ ′-computable, 2. i is the characteristic function of a set I such that I ≤m y (3), 3. g ′ ′ (e) = t(i(0),..., i(e)) for all e ∈ ω, 4. [0, g] is isomorphic to L, and 5. g is hyperimmune-free 1 Proof. The proof in [5] must be modified to employ the lattice tables of Proposition