Computable Scott Sentences and the Friedman-Stanley embedding

TL;DR

This paper studies a specific embedding of structures into labeled trees, analyzing its preservation of Scott complexity and the Borel complexity of subclasses based on Scott rank.

Contribution

It refines existing results by showing the preservation of Scott sentences and complexities under a particular computable embedding into labeled trees.

Findings

The embedding preserves Scott complexity and matching infinitary Scott sentences.

The class of labeled trees in the embedding range has non-Borel complexity.

Subclasses of trees with bounded Scott rank are Borel and complete at certain levels.

Abstract

Friedman and Stanley developed the notion of Borel reducibility and illustrated its use in comparing classification problems for some familiar classes of countable structures. For many embeddings, the fact that the embedding is $1-1$ on isomorphism types is explained by the existence of simple formulas that, uniformly, interpret the input structure in the output structure. For the embeddings of graphs in trees, and in linear orderings, there is no uniform interpretation. We focus on a version of the Friedman-Stanley embedding introduced by Harrison-Trainor and Montalban that takes each structure $A$ for the language of graphs to a labeled tree $T_A$. Gonzalez and Rossegger showed that this embedding preserves Scott complexity. We refine this result, showing that for an $X$-computable ordinal, if one of $A$, $T_A$ has a computable infinitary Scott sentence, then so does the other, and the complexities match. Let $\mathbb{T}$ be the class of labeled trees isomorphic to those in the range of the embedding, and let $\mathbb{T}^\alpha$ be the subclass consisting of structures of Scott rank at most $\alpha$. It follows from results of Gao that $\mathbb{T}$ is not Borel. We show that for each $\alpha$, $\mathbb{T}^\alpha$ is Borel. In fact, if $\alpha$ is an $X$-computable ordinal, then $\mathbb{T}^\alpha$ is complete $X$-effective $\Pi_{2\alpha+2}$.