A tractable case of the Turing automorphism problem: bi-uniformly $E_0$-invariant Cantor homeomorphisms

TL;DR

This paper investigates bi-uniform $E_0$-invariant Cantor homeomorphisms and proves they can only induce trivial automorphisms of the Turing degrees, shedding light on the structure of such automorphisms.

Contribution

It establishes that bi-uniform $E_0$-invariant Cantor homeomorphisms induce only trivial Turing degree automorphisms, addressing a specific case of the Turing automorphism problem.

Findings

Bi-uniform $E_0$-homeomorphisms induce trivial Turing automorphisms.

Characterization of $E_0$-isomorphisms in Cantor space.

Insight into the Turing automorphism problem for specific automorphisms.

Abstract

A function $F:2^\omega\to 2^\omega$ is an $E_0$-isomorphism if for all $x,y\in 2^\omega$, we have $xE_0y\iff f(x)E_0 f(y)$, where $xE_0y\iff(\exists a)(\forall n\ge b) x(n)=y(n)$. If such witnesses $a$ for $xE_0 y$ and for $f(x)E_0 f(y)$ depend on each other but not on $x$, $y$, then $F$ is called bi-uniform. It is shown that a homeomorphism of Cantor space which is a bi-uniform $E_0$-isomorphism can induce only the trivial automorphism of the Turing degrees.

Equations

$${\theta }^{-1}(m+1)=k({\theta }^{-1}(m)).\text{qed}$$

$$(\forall n)(\forall \tau ⪰\sigma )(\exists \rho ⪰\tau )({\Phi }^{\rho }(n)\downarrow \text{ and }\rho \circ g(n)={\Phi }^{\rho }(n)).$$

$$\{B:B\circ g={\Phi }^B\}$$

$$(\forall \sigma )(\exists n)(\exists \tau ⪰\sigma )(\forall \rho ⪰\tau )({\Phi }^{\rho }(n)\downarrow \to \rho \circ g(n)\ne {\Phi }^{\rho }(n)).$$

$$(\exists \sigma )(\forall n)(\forall \tau ⪰\sigma )(\exists \rho ⪰\tau )({\Phi }^{\rho }(n)\downarrow \text{ and }\rho \circ g(n)={\Phi }^{\rho }(n)).$$

$${\Pi }_{{\theta }^{-1}}({\Pi }_{\theta }(X))=(X\circ \theta )\circ {\theta }^{-1}=X=(X\circ {\theta }^{-1})\circ \theta ={\Pi }_{\theta }({\Pi }_{{\theta }^{-1}}(X)).$$

$${\Pi }^{-1}(X){\equiv }_r{\Pi }^{-1}(Y)\iff X{\equiv }_{r}Y\leftrightarrow {\Pi }_{{\theta }^{-1}}(X){\equiv }_r{\Pi }_{{\theta }^{-1}}(Y),$$

$$[{\Pi }^{-1}(Y){]}_r=[X{]}_r$$

$$\iff [Y{]}_r=[\Pi (X){]}_r$$

$$\{B:B\circ g={\Phi }^B\}$$

$$\pi ([X{]}_r)\le [X{]}_r$$

$${\pi }^{-1}([X{]}_r)\le [X{]}_r$$

$$X{=}_n^{\ast }Y⟺X\upharpoonright [n,\infty )=Y\upharpoonright [n,\infty ).$$

$$d_H(F(X),F(Y))\le d_H(F(X),X)+d_H(X,Y)+d_H(Y,F(Y))\le a+2$$

$$X\circ f(n)=X(f(f^{-1}(m)))=X(m)=Y(m)=Y(f(f^{-1}(m)))=Y\circ f(n),$$

$$({\pi }_n^{\star }\circ S^{\star })(A)(u)$$

$$\Theta \circ \Phi$$

$$\text{hence}({\pi }_n^{\star }\circ \Theta )\circ \Phi$$

$$(\sigma \searrow X)(n)=\{\begin{array}{ll}\sigma (n)& \text{if n<∣σ∣,}\\ X(n)& \text{otherwise.}\end{array}$$

$$F^X(n)=F^{\sigma \searrow X}(n)={\Phi }^{\sigma \searrow X}(n).\text{qed}$$

$$\{B:\Gamma (B)={\Phi }^B\}$$

$$\pi ([X{]}_r)\le [X{]}_r$$

$${\pi }^{-1}([X{]}_r)\le [X{]}_r$$

$${\Theta }^A(1)={\Theta }^{{\Phi }^A}(0)$$

$${\Theta }^A(2)$$

Full text

Dedicated to the celebration of the work of Theodore A. Slaman and W. Hugh Woodin *

1. 1 Introduction
2. 2 Excluding permutations by recursion
3. 3 Excluding bi-uniformly $E_{0}$-invariant homeomorphisms

1 Introduction

Let $\mathscr{D}_{\mathrm{T}}$ denote the set of Turing degrees and let $\leq$ denote its ordering. This article gives a partial answer to the following famous question.

Question 1*.*

Does there exist a nontrivial automorphism of $\mathscr{D}_{\mathrm{T}}$?

Definition 2**.**

A bijection $\pi:\mathscr{D}_{\mathrm{T}}\to\mathscr{D}_{\mathrm{T}}$ is an automorphism of $\mathscr{D}_{\mathrm{T}}$ if for all $\mathbf{x},\mathbf{y}\in\mathscr{D}_{\mathrm{T}}$, $\mathbf{x}\leq\mathbf{y}$ iff $\pi(\mathbf{x})\leq\pi(\mathbf{y})$. If moreover there exists an $\mathbf{x}$ with $\pi(\mathbf{x})\neq\mathbf{x}$ then $\pi$ is nontrivial.

Question 1 has a long history. Already in 1977, Jockusch and Solovay [3] showed that each jump-preserving automorphism of the Turing degrees is the identity above $\mathbf{0}^{(4)}$. Nerode and Shore [5] showed that each automorphism (not necessarily jump-preserving) is equal to the identity on some cone $\{\mathbf{a}:\mathbf{a}\geq\mathbf{b}\}$. Slaman and Woodin [6, 7] showed that each automorphism is equal to the identity on the cone above $\mathbf{0}^{\prime\prime}$ and that $\operatorname{Aut}(\mathcal{D})$ is countable.

There is an obstacle to reducing the base of the cone to $\mathbf{0}^{\prime}$ and ultimately $\mathbf{0}$: Turing reducibility is $\Sigma^{0}_{3}$, but not $\Pi^{0}_{2}$ or $\Sigma^{0}_{2}$ in the sense of descriptive set theory.

In the other direction, S. Barry Cooper [1] claimed to construct a nontrivial automorphism, induced by a discontinuous function on $\omega^{\omega}$, itself induced by a function on $\omega^{<\omega}$. That claim was not independently verified. In [4] we attacked the problem by ruling out a certain simple but natural possibility: automorphisms induced by permutations of finite objects. We showed that no permutation of $\omega$ represents a nontrivial automorphism of the Turing degrees. That proof was too complicated, in a way, and did not extend from $D_{T}$ down to $D_{m}$. Here we give a more direct proof using the shift map $n\mapsto n+1$. Our proof here will generalize to a certain class of homeomorphisms, distinct from the class of such homeomorphisms that the result in [4] generalizes to.

2 Excluding permutations by recursion

Lemma 3**.**

Suppose $\theta:\omega\to\omega$ is a bijection such that $\theta^{-1}\circ S\circ\theta$ is computable, where $S$ is the successor function given by $S(n)=n+1$. Then $\theta$ is computable.

Proof.

Let $k(n)=(\theta^{-1}\circ S\circ\theta)(n)$. Then for any $m$, $k(\theta^{-1}(m))=\theta^{-1}(S(m))$ and so we compute $\theta^{-1}$ by recursion:

[TABLE]

We write $\sigma\preceq\tau$ if $\sigma$ is a prefix of $\tau$.

Lemma 4**.**

Suppose $\sigma\in 2^{<\omega}$, $g:\omega\to\omega$, and $\Phi$ is a Turing functional, satisfying

[TABLE]

For any $\rho\succeq\sigma$ and $n$, if $\Phi^{\rho}(n)\downarrow$, then $g(n)<|\rho|$.

Proof.

If instead $g(n)\geq|\rho|$ then $\rho(g(n))$ is undefined. So let $\tau\succeq\rho$, $\tau(g(n))=1-\Phi^{\rho}(n)$. This $\tau$ violates (1). ∎

Lemma 5**.**

If $g:\omega\to\omega$ is injective and $\Phi$ is a Turing functional such that

[TABLE]

is nonmeager, then $g$ is computable.

Proof.

By assumption, it is not the case that

[TABLE]

So we have

[TABLE]

Pick such a $\sigma$: then $\Phi$ cannot make a mistake above $\sigma$, and we can always extend to get the right answer.

As finite data we assume we know the values of $n$ and $g(n)$ for which $g(n)<|\sigma|$.

We compute the value $g(n)$ as follows.

Check the finite database of $\{(k,g(k)):g(k)<|\sigma|\}$, and output $g(n)$ if found. Otherwise we know $g(n)\geq|\sigma|$.

By dovetailing computations, find a $\rho_{0}\succeq\sigma$ such that $\Phi^{\rho_{0}}(n)\downarrow$. By Lemma 4 we have that $g(n)<|\rho_{0}|$. Thus, $g(n)\in I$ where $I$ is the closed interval $[|\sigma|,|\rho_{0}|-1]$. Let $a\in I,b\in I,a<b$. It suffices to show how to eliminate either $a$ or $b$ as a candidate for being equal to $g(n)$.

Let $\tau\succ\sigma$ be such that $\tau(a)\neq\tau(b)$ and let $\rho\succeq\tau$ be such that $\Phi^{\rho}(n)\downarrow$. Then mark as eliminated whichever $c\in\{a,b\}$ makes $\rho(c)\neq\Phi^{\rho}(n)$. Thus we one-by-one eliminate all $a\in I$ until only one candidate remains. ∎

Definition 6**.**

We say that a permutation $\theta:\omega\to\omega$ induces the automorphism $\pi$ of $\mathcal{D}_{r}$ if, letting $\Pi=\Pi_{\theta}$ be defined by $\Pi(A)=A\circ\theta$ for all $A$, we have $\pi([X]_{r})=[\Pi(X)]_{r}$ for all $X$.

Lemma 7**.**

If $\theta$ is a permutation of $\omega$ and induces an automorphism $\pi$ of $\mathcal{D}_{r}$ then $\theta^{-1}$ induces $\pi^{-1}$.

Proof.

We must show that $\pi^{-1}([X]_{r})=[\Pi_{\theta^{-1}}(X)]_{r}$. We have $\Pi_{\theta}^{-1}=\Pi_{\theta^{-1}}$, since

[TABLE]

We have the well-definedness condition $X\equiv_{r}Y\Rightarrow\Pi(X)\equiv_{r}\Pi(Y).$ Moreover, since $\pi$ is injective, it must be that $X\equiv_{r}Y\Leftrightarrow\Pi(X)\equiv_{r}\Pi(Y).$ Since $\Pi_{\theta}^{-1}=\Pi_{\theta^{-1}}$, we also have

[TABLE]

so that $\theta^{-1}$ also induces the automorphism $\pi^{-1}$:

[TABLE]

Theorem 8**.**

No permutation of the integers can induce a nontrivial automorphism of $\mathcal{D}_{r}$, for any reducibility $\leq_{r}$ between $\leq_{1}$ and $\leq_{T}$.

Proof.

Suppose $\theta:\omega\to\omega$ is a permutation (bijection) and that $\theta$ induces an automorphism of $\mathcal{D}_{r}$. Thus, letting $\Pi(A)=A\circ\theta$ and $\pi([A]_{r})=[\Pi(A)]_{r}$, $\pi$ is well-defined and is an automorphism of $\mathcal{D}_{r}$.

Recall that $S$ from Definition 3 is simply the successor function given by $S(n)=n+1$. For any $B$, let $A=B\circ\theta^{-1}\circ S$ (so $x\in A$ iff $S(x)\in B\circ\theta^{-1}$). We have $A\leq_{1}{B\circ\theta^{-1}}$ and so by assumption $A\circ\theta\leq_{T}B\circ\theta^{-1}\circ\theta=B$. This gives $(B\circ\theta^{-1}\circ S)\circ\theta=\Phi^{B}$ for some Turing functional $\Phi$. Let $g=\theta^{-1}\circ S\circ\theta$. Since for each $B$ there exists such a $\Phi$, there must be some $\Phi$ such that the $G_{\delta}$ set

[TABLE]

is nonmeager. By Lemma 5, $g$ is computable. By Lemma 3, $\theta$ is computable.

But this means that for any $A$, $A\circ\theta\leq_{1}A$, so that the map $\Pi$ is everywhere-decreasing: $\Pi(X)\leq_{1}X$, and in particular

[TABLE]

where $\leq$ is the ordering of $\mathcal{D}_{r}$. By Lemma 7, $\pi^{-1}$ is also induced by a permutation, namely $\theta^{-1}$. Applying (2) to $\theta^{-1}$ we get

[TABLE]

for all $X$, and so $\pi([X])_{r}=[X]_{r}$ for all $X$. ∎

3 Excluding bi-uniformly $E_{0}$-invariant homeomorphisms

Making computability-theoretic uniformity assumptions is one way to rule out certain possible Turing automorphisms. We will instead focus on uniformity of a simpler, combinatorial kind.

Let $[a,\infty)=\{n\in\omega:a\leq n\}$. As usual we write $X=^{*}Y$ if $\{n\in\omega:X(n)\neq Y(n)\}$ is finite. The equivalence relation $=^{*}$ is also known as $E_{0}$. We refine this to

[TABLE]

As a relation, $=^{*}$ is the union $\bigcup_{n}=^{*}_{n}$. A map $F$ is $E_{0}$-invariant, or an $E_{0}$-endomorphism, if for all $X$, $Y$, if $X=^{*}Y$ then $F(X)=^{*}F(Y)$. In terms of the refinements $=^{*}_{n}$, this means that for all $X$, $Y$, and $a$, if $X=^{*}_{a}Y$ then there is an $b$ such that $F(X)=^{*}_{b}F(Y)$. If this $b$ only depends on $a$ we have a certain uniformity:

Definition 9**.**

A function $F:2^{\omega}\to 2^{\omega}$ is a uniformly $E_{0}$-invariant if for each $a$ there is a $b$ such that for all $X$, $Y$, if $X=^{*}_{a}Y$ then $F(X)=^{*}_{b}F(Y)$. If $F$ is invertible and both $F$ and $F^{-1}$ are uniformly $E_{0}$-invariant then $F$ is said to be a bi-uniformly $E_{0}$-invariant.

Some continuous maps $F:2^{\omega}\to 2^{\omega}$ induce maps $\tilde{F}:{{2^{\omega}}/{=^{*}}}\to{{2^{\omega}}/{=^{*}}}$ but do not have the uniform property:

Example 10*.*

Let $F(A)=A\setminus\min(A)$, if $A\neq\emptyset$, and $F(A)=A$ otherwise. We shall show that for $a=1$ there is no $b$ as required in Definition 9. Let $b$ be given. Let $Y$ be any set with $0<\min(Y)<\infty$ and $\min(Y)\geq b$, and let $X=Y\cup\{0\}$. Thus $X=^{*}_{1}Y$. We compute that $F(X)=Y$ and $F(Y)=Y\setminus\min(Y)$. Suppose $F(X)=^{*}_{b}F(Y)$. Then $\min(Y)<b$, a contradiction. Thus $F$ is not uniformly $E_{0}$-invariant.

On the other hand, $F$ is $E_{0}$-invariant. Indeed, let $d_{H}:2^{\omega}\times 2^{\omega}\to\omega+1$ denote the Hamming distance function, whereby $d_{H}(X,Y)=|\{n:X(n)\neq Y(n)\}|$. We have $d_{H}(F(X),X)\leq 1$, so that if $X=^{*}_{a}Y$ then

[TABLE]

and hence $F(X)=^{*}F(Y)$.

Lemma 11**.**

If $F(A)=A\circ f$ for a permutation $f:\omega\to\omega$ then $F$ is uniformly $E_{0}$-invariant.

Proof.

Let $a\in\omega$. Let $b=\max\{f^{-1}(m):m<a\}$. Suppose $X=^{*}_{a}Y$, i.e., $X(m)=Y(m)$ for all $m\geq a$. Let $n>b$ and let $m$ be such that $n=f^{-1}(m)$. By definition of $b$, we have $m\geq a$, so that

[TABLE]

giving $X\circ f=^{*}_{b+1}Y\circ f$. ∎

Define the ${\star}$ operator by $f^{\star}(A)(n)=A(f(n))$. Again, recall the successor function $S$ from Definition 3.

Lemma 12**.**

Suppose $\Theta:2^{\omega}\to 2^{\omega}$ is a homeomorphism such that the function $\Theta^{-1}\circ S^{\star}\circ\Theta$ is computable. Then $\Theta$ is computable.

Proof.

Let $\pi_{n}:\omega\to\omega$ be the constant $n$ function. For $\pi_{n}^{\star}:2^{\omega}\to 2^{\omega}$, note that $\pi_{n}^{\star}(A)(u)=(A\circ\pi_{n})(u)=A(n)$, so $\pi_{n}^{\star}(A)\in\{0^{\omega},1^{\omega}\}$. Note that

[TABLE]

so $\pi_{n}^{\star}\circ S^{\star}=\pi_{n+1}^{\star}$. Also, $S\circ\pi_{n}=\pi_{n+1}$. Let $\Phi=\Theta^{-1}\circ S^{\star}\circ\Theta.$ By assumption, $\Phi$ is computable. Then

[TABLE]

Since homeomorphisms have finite use, $\pi_{0}^{\star}\circ\Theta$ is just a finite amount of information. Thus, we can recursively compute $\pi_{n+1}^{\star}\circ\Theta$ this way. For a concrete case, the reader may wish to inspect Example 16. ∎

Definition 13**.**

For a real $X$ and a string $\sigma$ of length $n$,

[TABLE]

Lemma 14**.**

Suppose $F:2^{\omega}\to 2^{\omega}$ is a uniformly $E_{0}$-invariant continuous function. Suppose $F(X)=\Phi^{X}$ is forced above $\sigma\in 2^{<\omega}$, where $\Phi$ is a Turing functional. Then $F$ is computable.

Proof.

Indeed, let $\sigma$ force $F(X)=\Phi^{X}$. Let $a=|\sigma|$. Let $b$ be as in Definition 9. Let the truth tables for $F$ for $n<b$ be given as a finite database. For $n\geq b$,

[TABLE]

Theorem 15**.**

Let $\pi$ be an automorphism ot $\mathcal{D}_{r}$, for any reducibility $\leq_{r}$ between $\leq_{1}$ and $\leq_{T}$. Suppose that $\pi$ is induced by a homeomorphism $\Theta$ of $2^{\omega}$ that is a bi-uniform $E_{0}$-isomorphism111 The non-redundancy of the notion of bi-uniformly $E_{0}$-invariant Cantor homeomorphism has been pointed out by Salo [2]. . Then $\Theta$ is computable and $\pi$ is trivial.

Proof.

The overall proof strategy mirrors that for Theorem 8. Suppose that letting $\pi([A]_{r})=[\Theta^{A}]_{r}$ makes $\pi$ well-defined and makes $\pi$ an automorphism of $\mathcal{D}_{r}$.

Again, let $S$ be the successor function, $S(n)=n+1$. For any $B$, let $x\in A$ iff $S(x)\in\Theta^{-1}(B)$. We have $A\leq_{1}\Theta^{-1}(B)$ and so by assumption $\Theta(A)\leq_{T}\Theta(\Theta^{-1}(B))=B$. This gives $\Theta^{A}=\Phi^{B}$ for some Turing functional $\Phi$. Let $\Gamma=\Theta\circ S^{\star}\circ\Theta^{-1}$, which is also uniformly $E_{0}$-invariant. Since for each $B$ there exists such a $\Phi$, there must be some $\Phi$ such that the $G_{\delta}$ set

[TABLE]

is nonmeager. By Lemma 14, $\Gamma$ is computable. By Lemma 12, $\Theta$ is computable.

But this means that for any $A$, $\Theta(A)\leq_{tt}A$, and in particular

[TABLE]

where $\leq$ is the ordering of $\mathcal{D}_{r}$. Now by assumption, $\pi^{-1}$ is also induced by a homeomorphism, namely $\Theta^{-1}$ and so applying (3) to $\Theta^{-1}$ we get

[TABLE]

for all $X$, and so $\pi([X])_{r}=[X]_{r}$ for all $X$. ∎

Example 16* (The inductive procedure in Lemma 12.).*

Suppose $\Phi$ is the truth table reduction given by $\Phi^{A}(n)=A(2n)\cdot A(2n+1)$ for all $A$ and $n$. Suppose we know the first truth-table for $\Theta$, in that we know that $\Theta^{A}(0)=A(2)\to A(3)$ for all $A$. Then

[TABLE]

Next,

[TABLE]

Remark 17*.*

Woodin mentioned on June 6, 2019 that he and Slaman may have shown the following result in unpublished work from the 1990s: Each automorphism of $\mathcal{D}_{a}$, the degrees of arithmetical reducibility, is represented by a continuous function (outright). This gives some extra interest in a possible future $\mathcal{D}_{a}$ version of our results.