Reverse Mathematics on Lattice Ordered

Groups

Alexander S. Rogalski, Ph.D.

University of Connecticut, 2007

Several theorems about lattice-ordered groups are analyzed. RCA0 is sufficient to

prove the induced order on a quotient of `-groups and the Riesz Decomposition

Theorem. WKL0 is equivalent to the statement “An abelian group G is torsion

free if and only if it is lattice-orderable.” ACA0 is equivalent to the existence of

various substructures: the join of two convex `-subgroups, the convex closure of

an `-subgroup, the polar subgroup X⊥ of an `-subgroup X, and a sequence of

values {V (g): g 6= e}. The standard proof of Holland’s Embedding Theorem uses

ACA0. Holland’s Theorem is equivalent to the existence of a sequence of excluding prime subgroups {P (g): g 6= e}, and the existence of such a sequence is provable in WKL0 when G is abelian. Reverse Mathematics on Lattice Ordered

Groups

Alexander S. Rogalski

B.S., Marlboro College, Marlboro, VT, 2002

M.S., University of Connecticut, Storrs, CT, 2004

A Dissertation

Submitted in Partial Fullfilment of the

Requirements for the Degree of

Doctor of Philosophy

at the

University of Connecticut

2007 Copyright by

Alexander S. Rogalski

2007 APPROVAL PAGE

Doctor of Philosophy Dissertation

Reverse Mathematics on Lattice Ordered

Groups

Presented by

Alexander S. Rogalski, B.S., M.S.

Major Advisor

David Reed Solomon

Associate Advisor

Manuel Lerman

Associate Advisor

Joseph Miller

University of Connecticut

2007

ii ACKNOWLEDGEMENTS

First, I wish to acknowledge Old Joe, my banjo. I brought Old Joe to the annual

Math Department picnic, where Reed excitedly introduced himself and suggested that we play music sometime. We managed to accomplish this a few times, despite the pressing obligations of academia and graduate studies. Reed truly deserves my thanks for being a superb advisor. Tricia, you are wonderful, and I thank you for believing in me when I feared it would take another year to finish. I also thank our daughter, Georgia, for being born and giving me, among other numerous joys, a considerable motivation to graduate. Last but not least, I wish to thank my family for their considerable and constant support.

iii TABLE OF CONTENTS

1. Introduction ...... 1

1.1 Background ...... 1

1.2 Notation ...... 6

1.3 Reverse Math ...... 6

1.4 Lattices, Lattice-ordered groups ...... 7

1.5 Fundamental Examples ...... 8

2. Preliminary Results ...... 12

2.1 Definition of `-group ...... 13

2.2 Definition of an `-group for Reverse Math ...... 19

2.3 Basic computation in `-groups ...... 20

2.4 The Riesz Decomposition Theorem ...... 26

3. More Results ...... 30

3.1 Identifying groups which are lattice-orderable ...... 30

3.2 Convex `-subgroups and Right Cosets ...... 34

4. Convexity Results and Reversals for Substructure Existences . 38

4.1 Closure Operations ...... 38

4.2 Existence of the subgroup generated by A, B...... 44

4.3 Existence of the convex closure of an `-subgroup H ...... 45

iv 4.4 Existence of the convex `-subgroup generated by convex `-subgroups

A, B...... 47

4.4.1 Construction ...... 50

4.4.2 Verification ...... 52

4.5 Existence of the Polar X⊥...... 54

5. Prime Subgroups and Values ...... 62

5.1 Prime Subgroups ...... 62

5.2 Values ...... 65

5.3 Existence of a Sequence of Values ...... 67

5.4 Existence of a Sequence of Excluding Primes ...... 70

6. Holland’s Embedding Theorem ...... 78

6.1 Summary of Original Proof of Holland’s Theorem ...... 79

6.2 Proof of Holland’s Theorem using excluding primes ...... 81

6.3 A Reversal ...... 84

Bibliography 86

v LIST OF FIGURES

4.1 The Root System Γ ...... 56

5.1 Initial Subtree of T ∗ ...... 77

vi Chapter 1

Introduction

1.1 Background

It is a relatively common occurrence for a student of mathematics to read or write a proof which, at a key step, uses Zorn’s Lemma or an equivalent principle to prove the existence of a set with certain desired properties. Until one is used to doing so, it can seem odd to call on a result from set theory in the middle of a proof which otherwise only requires results in Algebra. One might wonder, “Does one really need something as strong as Zorn’s Lemma to build this set, or could it be done directly?”

Reverse Mathematics is a subfield of logic which tries to answer questions like this by finding exactly which set-theoretic axioms are truly necessary to prove a theorem. The usual axioms of set theory, ZFC or ZF, are quite strong. We can make finer distinctions by restricting ourselves to “countable” mathematics and axiom systems which, though weaker, are still able to prove many classical theorems of mathematics. More formally, the setting for Reverse Math is the

1 2

language of second order arithmetic Z2. In this language, we have symbols +, ·,

<, 0, and 1, and the usual axioms defining them in the natural numbers N, set

membership ∈, and two types of variables: number variables which are intended

to range over N and set variables which are intended to range over subsets X ⊂ N.

A typical investigation in Reverse Math goes like this:

1. Pick a theorem T hm to study.

2. Look at a textbook proof of T hm and find a set S of axioms for Z2 that are

suitable for the proof of T hm.

3. See if it is possible to prove the axioms of S using T hm. (This is why it’s

called Reverse Math.)

We need some amount of set theory to do such a proof as in step 3, and

we typically work in a weak base system called RCA0, which strikes a balance

between being strong enough to allow basic proofs and weak enough to keep the

set-theoretic impact reasonably minimal. Generally, one of two things happens.

If we are able to prove S from T hm, this tells us that S is the weakest axiom system capable of proving T hm. (If a strictly weaker system S0 proved T hm, S0 would then imply S, an impossibility!) In this case we have a proof of T hm from

S and a reversal of T hm to S, and we say the two are equivalent over RCA0. If,

on the other hand, attempts to prove S from T hm aren’t working out, we look

for a new proof of T hm that uses weaker axioms Sˆ and repeat the process, trying 3 to prove Sˆ using T hm.

The “full-blown” set existence axiom scheme for Z2 consists of axioms

∃X∀n(n ∈ X ↔ ϕ(n)), where ϕ is any formula in the language Z2 not men- tioning X. This scheme basically says: if we have a formula in Z2, the set of numbers which satisfy it exists. This full collection is too strong to be interesting for Reverse Math, but it contains five particular subsystems (subcollections of axioms) and most theorems successfully analyzed are equivalent to one of them.

1 In order of increasing strength, they are: RCA0, WKL0, ACA0, ATR0, and Π1-CA0.

In this dissertation, only the first three come into play.

RCA0 is the usual base system over which we prove equivalences, and its set

0 comprehension scheme is limited to ∆1 formulas. (We also include axioms allow-

0 ing Σ1 induction.) The next strongest, WKL0, consists of all the axioms of RCA0 plus the Weak K¨onig’s Lemma axiom saying “If T is an infinite binary tree, then

T has an infinite path”. WKL0 is sometimes sufficient where a standard proof uses Zorn’s Lemma, though there are notable cases where it does not suffice. For example, in the theory of commutative rings, the existence of a nontrivial prime ideal is equivalent to WKL0. The existence of a prime ideal is usually proved as a corollary to the existence of a maximal ideal. Since the existence of a maximal ideal requires ACA0, this is a case where a set-theoretically “simpler” existence proof for a prime ideal needed to be found. ACA0, or Arithmetical Comprehension, is the strongest subsystem used in this study, obtained by allowing set comprehen- 4 sion using arithmetical formulas – those which may have any number of number quantifiers in its definition but no set quantifiers.

The subsystem of axioms to which a given theorem is equivalent is a measure of the set-theoretic complexity of the theorem. A theorem which requires only weak set existence axioms is considered less complex than one which requires a lot of complicated assumptions, i.e., a very strong axiom system. Knowing which axioms are necessary is particularly of interest to those working on related problems in Computable Algebra or Combinatorics.

For instance, an effective analogue of a theorem is likely true if the theorem is provable in RCA0. As an example, RCA0 suffices to prove that the quotient of a commutative ring by a maximal ideal is a field. Thus, given such a ring and a maximal ideal in a computable presentation, one can effectively form the quotient field. In contrast, maximal ideals generally require ACA0 and it is not the case that, given a computable commutative ring, one can always effectively

find a maximal ideal. Knowing that maximal and prime ideals respectively require

ACA0 and WKL0 also enables one to make distinctions between the computational

“difficulty” of obtaining a maximal or prime ideal in a computable ring. That is, it is both set-theoretically easier and computationally easier to obtain a prime ideal than a maximal one.

This particular project in Reverse Math concerns theorems about lattice- ordered groups. These groups are not necessarily exotic. For example: an abelian 5 group G is lattice-orderable if and only if it is torsion-free. In Chapter 3, we see that this statement is equivalent to WKL0. Taking a step back, there are three main types of ordered groups, so called because they have both an algebraic group structure and a partial order structure which respects the group operations. The most general kind is a partially-ordered group, or p.o.-group. If the partial order is a lattice order, G is a lattice-ordered group, or `-group. If the order is a total

(linear) order, G is a totally ordered group, or o-group. We have the following class containments:

o-groups ( `-groups ( p.o.-groups

In [6], Solomon investigated theorems about p.o.-groups and o-groups. With `- groups in intermediate position between the other two, an immediate interest was what connections there were between the theory of lattice-ordered groups and the theories of the other two types, and whether or not there was a stronger resemblance to either side. There seem to be relatively few places where the theories of the three types of ordered groups can be compared directly, but these few are enough to suggest that `-groups are no more like o-groups than p.o.-groups.

For example, Theorem 3.7 is an instance of `-groups being like o-groups and unlike p.o.-groups, whereas in Corollary 4.10, the relationship is reversed.

The results presented follow an exploratory trend, beginning with an inves- tigation of how best to formalize `-groups in Z2, then moving on to the existence of the various types of subobjects which occur in the classical study of `-groups, 6 and culminating with an analysis of the more substantial Holland’s Embedding

Theorem.

Holland’s Embedding Theorem 1. Any `-group is `-isomorphic to an `- subgroup of the `-group of order-preserving permutations of a totally ordered set.

The standard proof of Holland’s Theorem makes use of a sequence of values: convex subgroups which are maximal with respect to excluding a specific element.

As shown in Chapter 5, the existence of such a sequence is equivalent to ACA0.

However, in Chapter 6, we show that Holland’s theorem is equivalent to an alter- native assumption and, when G is abelian, is provable in the weaker subsystem

WKL0.

1.2 Notation

We will use the symbols ∧, ∨ for lattice meet and join, and use the symbol “&” and the word “or” in formulas for conjunction and disjunction, respectively.

We fix an enumeration Φ0, Φ1,... of all partial computable functions.

1.3 Reverse Math

With the Reverse Math scope limited to RCA0, WKL0, and ACA0, we routinely use the following lemmas when proving equivalences over RCA0.

Lemma 1.1. The following are pairwise equivalent over RCA0 [5]. 7

1. WKL0

0 0 2. (Σ1 separation) Let φi(n), i = 0, 1 be Σ1 formulas in which X does not occur

freely. If ¬∃n(φ0(n)& φ1(n)) then

∃X∀n((φ0(n) → n ∈ X)&(φ1(n) → n 6∈ X)).

3. If f, g : N → N are one-to-one with ∀m∀n(f(m) 6= g(n)) then

∃X∀m(f(m) ∈ X & g(m) 6∈ X).

Lemma 1.2. The following are pairwise equivalent over RCA0 [5].

1. ACA0

0 0 2. Σ1 comprehension, i.e., ∃X∀n(n ∈ X ↔ φ(n)) restricted to Σ1 formulas

φ(n) in which X does not occur freely.

3. For all one-to-one functions f : N → N there exists a set X ⊆ N such that

∀n(n ∈ X ↔ ∃m(f(m) = n)), i.e., X is the range of f.

1.4 Lattices, Lattice-ordered groups

Definition 1.3. A lattice is a partially ordered set such that every pair of ele- ments has a least upper bound (join) and greatest lower bound (meet). The meet and join of elements a, b are denoted a ∧ b and a ∨ b, respectively. 8

Definition 1.4. A partially-ordered group or p.o.-group is a group G that is also a partial order and satisfies a condition that the group operation preserves the order on G, that is

a ≤ b → ∀g(ag ≤ bg & ga ≤ gb).

If the partial order on G is a lattice order, then G is a lattice-ordered group or

`-group. If the order on G is a total (linear) order, then G is a totally-ordered group or o-group.

1.5 Fundamental Examples

Example 1.5. Continuous functions on a topological space

From [1]: let X be a topological space and C(X) the additive group of real- valued continuous functions. We make C(X) an `-group by providing it with its usual pointwise order: f ≤ g if and only if f(x) ≤ g(x), for all x ∈ X.

Example 1.6. (Restricted) Hahn groups

From [1]: let Γ be a root system. That is, Γ is a partially ordered set for which {α : α ≥ γ} is totally ordered, for any γ ∈ Γ. Let {Hγ : γ ∈ Γ} be a collection of o-groups indexed by Γ. Consider functions v on Γ for which v(γ) ∈ Hγ, for all γ ∈ Γ. Given such a function v, the subset of Γ where v is 9

not zero is called its support. Let Σ(Γ,Hγ) be the set of all such functions whose support is finite. This is a group under addition. Furthermore, if we define an element to be positive if it is positive at each maximal element of its support, then

Σ(Γ,Hγ) is an `-group, called a restricted Hahn group on Γ. (The general Hahn group on a root system does not require finite support. Rather, the support of each element must have the ascending chain condition.)

Example 1.7. The group of finite sequences of integers

L Let G be the group ω Z. That is, G is the group of finite sequences of integers, with the group operation of componentwise addition. (If two strings differ in length, we append zeros to the shorter one until the lengths match.) We say an element of G is positive if each component is greater than or equal to zero. Thus, the meet and join of two elements f, g is computed by taking the componentwise minimum (maximum, resp.) of the two. G may be viewed as a restricted Hahn group on the root system Γ consisting solely of a countably infinite antichain, with each Hγ = Z equipped with its usual order as an o-group.

Of course, G may also be viewed as the additive group of finite-support functions from N to Z equipped with the pointwise order.

Example 1.8. Permutations of a Linear Order (and conventions)

From [4]: let L be a totally ordered set, and let Aut(L) be the set of order- preserving bijections from L to L. Then Aut(L) is an `-group under composition, 10 and lattice operations ∨, ∧ are defined by the rules:

[f ∨ g](l) = max{f(l), g(l)}

[f ∧ g](l) = min{f(l), g(l)}

In terms of the induced ≤ relation, we have f ≤ g ⇐⇒ f(l) ≤ g(l) for all l.

Example 1.9. Torsion-Free abelian group generated by {xi, yi}i∈N.

The formal presentation for this group is from Solomon as in [6], which is summarized below for reference.

Let G be the free abelian group on the generators {xi, yj}i∈N. Formally, elements of G are quadruples hI, q, J, pi where I and J are finite subsets of N and p and q represent functions

q : I → Z\0 and p : J → Z\0

X X The element hI, q, J, pi is denoted cixi + djyj. The elements represented i∈I j∈J by hI, q, J, pi and hI0, q0,J 0, p0i are equal if and only if the four components are equal. The sum

! ! X X X X qixi + pjyj + rkxk + slyl i∈I j∈J k∈K l∈L 11

X X is tmxm + unyn where M = (I ∪ K)\{z ∈ I ∩ K | qz + rz = 0} and m∈M n∈N   q if m ∈ I\K  m   tm = r if m ∈ K\I  m    qm + rm if m ∈ I ∩ K.

N and un are defined similarly. The identity 1G is represented by h∅, ∅, ∅, ∅i, and

−1 X X if g is represented by hI, q, J, pi then g is the sum −qixi + −pjyj. i∈I j∈J

Example 1.10. The group F in∗ Z

F in∗ is a formalization of Example 1.7 in RCA . With F in as the for- Z 0 Z malization of the set of all finite integer sequences in RCA , we let F in∗ = {α ∈ 0 Z

F inZ : α(|α| − 1) 6= 0}, i.e., the set of finite integer sequences which do not end in 0. The group operation +F in∗ consists of componentwise addition with the Z removal of any trailing zeros afterwards. If the strings are of different lengths, we pad the shorter one with zeros on the end until the lengths match, then add. The identity is the empty string, and the inverse of α ∈ F in∗ , denoted α−1 or −α, is Z obtained by changing the sign of every nonzero entry occurring in α.

F in∗ is an `-group under the pointwise order, and the meet and join can be Z calculated directly by (after appending zeros on the right until the lengths match) taking the minimum and maximum, respectively, in each component. Chapter 2

Preliminary Results

An important step in working with a class of objects in Z2 is to choose a work- able definition, especially when the object’s axioms involve quantifiers. We begin with the definitions of partially and linearly ordered groups, which have fairly straightforward axioms.

Definition 2.1. (RCA0) A group is a set G ⊆ N along with a constant, 1G (or

0G or e), and an operation, ·G, which obey the usual group axioms.

∀a, b, c ∈ G(a ·G (b ·G c) = (a ·G b) ·G c)

∀a ∈ G(1G ·G a = a ·G 1G = a)

−1 −1 −1 ∀a ∈ G∃a ∈ G(a ·G a = a ·G a = 1G)

Definition 2.2. (RCA0) A partial order is a set X with a binary relation ≤X

12 13 satisfying the following axioms.

∀x ∈ X(x ≤X x)

∀x, y ∈ X(x ≤X y & y ≤X x → x = y)

∀x, y, z ∈ X(x ≤X y & y ≤X z → x ≤X z)

Definition 2.3. (RCA0) A partially ordered group is a pair (G, ≤G) where

G is a group, ≤G is a partial order on G and, the group operation preserves the partial order on G. That is, for any a, b, c ∈ G, if a ≤G b then

(a ·G c ≤G b ·G c)&(c ·G a ≤G c ·G b).

If the order on G is a linear order, then G is an ordered group or o-group.

2.1 Definition of `-group

Classically, a lattice-ordered group or `-group is defined as a partially ordered group (G, ≤) where each pair of elements has unique greatest lower bound (meet) and least upper bound (join), i.e., the partial order is a lattice order.

0 Syntactically, the meet of a and b can be defined by a Π1 formula:

x = (a ∧ b) ↔ ∀z([z ≤ a & z ≤ b] → z ≤ x).

So can the join:

x = (a ∨ b) ↔ ∀z([z ≥ a & z ≥ b] → z ≥ x). 14

0 To say that the meet and join exist for every pair a, b is then a Π3 statement:

∀a, b∃x∀z([z ≤ a & z ≤ b] → z ≤ x)

∀a, b∃y∀z[z ≥ a & z ≥ b] → z ≥ y).

The obvious concern, then, when deciding on a definition for `-groups in

RCA0, is whether or not these usual axioms for meet and join will be useful in actually calculating meet and join. Will knowing that every pair of elements has

0 a meet, say, enable us to calculate it by a ∆1 function? Or is it necessary to explicitly require functions for meet and join as part of the definition in RCA0?

Considering (N,REC), the standard model of RCA0, the following theorem shows

0 that it is not possible in general to define meet and join by ∆1 functions.

Theorem 2.4. There is a computable p.o.-group (G, ≤) which is a lattice-ordered group, but for which the meet ∧ and join ∨ operations are not computable. Fur- thermore, they are as complicated as the halting problem.

Proof. Imitating a construction by Downey and Kurtz in [2], we construct our computable group G = ∪sGs in stages. Each Gs will be a finite set of integers with an associated representation function σs. For all elements n ∈ Gs, σs(n) will be an element of F in∗ , a finite string of integers with no trailing zeros. We will Z define n ◦ m = k if σ (n) + σ (m) = σ (k). The operation ◦ will be ∪ ◦ , s s F inZ s s s∈ω s that is, once n ◦s m = k is defined, we will have n ◦t m = k for all t ≥ s. Similarly, we define the partial order ≤s, whose union will be the partial order on G. The 15

partial order will satisfy the property n ≤G m ⇐⇒ σs(n) ≤s σs(m). That is, once σs(n), σs(m) are defined, the partial order is defined and will be preserved throughout the construction.

The construction of G will be done accomplishing the following:

• For some m(s) ∈ ω we will assign to all p ≤ m(s) a string σs(p).

• If t > s, then m(t) ≥ m(s).

• If p, q, r ≤ m(s) and σ (p) + σ (q) = σ (r), then for all t > s, we ensure s F inZ s s

that σ (p) + σ (q) = σ (r). t F inZ t t

• If p, q ≤ m(s) and σs(p) ≤ σs(q), then for all t > s, we ensure that σt(p) ≤

σt(q). We do the same for 6≤.

• For all p, lims∈ω σs(p) exists and is a finite string.

Additionally, G will have a computable sequence of pairs hae, bei, such that ae ∧ be 6= 0G ⇐⇒ Φe(e) ↓. We will meet the following requirements:

• Group Closure

Ce: If e = hp, qi, then p ◦ q is defined.

• Group Inverses

Ip: There exists a q such that p ◦ q = 0 (0 denotes the zero of ω and the

identity of G.) 16

• Coding Requirements

Ne: ae ∧ be can compute whether Φe(e) halts.

We order the requirements as follows: N0 ≥ C0 ≥ I0 ≥ N1 ≥ · · ·

Definition 2.5. We say

1. Ce requires attention at stage s + 1 if e = hp, qi for some p, q ≤ m(s) and

there is no r ≤ m(s) such that σ (p) + σ (q) = σ (r). s F inZ s s

2. Ip requires attention at stage s + 1 if p ≤ m(s) and there is no q ≤ m(s)

such that σ (p) + σ (q) = 0. s F inZ s

3. Ne requires attention at stage s + 1 if Ne is unrealized and either

3e 3e (a) for no t ≤ s are there a p, q such that σs(p) = 0 1, σs(q) = 0 01 (in

which case Ne is inactive); or

(b) there exists p, q as above and Φe,s(e) ↓. (we say Ne is active)

These p, q are the pair hae, bei.

The requirements Ne will achieve their goal as follows. When Ne is active and unrealized, we have introduced elements ae, be whose meet corresponds to the identity 0 ∈ F in∗ . Once Φ (e) halts, we change the mapping on a and b in a Z e e e way that’s compatible with all the p.o.-group requirements so that ae ∧be becomes a new nonidentity element. 17

Construction.

Stage 0. Set p(0) = 0 and σ (0) = 0 ∈ F in∗ . The zero of is to be the 0 Z N identity of hG, ◦i.

Stage s + 1. Let R be the requirement of highest priority that requires attention.

Case 1: R is Ce with e = hp, qi. Define m(s + 1) = m(s) + 1. Define σs+1(r) for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  σ (p) + σ (q) if r = m(s + 1).  s F inZ s

Case 2: R is Ip. Let m(s + 1) = m(s) + 1. Define σs(r) for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  −σs(p) if r = m(s + 1).

Case 3: R is Ne and Ne is inactive. Let m(s + 1) = m(s) + 2 and define

σs+1 for r ≤ m(s + 1) via:   σ (r) if r < m(s),  s   σs+1(r) = 03e1 if r = m(s + 1),     3e 0 01 if r = m(s + 2).

At this point, Ne becomes active. 18

Case 4: R is Ne and Ne is active. Thus, there are p, q ≤ m(s) such that

3e 3e σs(p) = 0 1, σs(q) = 0 01 and Φe,s(e) ↓. Let m(s + 1) = m(s) + 1, and define

σs+1 for r < m(s + 1) via:   σs(r)(x) if x 6= 3e + 2 σs+1(r)(x) =  σs(r)(3e) + σs(r)(3e + 1) if x = 3e + 2 and define σs+1 for r = m(s + 1) by

3e σs+1(r) = 0 001.

Ne is now realized.

This has the effect of re-mapping the generators 03e1 → 03e101 and 03e01 →

03e011, and defining an element mapped to the new meet, 03e001. This of course means that the lengths of strings may change at some stage, but each string can only get at most two digits longer.

3e For example, suppose we have σs(r) = 0 23, where Φe,s ↑. If there is a

0 3e t > s such that Φe,t ↓, then for some stage t ≥ t we will have σt0 (r) = 0 235, as

a linear combination of the redefined generators.

0 This will allow us to compute 0 from a meet function, since ae ∧ be 6=

0G ⇐⇒ Φe(e) ↓.

Verification.

k The limit of σs exists, because the map will change at most d 3 e times for a

string of length k. The partial order and group operations are preserved by the 19

redefinition in Case 4, because it is a linear mapping. The injectivity of lims σs follows by construction and is preserved by the redefinition in Case 4.

Thus we have an injection of G into the finite sequences of integers as a p.o. group. G is a computable p.o-group by the construction. Nonuniformly, we see that G is isomorphic to the subgroup of F in∗ generated by the strings Z

{03e1, 03e01 : e 6∈ HALT } ∪ {03e1, 03e01, 03e001 : e ∈ HALT }.

This is also an `-subgroup of F in∗ , so the order ≤ on G ends up being a lattice Z order in the classical sense. Because of the way that G is constructed, no steps need to be taken to ensure that meets and joins exist, although requirements to that effect could be easily added.

2.2 Definition of an `-group for Reverse Math

Because of the connection between RCA0 and computable mathematics, the above result means that in order to work with meet and join in `-groups in RCA0, we need to explicitly include them in the formal definition. Thus, we consider an

`-group to be given by a group G together with a lattice order ≤ and functions ∧ and ∨ giving the meet and join operations.

Definition 2.6. An `-group is a tuple (G, e, ·, ∧, ∨) such that under the partial order defined by x ≤ y ⇔ x∨y = y, (G, e, ·, ≤) satisfies the axioms of a p.o.-group 20 and the following identities:

L.1. x ∨ x = x x ∧ x = x

L.2. x ∨ y = y ∨ x x ∧ y = y ∧ x

L.3. (x ∨ y) ∨ z = x ∨ (y ∨ z)(x ∧ y) ∧ z = x ∧ (y ∧ z)

L.4. (x ∨ y) ∧ x = x (x ∧ y) ∨ x = x

We sometimes take the partial (lattice) order ≤ as included, though it is computed directly from ∨. It is easy to give direct proofs that the lattice ax- ioms imply the poset axioms of reflexivity, antisymmetry, and transitivity via the equivalence x ≤L y ⇔ x ∨L y = y using the lattice axioms L.1, L.2, and L.3, respectively. Verifying that the join is the least upper bound uses L.3. Verifying the meet requires using L.4. to redefine the above equivalence using ∧, then uses

L.3 as in the join. Thus, it is possible to prove in RCA0 that if (G, ∨, ∧) satisfies

L.1 through L.4, then it is a lattice under the induced partial order.

2.3 Basic computation in `-groups

The following theorems have proofs that can be easily formalized in RCA0. (See pages 15–18 of [4].)

Theorem 2.7. (RCA0) [Multiplication Across Meet and Join] Let (G, ·, e, ∧, ∨) be such that (G, ·, e) is a group, (G, ∧, ∨) is a lattice, and in G the identities

u(x ∨ y)v = uxv ∨ uyv , u(x ∧ y)v = uxv ∧ uyv 21

hold. Then G is a lattice-ordered group under the partial order of the lattice

(G, ∧, ∨). If G is a lattice-ordered group, then the above identities hold.

Proof. See proof of Theorem 2 on pages 14–15 of [4].

Definition 2.8. An element g of an `-group is said to be positive if e ≤ g, or

equivalently, g ∧ e = e. Negative elements are defined similarly.

Theorem 2.9. (RCA0) In any `-group G the inequality a ∧ bc ≤ (a ∧ b)(a ∧ c) is

satisfied for any positive elements a, b, c. Furthermore, if a ∧ b = e then a ∧ bc =

a ∧ c.

Proof. From page 15 of [4]. If a, b, c ≥ e then a ≤ a2, a ≤ ac, a ≤ ba and

(a ∧ b)(a ∧ c) = a2 ∧ ba ∧ ac ∧ bc ≥ a ∧ bc. The second statement then follows from

the first.

n m Corollary 2.10. (RCA0) For all m, n ∈ N, a ∧ b = e → a ∧ b = e

Proof. We use the special cases where c = a or c = b to build up by induction.

Theorem 2.11. (RCA0) In any `-group, the following hold:

(x ∨ y)−1 = x−1 ∧ y−1 , (x ∧ y)−1 = x−1 ∨ y−1

x(x ∧ y)−1y = x ∨ y , x(x ∨ y)−1y = x ∧ y

Proof. See proof of Proposition 1 on page 16 of [4].

n Theorem 2.12. (RCA0) In an `-group G, the following holds: ∀n ≥ 1(x ≥ e →

x ≥ e). 22

To prove this universal statement in RCA0 requires several lemmas formaliz- ing the standard proof. We define, by primitive recursion, a function f(x, n) = xn, which we will use transparently by writing exponents, and two functions m, and m+ with the following inductive definitions.

m(x, 1) = x ∧ e, m(x, n + 1) = xn+1 ∧ m(x, n) for n ≥ 1

m+(x, 1) = x, m+(x, n + 1) = xn+1 ∧ m+(x, n) for n ≥ 1

Intuitively, m(x, n) = xn ∧ xn−1 ∧ · · · ∧ x ∧ e, and m+(x, n) = xn ∧ xn−1 ∧ · · · ∧ x.

+ Lemma 2.13. (RCA0) ∀x∀n ≥ 1(m(x, n) = m (x, n) ∧ e)

Proof. The proof is by induction on n. When n = 1, m(x, 1) = m+(x, 1) ∧ e by definition. Suppose the lemma holds for n.

m(x, n + 1) = xn+1 ∧ m(x, n)

= xn+1 ∧ m+(x, n) ∧ e (by induction hypothesis)

= m+(x, n + 1) ∧ e (by definition of m+)

+ Lemma 2.14. (RCA0) ∀x∀n ≥ 1(x · m(x, n) = m (x, n + 1)) 23

Proof. Again, proof by induction on n. When n = 1,

x · m(x, 1) = x · (x ∧ e)

= x2 ∧ x

= x2 ∧ m+(x, 1)

= m+(x, 2).

Supposing the lemma for n,

x · m(x, n + 1) = x(xn+1 ∧ m(x, n))

= xn+2 ∧ x · m(x, n)

= xn+2 ∧ m+(x, n + 1) (by induction hypothesis)

= m+(x, n + 2).

n Lemma 2.15. (RCA0) ∀x∀n ≥ 1((x ∧ e) = m(x, n))

Proof. Again, a proof by induction on n. The case n = 1 is trivial. 24

Suppose we have the lemma for n.

(x ∧ e)n+1 = (x ∧ e)(x ∧ e)n

= (x ∧ e) · m(x, n) (by induction hypothesis)

= (x · m(x, n)) ∧ (e · m(x, n))

= (m+(x, n + 1) ∧ m(x, n) (by Lemma 2.14)

= xn+1 ∧ m+(x, n) ∧ m(x, n)

= xn+1 ∧ m+(x, n) ∧ m+(x, n) ∧ e (by Lemma 2.13)

= xn+1 ∧ m+(x, n) ∧ e

= xn+1 ∧ m(x, n) (by Lemma 2.13)

= m(x, n + 1)

Now we return to proof of the theorem: ∀x∀n ≥ 1(xn ≥ e → x ≥ e)

Proof. The proof is now direct, and the case n = 1 is trivial. Otherwise, suppose xn+1 ≥ e. 25

(x ∧ e)n+1 = m(x, n + 1) (by Lemma 2.15)

= xn+1 ∧ m(x, n)

= xn+1 ∧ m+(x, n) ∧ e

= (xn+1 ∧ e) ∧ m+(x, n)

= m+(x, n) ∧ e (By hypothesis, xn+1 ≥ e, so xn+1 ∧ e = e.)

= m(x, n) = (x ∧ e)n

Thus, if we suppose (x ∧ e)n+1 ≥ e, then the above calculation shows that

(x ∧ e)n+1 = (x ∧ e)n. Multiplying both sides by (x ∧ e)−n, we obtain x ∧ e = e, so x ≥ e.

Definition 2.16. For x ∈ G, we define x+ = (x ∨ e), x− = (x ∧ e), and |x| =

(x ∨ x−1).

+ − + − −1 + − −1 Theorem 2.17. (RCA0) In an `-group G, x = x x , x ∧(x ) = e, x (x ) =

(x−)−1x+.

Proof. See proof of Proposition 5 on page 17 of [4].

Theorem 2.18. (RCA0) In an `-group G, the following hold:

(xy)+ ≤ x+y+ , (xy)+ = x(x ∧ y−1)−1

∀n((xn)+ = (x+)n) , ∀n((xn)− = (x−)n) 26

Proof. See proof of Proposition 6 on page 18 of [4].

Theorem 2.19. (RCA0) In an `-group G, the following hold:

|x| = x+(x−)−1

∀n(|xn| = |x|n)

Proof. See proof of Proposition 7 on page 18 of [4].

Note that since x+ is positive and x− is negative, Theorem 2.19 implies that absolute values are positive.

Theorem 2.20. (RCA0) In an `-group, |xy| ≤ |x||y||x|, and |x ∨ y| ≤ |x||y|.

Proof. From page 18 of [4]. Both are brief. First, |x|−1|y|−1|x|−1 ≤ |x|−1|y|−1 ≤ xy ≤ |x||y| ≤ |x||y||x|. This shows that xy and (xy)−1 are both bounded above by |x||y||x|. Then |xy| = (xy) ∨ (xy)−1 ≤ |x||y||x|. Second, |x ∨ y| = (x ∨ y) ∨

(x ∨ y)−1 = (x ∨ y) ∨ (x−1 ∧ y−1) ≤ x ∨ y ∨ x−1 ∨ y−1 = |x| ∨ |y| ≤ |x||y|. (To see this last inequality, just consider the fact that |x| ≤ |x||y| and |y| ≤ |x||y|.)

2.4 The Riesz Decomposition Theorem

Riesz Decomposition Theorem 1. Suppose h1, . . . , hn are positive elements of the `-group G. If e ≤ g ≤ h1h2 ··· hn, then g = g1g2 ··· gn, where e ≤ gi ≤ hi.

Proof. A proof of this Theorem (see page 3 of [1]) goes by induction on n. The base case n = 1 is trivial. For the induction step, suppose e ≤ g ≤ h1h2 ··· hn. 27

Let g1 = g ∧ h1 and note that e ≤ g1 ≤ h1. Then

−1 −1 −1 −1 k = g1 g = (g ∨ h1 )g = e ∨ (h1 g) ≤ h2 ··· hn,

Now, by induction k = g2 ··· gn with e ≤ gi ≤ hi. But g = g1k, and e ≤ g1 ≤ h1.

Theorem 2.21. The Riesz Decomposition Theorem is provable in RCA0

Proof. We formalize the proof given above in RCA0. Suppose we are given an

`-group G. Let R = {hn, g, xi : x is the code for an n-tuple hh1, . . . , hni s.t. each hi ≥ e and g ≤ h1 · h2 ··· hn}. Let Q = {hn, g, x, yi : hn, g, xi ∈ R & y codes an n-tuple hg1, . . . , gni s.t. e ≤ gi ≤ hi for each i and g = g1g2 ··· gn}.

By primitive recursion on n, we define a function f : R → N such that for all hn, g, xi ∈ R, hn, g, x, f(n, g, x)i ∈ Q.

Case n = 1. If h1, g, xi ∈ R, then x = hgi and e ≤ g so h1, g, x, gi ∈ Q. So we define f(h1, g, xi) = g.

−1 Case n > 1. We define f(hn, g, xi) = h(g ∧h1)i?f(n−1, (g ∧h1) g, hh2, . . . , hni), where x = hh1, . . . , hni and ? denotes concatenation and subsequent coding of lists of elements.

−1 By the induction hypothesis, f(n − 1, (g ∧ h1) g, hh2, . . . , hni) returns a

−1 yˆ such that (g ∧ h1) g = g2g3 ··· gn wherey ˆ = hg2, . . . , gni. Then g = (g ∧ h1)g2 ··· gni, so the code h(g ∧ h1), g2, . . . , gni returned by this process is correct.

Let Π(x) be the product of the elements coded by x, and let P (n) be the 28 formula ∀g∀x(hn, g, xi ∈ R → g = Π(f(n, g, x))) By our definition of f, and by calculations similar to those in the first proof, we have ∀n(P (n) → P (n + 1)), so

0 by Π1 induction:

∀n∀g∀x(hn, g, xi ∈ R → g = Π(f(n, g, x)))

The following criterion for an `-subgroup will be convenient for later verifi- cations.

Theorem 2.22. (RCA0) A subgroup H of the `-group G is an `-subgroup iff

∀a ∈ H(a ∨ e ∈ H).

Proof. From [4]: The first implication is obvious. Conversely, let H be a subgroup of G and suppose ∀a ∈ H(a ∨ e ∈ H). Then x−1y ∨ e ∈ H for any x, y ∈ H and x(x−1y ∨ e) = y ∨ x ∈ H. Now that we have joins, x ∧ y = (x−1 ∨ y−1)−1 ∈ H.

The following Lemma will be referred to in Chapter 5, and we put it here with the other results about basic computations.

Lemma 2.23. (RCA0) |a| ∧ |b| ≤ |a ∧ b|

Proof. By writing out the absolute values and distributing, we obtain the equali- 29 ties

|a| ∧ |b| = (a ∧ b) ∨ (a ∧ b−1) ∨ (a−1 ∧ b) ∨ (a−1 ∧ b−1)

|a ∧ b| = (a ∧ b) ∨ (a−1 ∨ b−1).

Note that each is a join of (a ∧ b) with another element. Furthermore,

(a ∧ b−1) ≤ (a−1 ∨ b−1)

(a−1 ∧ b) ≤ (a−1 ∨ b−1)

(a−1 ∧ b−1) ≤ (a−1 ∨ b−1)

where the first is justified by noting that (a−1 ∨ b−1) ≥ b−1 ≥ (a ∧ b−1), and the rest are similarly justified. The conclusion then follows by the principle x ≤ y → a ∨ x ≤ a ∨ y. Chapter 3

More Results

3.1 Identifying groups which are lattice-orderable

Since an `-group is first and foremost a group, it is worth asking the question

“Can I tell if a given group G can be made into an `-group?” This question is

partially answered by the following theorem about o-groups from [1].

Theorem 3.1. For an abelian group G, the following are equivalent:

1. G is totally orderable

2. G is lattice orderable

3. G is torsion-free

In the context of Reverse Math, the equivalences of the orderability state- ments with the torsion-free statement are somewhat unbalanced in that one di- rection can be done in RCA0 and the other cannot.

Lemma 3.2. (RCA0) If G is a (potentially non-abelian) group, then “lattice- orderable → torsion free”

30 31

Proof. This follows from Theorem 2.12. Suppose we have some torsion element x with xn = e Then, in particular xn ≥ e, so by Theorem 2.12, x ≥ e. On the other hand, we must also have x−n = e, and a similar argument shows that x−1 ≥ e, and thus x = e.

The equivalence of “t.f. abelian → totally orderable” and WKL0 was shown

in a paper by Hatzikiriakou and Simpson [3]. Since totally ordered groups are auto-

matically lattice ordered groups, we have WKL0 ` “t.f. abelian → lattice orderable.”

Since we already know that the implication for totally-orderable groups is equiv-

alent to WKL0, the remaining question is then: Does the weaker statement

“t.f. abelian → lattice orderable” imply WKL0? As it turns out, this is the case.

Theorem 3.3. (RCA0) The following are equivalent.

1. Every torsion-free abelian group is lattice-orderable.

2. WKL0

Proof. By the comments above, we have 2 → 1. Now, we prove 1 → 2.

Let u, v be one-to-one functions with disjoint ranges. We will form a sep-

arating set. Let G be the free abelian group on generators ai, bi modulo the

relations (2k + 3) · au(k) = bu(k) and (2k + 3) · bv(k) = av(k). (We use 2k + 3 so that we do not have equality when k = 0.) Since torsion free and abelian implies lattice-orderable, there is some lattice order with ∧, ∨, ≤ defined on G. We modify

the formal definition from Example 1.9 to get a formal representation of G. The 32 technique of introducing relations is adapted from [3]. Any g ∈ G has a unique normal form X X g = qiai + pjbj i∈I j∈J where the coefficients are reduced according to the relations above. In terms of the quadruple hI, q, J, pi which formally represents g, we require that

∀i ∈ I¬∃k ≤ (|qi| − 2)(i = u(k))

&

∀j ∈ J¬∃k ≤ (|pj| − 2)(j = v(k))

Since these are bounded-quantifier formulas, set of these normal forms exists in RCA0. To see why these requirements give us a unique normal form, suppose that u(k) = i. Then we can start generating normal forms like

ai, 2ai, . . . , kai, (k + 1)ai

−ai, −2ai,..., −kai, −(k + 1)ai.

With positive coefficients, once we get to (k + 2)ai, we have the relation (2k +

3)ai = bi which gives us the equality (k + 2)ai = bi − (k + 1)ai. The bounds on the coefficients appearing in the requirements above force us to use the latter representation for the normal form. 33

We formally work with G0, defined as the set of all elements of G that are

0 0 0 in normal form. Then, for the G group sum on normal forms g +G0 h , we take

0 0 0 the G-sum g +G h and, if necessary, search for the normal form in G obtained by

0 0 0 0 reducing g +G h according to the relations. Since g and h were in normal form before adding, this process may only change a given qi by 2k+3, where u(i) = k, and change the corresponding pi by one, and similarly for the relations involving the function v.

0 0 P That is, after adding the normal-form elements g +G h to get i∈I qiai +

P 0 0 0 0 j∈J pjbj, the associated normal form has components hI , q ,J , p i, where each component can be calculated as demonstrated below.

If, for example, ∃k ≤ (|qi| − 2)(i = u(k)), then we reduce to a normal form by cases. 0G z }| { Case 1: qi > 0. Then we take qiai+(bi − (2k + 3)ai) to get (qi−2k−3)ai+(pi+1)bi.

Since the two summands were in normal form, we have qi ≤ 2k + 2. Thus k + 3 ≤ qi ≤ 2k + 2, so −k ≤ qi − 2k − 3 ≤ −1, and the result is in normal form.

Coincidental cancellation may result in differences between I,J and I0,J 0.

Case 2: qi < 0 is handled similarly.

Because this process uses only bounded quantification, including quantifying over (codes for) finite sets, the function taking hI, q, J, pi to the normal form

0 0 0 0 0 0 hI , q ,J , p i may be defined by a ∆1 formula, so G exists as a group in RCA0.

Having done this, we refer to G0 as G. 34

Claim. G is torsion free.

The claim follows from the fact that u and v have disjoint ranges. Essen- tially, a reduction of m · ai to bi may only happen if i ∈ ran(u) and a reduction of m · bi to ai may only happen if i ∈ ran(v).

Now, we return to the proof of a reversal. The theorem asserts that there is a lattice order on G, which implies there are meet and join functions ∨, ∧.

This, in turn, lets us work with the absolute value |x| = x ∨ x−1. Then the set

{n : |an| ≤ |bn|} separates the ranges of u, v. Suppose u(k) = n. Then we have

2k+3 2k+3 an = bn and |an| = |bn|. Since |x| ≥ e for all x, it follows that |an| ≤ |bn|. It is easy to show that for every n, |an| 6= |bn|, so the complement of the separating set contains {n : |an| > |bn|}, which is the range of v. Note that |an| and |bn| are only related by < or > if n is in the range of u or v. By Lemma 1.1, the existence of a separating set implies WKL0.

3.2 Convex `-subgroups and Right Cosets

In the context of `-groups, a type of `-subgroup called a convex `-subgroup plays an important role, analogous to that of normal subgroups in the context of general group theory. Cosets of a normal subgroup inherit the group structure in a natural way and, similarly, cosets of a convex `-subgroup inherit the lattice structure in a natural way. 35

Definition 3.4. An `-subgroup C is convex if for every c1, c2 ∈ C and every x, c1 ≤ x ≤ c2 → x ∈ C.

Theorem 3.5. (RCA0) The subgroup C is a convex `-subgroup iff ∀c ∈ C(|x| ≤

|c| → x ∈ C).

Proof. On page 23 of [4] there is a quick proof which can be done in RCA0.

Theorem 3.6. If C is a convex `-subgroup of the `-group G, then the right cosets of C in G form a lattice under the partial order given by Cg ≤ Ch ↔ ∃c ∈ C(g ≤ hc).

This same condition gives us the induced order in the context of cosets of a convex subgroup of a p.o.-group or an o-group. As shown in [6], the existence of the induced order on G/C is equivalent to ACA0 when G is a p.o.-group, and provable in RCA0 when G is an o-group. In this regard, it turns out that `-groups are closer to o-groups than p.o.-groups.

Theorem 3.7. (RCA0) Let G be an `-group. Let H be a convex `-subgroup of G.

The induced order on the set of right cosets G/H exists.

Proof. Let a, b be elements of G. The induced order on G/H as originally given is: Ha ≤ Hb ↔ (∃h ∈ H)a ≤ hb. Note that by choosing h = e, it is apparent that a ≤ b → Ha ≤ Hb. Also note the following identity, which holds for all elements x, y.

x = (e ∨ xy−1)(x ∧ y) 36

To verify this identity, we distribute. First,

(e ∨ xy−1)(x ∧ y) = [(e ∨ xy−1)x] ∧ [(e ∨ xy−1)y]

= (x ∨ xy−1x) ∧ (y ∨ x)

≥ x

Then, distributing the other way,

(e ∨ xy−1)(x ∧ y) = [e(x ∧ y)] ∨ [xy−1(x ∧ y)]

= (x ∧ y) ∨ (xy−1x ∧ x)

≤ x

Claim. (e ∨ ab−1) ∈ H → Ha ≤ Hb

Proof. Let (e ∨ ab−1) ∈ H. Then a = (e ∨ ab−1)(a ∧ b), by the identity above.

Therefore, Ha = H(a ∧ b) ≤ Hb.

Claim. Ha ≤ Hb → (e ∨ ab−1) ∈ H

Proof. Fix h ∈ H s.t. a ≤ hb. Then we have

(a ∧ b) ≤ a ≤ hb

(a ∧ b)b−1 ≤ ab−1 ≤ h

(ab−1 ∧ e) ≤ ab−1 ≤ h

e ∨ (ab−1 ∧ e) ≤ (e ∨ ab−1) ≤ e ∨ h

e ≤ (ab−1 ∨ e) ≤ e ∨ h 37

Since e, h ∨ e belong to H and H is convex, we have ab−1 ∨ e ∈ H.

By these two claims, we can decide the induced order on cosets by the quantifier free condition aH ≤ bH ↔ (e ∨ ab−1) ∈ H.

If C is normal then G/C has the familiar quotient group structure described by Ca · Cb = C(ab). Even when C is not normal, by convexity alone, we have the induced order on the right cosets, as defined above, and this gives us the properties

Ca ∨ Cb = C(a ∨ b) and Ca ∧ Cb = C(a ∧ b) as discussed in the following theorem from [4].

Theorem 3.8. (RCA0) Let C be a convex subgroup of G. Then C(x∧y) = Cx∧Cy and C(x ∨ y) = Cx ∨ Cy under the induced order.

Proof. We prove the case for joins, and the case for meets is similar.

By definition of the induced order, it follows that Cx ≤ C(x∨y) and Cy ≤ C(x∨y).

Let Cz ≥ Cx, Cz ≥ Cy. for some z ∈ G. Then there exist c1, c2 ∈ C such that c1z ≥ x, c2z ≥ y. Let c = c1 ∨ c2 ∈ C. Then cz ≥ c1z ≥ x and cz ≥ c2z ≥ y, so cz ≥ x ∨ y. Thus Cz ≥ C(x ∨ y) and C(x ∨ y) is the least upper bound of Cx, Cy in G/C. Chapter 4

Convexity Results and Reversals for Substructure

Existences

4.1 Closure Operations

To each kind of subgroup we associate a closure operation.

Definition 4.1. Let A be a subset of the `-group G. Let

• hAi denote the subgroup of G generated by A,

• hAi` denote the `-subgroup of G generated by A,

• and CL(A) denote the convex `-subgroup of G generated by A.

To form hAi, one must close under inverses and group composition. To form

−1 −1 −1 hAi`, it is sufficient to further close under join, since x ∧ y = (x ∨ y ) . From this, one may form CL(A) by adding all elements bounded above and below by elements of hAi`.

To work with these closures in RCA0, we define a mechanism of generation

0 by primitive recursion. First we define a Σ1(A) mechanism for generating hAi. Let

38 39

A be a subset of the `-group G, and suppose G is enumerated as {g0 = e, g1,...}.

Let

• hAi0 = {0}

• hAit+1 = {gk ∈ G : gk ≤N t & ∃gi, gj ∈ hAit s.t.

( gk ∈ hAit

or gk ∈ A

−1 or gi = gk

or gk = gigj)}

0 Then we define hAi as the set {g ∈ G : ∃n(g ∈ hAin)}, which is Σ1(A). Thus, given an `-group G and a subset A ⊂ G, ACA0 is sufficient to prove the existence of hAi.

By adding the disjunction “or gk = gi ∨ gj”, one defines hAi`. By further adding the disjunction “or gi ≤ gk ≤ gj”, one defines CL(A).

0 Note that the definition of the closure is Σ1 in all three cases. While ACA0

is equivalent to the existences of hAi, hAi`, and CL(A), given A, many of the

interesting properties of these sets can be verified in RCA0 if they exist. For the

next lemma, we will use the following notation which generalizes to the other

forms of closure.

• g ∈ hAi means ∃n(g ∈ hAin) 40

• hAi ⊆ hBi means ∀g(∃n(g ∈ hAin) → ∃n(g ∈ hBin)).

• hAi is a subgroup of G means ∀g, h(g, h ∈ hAi → (g−1 ∈ hAi & g ·h ∈ hAi)).

Lemma 4.2. (RCA0) Using the abbreviations above, for all A, B ⊆ G,

1. hAi is a subgroup of G.

2. If A is a subgroup of G then hAi = A.

3. If A ⊆ B then hAi ⊆ hBi.

4. If A ⊆ B and B is a subgroup of G then hAi ⊆ B.

Moreover, the Lemma is also true if one replaces all instances of the word “sub- group ” with “`-subgroup ” or “convex `-subgroup” and uses the appropriate closure operation.

Proof. Statement (1) is clear by the construction of hAi. For (2), the containment

A ⊆ hAi follows from the construction. The reverse containment is shown by the

0 statement ∀n(hAin ⊆ A) which is proved by Σ0 induction. There are three cases

th to consider for an element added at the (n + 1) step. If gk is added because it

belongs to A, then it is obviously in A already. If gk is added because it is the

inverse of an element in hAin, then that element, by the induction hypothesis,

belonged to A, and thus so did gk since A is a subgroup. The third case, that gk

is a product of elements in hAin, is similar. 41

Similarly, to prove (3) one uses the statement “A ⊆ B → ∀n(hAin ⊆ hBin)”

0 which is proved by Σ0 induction. Finally, (4) follows from (2) and (3).

The generalizations to `-subgroup closure and convex closure are proved in the same way.

We also state the following lemma using the abbreviations described above.

Lemma 4.3. (RCA0) Let S be a subset of the `-group G. Then we have the

following:

• y ∈ S → hSi = hS ∪ {y−1}i

• x, y ∈ S → hSi = hS ∪ {xy}i

• x, y ∈ S → hSi` = hS ∪ {x ∨ y}i`

• x, y ∈ S & x ≤G z ≤G y → CL(S) = CL(S ∪ {z})

Furthermore, the first three implications are also valid for the more specific closure

operations.

Proof. We prove only the first statement. The rest are similar. First, suppose

y ∈ S. Then we have the containments S ⊆ S ∪ y−1 ⊆ hSi. To be explicit about

−1 −1 the second containment, suppose y is not already in S. If y = ga and y = gb,

with a < b, then the latter will be a member of hSib by construction. If the

inequality is reversed, then the latter will be a member of hSia+1. By arguments

like those in Lemma 4.2, we want to show that hSi ⊆ hS ∪ y−1i ⊆ hhSii = hSi. 42

−1 Formally, the containments can be stated as ∃n(x ∈ hSi) → ∃n(x ∈ hS ∪ y in)

−1 and ∃n(x ∈ hS ∪ y in) → ∃m∃n(x ∈ hhSinim), while the equality can be stated

∃m∃n(x ∈ hhSinim) ↔ ∃n(x ∈ hSin).

Lemma 4.4. (RCA0) Let g be an element of the `-group G. Then CL({g}) =

n n {h ∈ G : ∃n > 0(|h| ≤ |g| )}, i.e. ∀h[∃n(h ∈ CL({g})n ↔ ∃n > 0(|h| ≤ |g| ))].

n Proof. Fix h ∈ G. The statement ∃n > 0(|h| ≤ |g| ) → ∃n(h ∈ CL({g})n) follows from the definition of CL({g}).

For the other direction, instead of proving the implication ∃n > 0(|h| ≤

n 0 0 |g| ) → ∃n(h ∈ CL({g})n), which is neither Σ1 nor Π1, we use a bounded quan- tifier in the second part. Let f(n) = 3n, so that f(n + 1) = 3 · f(n). As in a

 0 classic “ 3 ” argument, this choice will make sense in the future. We show by Π1

f(n) induction on n that ∃n(h ∈ CL({g})n) → |h| ≤ |g| ). For n = 0, we have

1 CL({g})0 = {0}, |0| = 0 ≤ |g| , and 0 < 1 ≤ f(0) = 1. For the induction step, suppose h ∈ CL({g}n+1. Then we have several cases, corresponding to the ways an element can enter the convex closure.

• If h = g then |h| ≤ |g|1 and 1 ≤ f(n + 1).

k • If h ∈ CL({g})n, then |h| ≤ |g| for some 0 < k ≤ f(n) < f(n + 1).

−1 −1 k • If ∃gi ∈ CL({g})n with h = gi , then |h| = |gi | = |gi| ≤ |g| for some

0 < k ≤ f(n) < f(n + 1). 43

• If ∃gi, gj ∈ CL({g})n such that either h = gigj, h = gi ∨ gj, or gi < h < gj,

then we have 3 subcases.

f(n) f(n) – If h = gigj then, by Theorem 2.20, |h| ≤ |gi| · |gj| · |gi| ≤ |g| · |g| ·

|g|f(n) = |g|3f(n) = |g|f(n+1).

2f(n) f(n+1) – If h = gi ∨ gj then |h| ≤ |gi| · |gj| ≤ |g| < |g| .

−1 −1 – Suppose gi < h < gj. Then |h| = (h ∨ h ) ≤ (gj ∨ h ) ≤ (gj ∨

−1 f(n) gi ) ≤ (|gi| ∨ |gj|). Since gi, gj ∈ CL({g})n, |gi| ∨ |gj| ≤ |g| . Thus,

|h| ≤ |g|f(n) < |g|f(n+1).

Lemma 4.5. (RCA0). Let g1, . . . gn be positive elements of the `-group G. Then Y CL({g1, . . . , gn}) = CL({ gj}). 1≤j≤n

Proof. Both containments are proved using an n-fold application of Lemma 4.3.

Q First, since each gj is positive, e ≤ gi ≤ j gj. This proves the “⊆” containment. Y For the other direction, observe that the product gj must be contained in 1≤j≤n

CL({g1, . . . , gn}), since the latter is a subgroup containing each factor.

Theorem 4.6. (RCA0) If A, B are convex `-subgroups and hA, Bi exists, then

CL(A ∪ B) = hA, Bi.

Proof. The method of proof on page 8 of [1] uses the Riesz Decomposition The- orem and Theorem 2.20, and can be adapted to work in RCA0. We sketch the 44

proof. Let C = hA, Bi := hA ∪ Bi. By Lemma 4.2, C is a subgroup. We use the criterion in Theorem 3.5 to show C is a convex `-subgroup. Let c ∈ C, g ∈ G, and |g| ≤ |c|. Then there is a finite collection of elements ci, i ≤ n which belong

to either A or B such that

c = c0c1 ··· cn.

+ − −1 Then g ∨ (g ) = |g| ≤ |c0c1 ··· cn|, and by repeated use of Theorem 2.20 one

obtains the inequality

|c0c1 ··· cn| ≤ |c0||c1| · · · |cn||cn−1| · · · |c1| · · · |c0|.

Then, by the Riesz Decomposition Theorem, g+ and (g−)−1 (and hence g−) can

be expressed as products of elements which belong to A or B by convexity. Since

g = g+g−, g ∈ C.

4.2 Existence of the subgroup generated by A, B.

Theorem 4.7. (RCA0) The following are equivalent.

1. For any group G and subgroups A, B ⊆ G, the subgroup hA, Bi exists.

2. ACA0

0 Proof. hA, Bi can be formed in ACA0 since it has a Σ1 definition. To show

the reversal, let f be a one-to-one function. Let G be the torsion free abelian

group generated by {xi, yi}i∈N. The formal presentation for this group as sums 45

P P i∈I qiai + j∈J pjbj was defined in Example 1.9 and we subsequently refer to that notation below.

P Let A be the set of elements of the form i∈I qixi + 0, i.e., those for which

J = p = ∅. Clearly, A is a subgroup of G. Then let B be the subgroup of G generated by all elements of the form xi + yf(i). The exact condition for a formal

sum to belong in B can be stated as the following conjunction:

∀i ∈ I∃j ∈ J(j = f(i)& pi = qj)

&

∀j ∈ J∃i ∈ I(j = f(i)& pi = qj).

Since I and J are finite, the quantifiers are bounded, so both A and B may be formed in RCA0.

Clearly, if ∃i(f(i) = n) then the elements xi and xi + yn belong to A and

B, respectively. In particular, the group sum −1xi + (xi + yn) = yn belongs to hA, Bi. On the other hand, assume that yn = a + b for some a ∈ A, b ∈ B. Then P P P yn = i0∈I0 pi0 xi0 + ( i∈I pixi + i∈I piyf(i)). The parts involving x terms must

0 cancel, leaving I = I. Since only yn appears in the left hand side, we must have

I = {m} where f(m) = n and pn = 1.

Then yn ∈ hA, Bi ↔ ∃i(f(i) = n), and ran(f) = {n : yn ∈ hA, Bi}.

4.3 Existence of the convex closure of an `-subgroup H

Theorem 4.8. (RCA0) 46

1. For any `-group G and `-subgroup H ⊆ G, CL(H) exists.

2. ACA0.

0 Proof. Since CL(H) has Σ1 definition with H as a parameter, it is clear that

2 → 1. For the other direction, let G be F in∗ , the `-group of finite-support Z functions from N to Z in the pointwise order. Let f be a one-to-one function.

The set A of all strings σ such that all three of the following hold

1. ∀x < |σ|(σ(x) = 0 or 1 or 2)

2. ∀x < |σ|(σ(x) = 1 → ∃i ≤ x[x = 2i & 2f(i) + 1 < |σ| & σ(2f(i) + 1) = 2])

3. ∀x < |σ|(σ(x) = 2 → ∃i[2i < |σ| & x = 2f(i) + 1 & σ(2i) = 1])

is definable by a bounded-quantifier formula, so exists in RCA0.

2i−1 2n In other words, if ai = 0 1 and bn = 0 1, then the generators in A are

all strings of the form ai + 2bf(i). The subgroup H = hAi may also be described X X as all elements of the form ci · ai + 2ci · bf(i) for some finite set I, and the i∈I i∈I

set A represents elements for which ci = 1. Thus H exists in RCA0. Clearly, if

f(i) = n then 0 < bn < 2bn < ai + 2bf(i), so bn must belong to CL(H). On the other hand, suppose bn ∈ CL(H). Then there is some finite set I such that X X 0 ≤ b ≤ c · a + 2c · b . Since F in∗ has the pointwise order and n i i i f(i) Z i∈I i∈I 2n bn has the form 0 1, the sum bounding bn must have a positive number in the

th X (2n + 1) place. This can’t come from the ci · ai part, so must come from the 47

X the 2ci · bf(i) part. This means ∃i(f(i) = n). Then bn ∈ CL(H) if and only if n ∈ ran(f).

4.4 Existence of the convex `-subgroup generated by convex

`-subgroups A, B.

In the setting of totally ordered groups, the question is trivial, since CL(A, B) is simply the larger of the two or, equivalently, their union.

The first version of this proof uses computable algebra, and is followed by a more direct proof typical of Reverse Math. In all but the easiest reversals, it seems easier to work in the computable setting, and build an computable encoding object by a step-by-step construction, and to then use the process of the construction

0 to help define the object in more general terms using ∆1 formulas. Theorem 4.6 allows us to simplify both versions of the proof by looking at hA, Bi instead of

CL(A, B).

Theorem 4.9. There is a computable abelian `-group G with computable convex

0 `-subgroups A, B such that A + B ≥1 0 .

Corollary 4.10. (RCA0) The following are equivalent.

1. For any convex `-subgroups A, B of the `-group G, the subgroup hA, Bi exists.

2. ACA0. 48

Proof. As in the proof of Theorem 2.4, we construct an injection from a com-

L putable `-group G into the `-group D = ω Z, equipped with the pointwise

order. Elements of D are finite sequences of integers with the last entry nonzero.

The image of G in D will be a (nonconvex) `-subgroup of D, yet classically iso-

morphic to the full group. G will contain computable convex `-subgroups A, B

0 such that A + B=hA, Bi is ≥1 0 .

We construct our computable group G = ∪sGs in stages. Each Gs will be a finite set of integers, and representation function σs : Gs → D. For all elements n ∈ Gs, σs(n) will be a finite string of integers with no trailing zeros. We will define n ◦s m = k if σs(n) + σs(m) = σs(k). The operation ◦ will be ∪s∈ω◦s.

Similarly, we define ∧s, ∨s whose union will be the usual meet and join on G. The partial order on G can be recovered from ∨ and ∧ since a ∧ b = b ↔ b ≤ a.

The construction of G will be done accomplishing the following:

• For some m(s) ∈ ω we will assign to all p ≤ m(s) a string σs(p).

• If t > s, then m(t) ≥ m(s).

• If p ≤ m(s), then σs(p) = σt(p) for all t ≥ s.

Additionally, G will have computable convex `-subgroups A, B. Once σs(p) is defined, the membership of p in A or B will be computable from σs(p). We will have p ∈ A ⇐⇒ σs(p)(n) = 0 for each odd n up to |σs(p)|, and p ∈ 49

B ⇐⇒ σs(p)(n) = 0 for each even n up to |σs(p)|. We will meet the following requirements:

• Group Closure

Ce: If e = hp, qi, then p ◦ q is defined.

• Group Inverses

Ip: There exists a q such that p ◦ q = 0 (0 denotes the zero of ω and the

identity of G.)

• Meet and Join

Me,Je: If e = hp, qi, then p ∧ q, p ∨ q are defined.

• A∨B non-computability requirements

Ne: hA, Bi can compute whether φe(e) halts.

We order the requirements as follows: N0 ≥ C0 ≥ M0 ≥ J0 ≥ I0 ≥ N1 ≥ · · ·

Definition 4.11. We say

1. Ce requires attention at stage s + 1 if e = hp, qi for some p, q ≤ m(s) and

there is no r ≤ m(s) such that σs(p) + σs(q) = σs(r).

2. Me requires attention at stage s + 1 if e = hp, qi for some p, q ≤ m(s) and

there is no r ≤ m(s) such that σs(p) ∧ σs(q) = σs(r), and similarly for Je. 50

3. Ip requires attention at stage s + 1 if p ≤ m(s) and there is no q ≤ m(s)

such that σs(p) + σs(q) = 0

4. Ne requires attention at stage s + 1 if Ne is unrealized and either

2e (a) for no t ≤ s is there a p such that σs(p) = 0 11 (in which case we say

Ne is inactive); or

(b) there exists p as above and φe,s(e) ↓ (in which case we say Ne is active).

4.4.1 Construction

Stage 0. Set p(0) = 0 and σ0(0) = 0. The zero of ω is to be the identity of hG, ◦i.

Stage s + 1. Let R be the requirement of highest priority that requires attention.

Case 1: R is Ce with e = hp, qi. Define m(s + 1) = m(s) + 1. Define σs+1(r) for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  σs(p) + σs(q) if r = m(s + 1).

Case 2: R is Ip. Let m(s + 1) = m(s) + 1. Define σs(r) for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  −σs(p) if r = m(s + 1). 51

Case 3: R is Me with e = hp, qi. Define m(s+1) = m(s)+1. Define σs+1(r) for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  σs(p) ∧ σs(q) if r = m(s + 1).

Case 4: R is Je. Similar to case 3.

Case 5: R is Ne and Ne is inactive. Let m(s + 1) = m(s) + 1 and define

σs+1 for r ≤ m(s + 1) via:   σs(r) if r < m(s + 1), σs+1(r) =  2e 0 11 if r = m(s + 1).

At this point, Ne becomes active.

Case 6: R is Ne and Ne is active. Thus, there is a p ≤ m(s) such that

2e σs(p) = 0 11 and φe,s(e) ↓. Let m(s + 1) = m(s) + 2, and define σs+1 for r ≤ m(s + 1) via:   σ (r) if r ≤ m(s)  s   σs+1(r) = 02e1 if r = m(s) + 1     2e+1 0 1 if r = m(s) + 2 This creates elements of G in A and B, respectively, whose group sum is p, thereby

0 putting p ∈ hA, Bi. Ne is now realized. Notice that this will ensure hA, Bi ≥1 0 :

To determine if e ∈ 00, find p such that σ(p) = 02e11. Then e ∈ 00 ⇐⇒ p ∈ hA, Bi. 52

4.4.2 Verification

G is clearly a computable `-group by the construction. A, B, and hA, Bi are easily verified as convex `-subgroups. For example, A contains the elements p such that

σ(p) is zero in all odd-numbered components. The operations of group sum, meet and join on two elements of A cannot introduce a nonzero entry into an odd- numbered component with respect to the map σ, so A is an `-subgroup. As for the convexity of A: suppose a ∈ A and 0 ≤ |g| ≤ |a|. It follows that σ(g) is zero in all odd-numbered components, and so g ∈ A.

The verification for B, hA, Bi is very similar, except that in the case of hA, Bi, we say p ∈ hA, Bi ⇐⇒ σ(p) has nonzero entries only in components numbered 2e or 2e + 1 where φe(e) ↓.

Now, we give a formal proof of the reverse math corollary.

Theorem 4.12. (RCA0) The following are equivalent.

1. ACA0

2. If A, B are convex `-subgroups of G, then

∃X∀t(t ∈ X ↔ ∃n(t ∈ hA ∪ Bin)).

0 Proof. Since X has a Σ1 definition, 1 → 2. Now we show 2 → 1. Fix a one-to-one function f. We define an injection σ : → F in∗ by primitive recursion. The N Z 53

group G will be with +G given by g +G h = k ↔ σ(g) +F in∗ σ(h) = σ(k), and N Z

2e 2e 2e+1 we define αe = 0 11, βe = 0 1, andγe = 0 1.

• Let σ(0) = 0 (as a string)

• For r = 5m+1, set σ(r) = αe where e is least such that ∀k < r(σ(r) 6= αe).

(This will be the default action, if the test condition fails in other stages.)

• For r = 5m + 2, let hi, ji be N-least such that i, j < r and ∀k < r(σ(i) +

σ(j) 6= σ(k)). If such a pair exists, define σ(r) = σ(i) +F in∗ σ(j), otherwise Z perform the default action.

• For r = 5m + 3, let hi, ji be N-least such that i, j < r and ∀k < r(σ(i) ∧

σ(j) 6= σ(k)). If such a pair exists, define σ(r) = σ(i) ∧ σ(j), otherwise perform the default action.

• For r = 5m+4, let i be -least such that i < r and ∀k < r(σ(i)+F in∗ σ(k) 6= N Z 0. If such i exists, define σ(r) = σ(i)−1, otherwise perform the default action.

• For r = 5m + 5 look for N-least e < r s.t. ∃k < r(σ(k) = αe)& ¬∃k < r(σ(k) = βe or σ(k) = γe)& ∃k < r(f(k) = e). Then we define σ(r) = βe.

Then we define the sets

A = {n : ∀s(2s + 1 ≤ |σ(n)| → σ(n)[2s + 1] = 0)}

B = {n : ∀s(2s ≤ |σ(n)| → σ(n)[2s] = 0)}

It is easy to verify that A, B are both convex `-subgroups. Let e ∈ N. By definition of σ, there is some k s.t. σ(k) = αe. If e ∈ ran(f), then by definition of σ, there are m, n s.t. σ(n) = βe, σ(m) = γe. Then m ∈ A and n ∈ B. Then σ(n)+σ(m) = 54

αe = σ(k), so k ∈ hA, Bi. On the other hand, suppose k ∈ hA, Bi = A + B. Then there exist m ∈ A, n ∈ B s.t. m + n = k i.e., σ(m) + σ(n) = αe. But, by the definitions of αe and A, B, this means that all entries of σ(m), σ(n) must be zero except at 2e + 2 and 2e + 1, respectively. This implies that σ(n) = βe, which can only happen if ∃j < r(f(j) = e), by definition of σ. Thus we define a function

φ(e) by φ(e) = k ↔ k ≤ 5e + 1 & σ(k) = αe. By the construction, φ is total and definable in RCA0. By the verification, ran(f) = {e : φ(e) ∈ hA ∪ Bi}.

4.5 Existence of the Polar X⊥.

Definition 4.13. Two elements a, b of an `-group are said to be orthogonal if a ∧ b = e.

Definition 4.14. Let M be a subset of the `-group G. The polar of M, denoted

M ⊥, is defined as the set {g ∈ G : ∀x ∈ M(|x| ∧ |g| = e)}.

⊥ Theorem 4.15. (RCA0) If it exists, M is a convex subgroup of G.

Proof. Adapted from [1]. We verify that M ⊥ is a subgroup and is convex. Suppose g, h are positive elements of M ⊥. Let m ∈ M. Then

e = |m| ∧ g = |m| ∧ (|m| ∧ h)g = |m| ∧ (|m|g) ∧ (hg) = |m| ∧ hg.

Since the definition of M ⊥ uses absolute value, inverses are automatically included, and the identity satisfies the definition. Thus, checking composition is enough to 55

show that M is a subgroup. For convexity, suppose b ∈ M ⊥ and |a| ≤ |b|. Then

|a| ≤ |b| → e ≤ |m| ∧ |a| ≤ |m| ∧ |b|.

Since b ∈ M ⊥, |m| ∧ |b| = e, and therefore |m| ∧ |a| = e. So a ∈ M ⊥.

⊥ 0 Note that if M is finite, M has a Π0 definition, so exists in RCA0. In

⊥ ⊥ particular, for each g ∈ G, g = hgi exists in RCA0. Such polar subgroups are

⊥ 0 called principal polars. In general, though, M is Π1(M) and can be complicated.

When seeking a reversal for the existence of M ⊥, it is worth considering

how much structure M should have – when M is finite, the case is simple. When

M is infinite, is proving the existence of M ⊥ any simpler if M is already, say, a

subgroup, `-subgroup, or even convex `-subgroup? The following shows that, in

general, it is not.

Theorem 4.16. (RCA0) Let G be an `-group. The following are equivalent.

1. For every M ⊆ G, M ⊥ exists.

2. For every convex `-subgroup M ⊆ G, M ⊥ exists.

3. ACA0

Proof. 1 ⇒ 2 is trivial, and 3 ⇒ 1 by the arithmetical definition of M ⊥. It

remains to show 2 ⇒ 3. We will construct an abelian `-group G with a convex

`-subgroup M such that M ⊥ computes the range of a given one-to-one function.

Group theoretically, G is a restricted Hahn group on the root system Γ which may 56

be described as ω incomparable copies of an ω∗ ordering.

!

...

Fig. 4.1: The Root System Γ

We formalize this group in RCA0: let G be the free abelian group on gen- X erators aij. That is, an element of G is a reduced finite sum cijaij, 1≤i≤m,1≤j≤n where cij ∈ Z and ∃i∃j(cmj 6= 0 & cin 6= 0.) That is, the normal form for an element has coefficients indexed by an m × n array, where m, n are as small as possible in the sense that the last column and last row must each contain some

th nonzero coefficient. We may think of each aij as representing the j element of the ith column of Γ. The underlying formal representation uses triples hm, n, ci where c is a function from {hi, ji : 1 ≤ i ≤ m, 1 ≤ j ≤ n} to Z. As with Example

1.9, the group operation is componentwise addition, with cancellation. (If adding

elements of different sizes, one may perform the operation with oversized arrays

(using extra 0s) and then put the result in normal form by eliminating unneces- X sary rows and columns.) Thus, the inverse of the element cijaij is given by i≤m,j≤n X −cijaij, and the group identity is represented by h∅, ∅, ∅i. i≤m,j≤n

Let g = hm, n, ci be an element of G and let Sg = {i ≤ m : ∃j ≤ n(cij 6= 0)}, 57

th that is, Sg is the set of indices i such that g has a nonzero coefficient in the i column. An element in a Hahn group is positive if it is so at each maximal element of its support. Thus g = hm, n, ci is positive iff the set

max Sg = {cij : i ∈ Sg, j ≤ n & cij 6= 0 & ∀k < j(cik = 0)}

contains only positive numbers. This allows us determine the partial order ≤G.

−1 max For example, a ≤ b ↔ e ≤ ba , i.e., Sba−1 only contains positive numbers.

Equivalently, suppose we have two elements c, d with coefficients cij, dij defined

on an x × y array. Then

c ≤ d ↔ ∀i ≤ x[∃j ≤ y(cij 6= dij) → [dik > cik where k = µs(cis 6= dis)]].

This reasoning behind this is that when attempting to compare the two elements

c, d by testing to see if cd−1 is positive, the maximal elements of the support of

cd−1 correspond to the least-numbered row in each column such that differing

entries can be found.

So far, G is a p.o.-group in RCA0. Classically, we know it to be an `-group.

To verify that it is an `-group in RCA0, we need to be able to calculate the join of

elements a = hm, n, ci and b = hp, q, di. Putting some formality aside, we describe

an algorithm which can be used to calculate a ∨ b and claim that a function that

0 uses said algorithm to calculate joins is Σ0 definable.

First, letting x = max(m, p), y = max(n, q) we work with the “improper” 58 representations a∗ = hx, y, c∗i and b∗ = hx, y, d∗i where   cij if 1 ≤ i ≤ m & 1 ≤ j ≤ n ∗  cij :=  0 if m < i ≤ x or n < j ≤ y and d∗ is defined similarly.

We build the join a ∨ b by using c∗ and d∗ to construct a coefficient function h for an x × y array, and then, if necessary, we reduce the array size to eliminate unnecessary rows or columns of zeros. For each i ≤ x, one compares the ith columns of c∗ and d∗, and defines the ith coefficient column of h according to three cases.

∗ ∗ ∗ ∗ 1. If ∀j ≤ y(cij = dij), then ∀j ≤ y we define hij := cij.

∗ ∗ ∗ ∗ ∗ ∗ 2. If, at the least j such that cij 6= dij, cij > dij, then ∀j ≤ y we define hij := cij.

∗ ∗ ∗ ∗ ∗ ∗ 3. If, at the least j such that cij 6= dij, cij < dij, then ∀j ≤ y we define hij := dij.

Essentially, this procedure compares the ith columns as if they were elements

L of the lexicographically ordered direct sum ω Z, and takes the larger of the two for the ith column of the join. This follows from the Hahn order on the root system Γ, which in this case ends up being a “pointwise” order on the columns of

Γ, where each column is given a lexicographic ordering.

To see that the resulting h gives us a join for hx, y, c∗i and hx, y, d∗i, we consider each column. Fix some i, and consider the ith column. If c∗ = d∗ in 59 the ith column, then any “larger” column would be unsuitable for a least upper bound, and any “smaller” column wouldn’t work for an upper bound.

If c∗ > d∗ at the first disagreement in the ith column, then the ith column of c∗ − d∗ will be positive at the maximal element of its support. Since we use c∗ to define h in this case, we have h = c∗ and h > d∗ in the ith column. Any “larger” column would not contribute to a least upper bound, and any “smaller” column would result in an element that was no longer an upper bound of hx, y, c∗i.

The remaining case is similar. Thus, after reducing, one gets the join of a ∨ b in normal form with no extraneous rows or columns of zeros. This algorithm

0 gives rise to a Σ0 definition for the join function, so RCA0 proves that G exists and is an `-group.

Now, on to the reversal. Let f be a one-to-one function. Let M = {hm, n, ci ∈

G : ∀i ≤ m∀j ≤ n(cij 6= 0 → ∃k ≤ j(f(k) = i))}.

Claim. M is a convex `-subgroup of G.

Proof. The question of inverses is easy, since the support of the coefficient function is identical for an element and its inverse. The support of the coefficient function for a group sum is contained in the union of the supports of its summands’ co- efficient functions, all of which must satisfy the condition in the definition of M.

Thus, we have that M is at least a subgroup. Suppose that x, y ∈ M. Since the columns of x ∨ y are columns from either x or y, all of which satisfy the condition of M, the coefficients of x ∨ y must satisfy the conditions of M. So M is closed 60 under meet (and join), and is an `-subgroup.

To verify convexity, suppose that |x| ≤ |m| and m ∈ M. Suppose, for a contradiction, that x 6∈ M. Then ∃i, j(|x|ij 6= 0 & ∀k ≤ j(f(k) 6= i)). Without loss of generality, take j to be the least so that this is true for the given i, that is,

∀k < j(|x|ik = 0). Since |x| is positive, |x|ij, as the maximal element of the support

th of the i column, must be positive. Then, since |m| ≥ |x|, either ∀s(|m|is = |x|is),

(which is impossible, since it implies that m 6∈ M) or, at the first place they differ

th in the i column, |m|is > |x|is. This would imply that either |m|ij > |x|ij > 0, or for some k < j(|m|ik > 0). However, ∀k ≤ j(f(k) 6= i) → ∀k ≤ j(|m|ik = 0), so neither case is possible, and we have our contradiction.

Then the convex `-subgroup M is definable by a bounded-quantifier formula, so exists in RCA0.

Let αuv denote the element hu, v, ci, where   1 if i = u & j = v cij =  0 otherwise.

⊥ Suppose f(j) = i. Then αij ∈ M, so |αi0|∧|αij| = |αij|= 6 e and αi0 6∈ M . On the

⊥ other hand, suppose αi0 ∈ M . Then, for all m ∈ M, |αi0|∧|m| = e. In particular, no αij is in M, since |αi0| ∧ |αij| = |αij|. But [(∀j)αij 6∈ M] → i 6∈ ran(f). Thus

⊥ we have i ∈ ran(f) ⇐⇒ αi0 6∈ M .

The reversals of this chapter can be summarized as follows: 61

Theorem 4.17. (RCA0) Let G be an `-group. The following are equivalent:

1. ACA0

2. The existence of the subgroup generated by two subgroups A, B.

3. The existence of the convex `-subgroup generated by two convex `-subgroups

A, B.

4. The existence of the convex `-subgroup generated by a (nonconvex) `-subgroup

A.

⊥ 5. The existence of the polar subgroup A = {g ∈ G :(∀a ∈ A)|g| ∧ |a| = 1G} Chapter 5

Prime Subgroups and Values

5.1 Prime Subgroups

Definition 5.1. The convex `-subgroup P is prime if

∀x∀y[x ∧ y ∈ P → x ∈ P or y ∈ P ].

The following lemma from [4] will be useful.

Lemma 5.2. (RCA0) Let g be an element of the `-group G and P be a convex

`-subgroup of G. Then the set C = {x ∈ G : |g| ∧ |x| ∈ P } is a convex `-subgroup of G.

62 63

Proof. Let x, y ∈ C. Then e ≤ |x| ∧ |g| = h1 ∈ P and e ≤ |y| ∧ |g| = h2 ∈ P .

|xy−1| ∧ |g| ≤ |x||y||x| ∧ |g| (by Thm. 2.20)

= |x||y||x| ∧ |x||y||g| ∧ |g|( since |x||y||g| ≥ |g|)

= |x||y|(|x| ∧ |g|) ∧ |g|

= |x||y|h1 ∧ |g|

≤ |x||y|h1 ∧ |g|h1

= (|x||y| ∧ |g|)h1

= (|x||y| ∧ |x||g| ∧ |g|)h1( since |x||g| ≥ |g|})

= (|x|(|y| ∧ |g|) ∧ |g|)h1

= (|x|h2 ∧ |g|)h1

≤ (|x|h2 ∧ |g|h2)h1

= (|x| ∧ |g|)h2h1 = h1h2h1 ∈ P.

−1 −1 −1 Thus, we have e ≤ |xy | ∧ |g| ≤ h1h2h1, so |xy | ∧ |g| ∈ P and xy ∈ C.

C is therefore a subgroup of G. According to Theorem 2.22, the subgroup C is an `-subgroup iff ∀x ∈ C(x ∨ e ∈ C). By Theorem 2.20 |x ∨ e| ≤ |x|. Thus e ≤ |x ∨e| ∧ |g| ≤ |x| ∧ |g| ∈ P , so x ∨e ∈ C and C is an `-subgroup. To show C is convex, let e ≤ x ≤ h where h ∈ C. Thus |x| ≤ |h| so e ≤ |x| ∧ |g| ≤ |h| ∧ |g| ∈ P , so |x| ∧ |g| ∈ P and x ∈ C. 64

Theorem 5.3. (RCA0) For a convex subgroup P of the `-group G, the following

are equivalent:

1. P is prime.

2. x ∧ y = 0 → x ∈ P or y ∈ P .

3. G/P is totally ordered under the induced order.

4. If A, B are convex `-subgroups containing P then A ⊆ B or B ⊆ A.

Note that RCA0 is sufficient to form the quotient group from (3). The proofs of 1 → 2, 2 → 3, 3 → 4 are in [1], whereas 4 → 1 is from [4].

Proof.

1 → 2: Trivial.

2 → 3: Given g, h ∈ G, we have

((g ∨ h)g−1) ∧ ((g ∨ h)h−1) = (g ∨ h)(g−1 ∧ h−1) = (g ∨ h)(g ∨ h)−1 = e.

WLOG, by 2, (g ∨ h)h−1 ∈ P . Then P h = P (g ∨ h) ≥ P g.

3 → 4: Suppose A, B are convex `-subgroups containing P s.t. A 6⊂ B and B 6⊂ A.

By taking absolute values if necessary, choose positive a ∈ A\B and positive 65

b ∈ B\A. Then, WLOG, P a ≥ P b, so there exists p ∈ P s.t. pa ≥ b. Since

P ⊂ A, pa ∈ A and by convexity we have b ∈ A, a contradiction.

4 → 1: We prove ¬1 → ¬4. Suppose P is not prime, so there exists a, b not in

P with a ∧ b ∈ P . Form the Σ0 sets A = {x ∈ G : |x| ∧ |b| ∈ P },B = {x ∈ G :

|x| ∧ |a| ∈ P }. Let x = a. By Lemma 2.23, |a| ∧ |b| ≤ |a ∧ b|. Since a ∧ b ∈ P ,

|a ∧ b| ∈ P , and |a| ∧ |b| ∈ P by convexity, and we have x ∈ A. On the other

hand, if x = a then x 6∈ B. Thus we have A 6⊆ B and the proof of B 6⊆ A is

similar. By the definitions of A and B and the convexity of P , it follows that

P ⊆ A and P ⊆ B. By Lemma 5.2, A and B are convex `-subgroups. Since

A and B contain P and are incomparable under inclusion, we have shown the

contrapositive, ¬1 → ¬4.

5.2 Values

Definition 5.4. Let g be a nonidentity element of an `-group G. A value of g,

denoted V (g), is a convex `-subgroup maximal w.r.t not containing g.

In general, values are not unique. The notation V (g) serves mainly to denote

that V is maximal with respect to excluding g.

The most natural way to describe the maximality of a value would be to

say V is a value of g if it is a convex subgroup not containing g such that for

1 every convex `-subgroup W ) V, g ∈ W . Unfortunately, this statement is Π1(V ). 66

We can describe the maximality, however, in terms of convex closure, which is arithmetical.

Definition 5.5. Let g be a nonidentity element of the `-group G. A value of g is a set V such that

• V is a convex `-subgroup (which is an arithmetical condition).

• g 6∈ V .

• ∀x 6∈ V ∃n(g ∈ CL(V ∪ {x})n), that is, the convex `-subgroup generated by

V ∪ {x} contains g.

Thus, whether or not a set V is a value of g is decidable in ACA0.

Theorem 5.6. (RCA0) Values are prime.

Proof. The proof is adapted from [4] : Let V be a value of g. Suppose there are elements a, b > e such that a ∧ b = e and a 6∈ V . Form the set C = {x ∈ G :

|x| ∧ |g| ∈ V }. By Lemma 5.2, C is a convex `-subgroup. Since |x| ∧ |g| ≤ |x| and

V is convex, every element of V belongs to C.

Suppose, for a contradiction, that there exists an element y ∈ C\V . By the definition of V as a value, g ∈ CL(V ∪ {y}). By Lemma 4.2, CL(V ∪ {y}) ⊂

CL(C) = C. Then g ∈ C so, by definition of C, |g|∧|g| = |g| ∈ V , a contradiction.

So, V = C.

Now form the set A = {x ∈ G : |x| ∧ b ∈ V }. By Lemma 5.2, A is a convex

`-subgroup. Since |x| ∧ b ≤ |x| and V is convex, every element of V belongs to 67

A. Furthermore, we have assumed a 6∈ V , but a ∈ A. So A strictly contains

th V , and since V is a value, there is an n s.t. g ∈ CL(V ∪ {a})n, the n -stage approximation to CL(V ∪ {a}). As above, by properties of the convex closure, A must contain g, so |g| ∧ b ∈ V . This implies that b ∈ C, and since C = V , we have b ∈ V . To recap, we have shown that (a ∨ b = e & a 6∈ V ) → b ∈ V , which implies V is prime.

5.3 Existence of a Sequence of Values

As mentioned above, values are not generally unique. However, in an o-group, the convex subgroups form a chain, so there is only one value for each element.

Furthermore, in an o-group, any two elements are comparable.

Definition 5.7. We say x  y if ∀n(|x|n < |y|), and say x ≈ y ↔ ¬(x  y or y  x). Equivalently, x ≈ y ↔ ∃n([|x|n ≥ |y|&|y|n ≥ |x|]).

Lemma 5.8. (RCA0) If a, b are elements of an o-group, then exactly one of the three relations a  b, a  b, a ≈ b must hold.

We will need the following:

Lemma 5.9. (RCA0) Let A be a subset of an o-group G such that ∀x ∈ A(x  y).

Then ∀x ∈ G∀n ∈ N(x ∈ CL(A)n → x  y).

Proof. In the context of o-groups, join is trivial to compute since a∨b = max≤G (a, b).

The result follows by an induction on the definition of CL(A). Since join is trivial, 68 the only interesting case is the group operation, which is also resolved easily.

Lemma 5.10. (RCA0) Let G be an o-group. Suppose V is a value of g. Then x ∈ V iff x  g.

Proof. Let x ∈ V . Suppose ¬(x  g). Then x ≈ g or x  g. Then, since V is a convex `-subgroup, x generates g ∈ V , a contradiction. For the other direction, suppose x  g and x 6∈ V . By Lemma 5.9, ∀h∀n(h ∈ CL(V ∪{x})n → h  g). In particular, ∀n(g 6∈ CL(V ∪{x})). Along with the fact that x 6∈ V , this violates the third condition of Definition 5.5, so V is not a value and we have our contradiction.

Corollary 5.11. (RCA0) Let G be an o-group. Let x, y be distinct elements of

G, and V (x),V (y) be values of x, y, respectively. Then x ≈ y ↔ x 6∈ V (y)& y 6∈

V (x).

Now, suppose one had a set K coding a sequence of values Vi for each nonidentity gi ∈ G. By the corollary above, K can compute the set of pairs R =

{hi, ji : gi ≈ gj} and thus also compute a set of archimedean class representatives

C = {i :(∀s

Theorem 5.12. (RCA0) The following are equivalent.

1. For every o-group G the sequence of values Vi = V (gi) exists. 69

2. ACA0

Corollary 5.13. (RCA0) The existence of a sequence of values for an `-group implies ACA0.

Together with this corollary, the following theorem constitutes an equiva- lence of the existence of a sequence of values to ACA0.

Theorem 5.14. (RCA0) ACA0 implies the existence of a sequence of values for an `-group.

Proof. Let G be enumerated as g0, g1 .... Let A be (the code for) a finite subset of

G. Using the mechanism of convex closure defined earlier, we define a bounded- quantifier relation Ψ(A, i, s) such that gi ∈ CL(A) ↔ ∃sΨ(A, i, s). Informally, s

represents a stage of the construction of CL(A) such that gi ∈ CL(A)s. In ACA0, we may form the set S of pairs hA, ii such that A is a (code for a) finite set, and

∀s¬Ψ(A, i, s). We then define a function f(j, n): N2 → {0, 1}. We define by primitive recursion on n:   1 if h{g0}, ji ∈ S Case n=0: f(j, 0) =  0 if h{g0}, ji 6∈ S.   1 if h({gk : k ≤ n & f(j, k) = 1} ∪ {gn+1}, ji ∈ S Case n+1: f(j, n + 1) =  0 if h({gk : k ≤ n & f(j, k) = 1} ∪ {gn+1}, ji 6∈ S.

Claim. For each j, the set W = {gk : f(j, k) = 1} is a value of gj. 70

Proof. Each successive element gk is included or excluded by f based solely on whether its inclusion would eventually cause gj to be generated. By Lemma 4.3, the convex closure of a set of elements is unaffected by adding group inverses, compositions, joins, and positive elements bounded above by a member of that

−1 set. For example, if f(j, 5) = 1, and g6 = g5 , then f(j, 6) = 1 because including inverses will not cause any new elements (particularly gj) to enter the convex closure. The other criteria for a convex `-subgroup are satisfied similarly. Thus

W is a convex `-subgroup. By the definition of f, gj 6∈ W . Also by the definition of f, any element not in W is excluded specifically because including it would generate gj.

5.4 Existence of a Sequence of Excluding Primes

Definition 5.15. An excluding prime for g, denoted P (g), is a prime convex

`-subgroup not containing g.

Since values are prime, a value V (g) is a maximal excluding prime for g.

However, it is easy to show that there are, in general, non-maximal excluding primes. Take the o-group obtained by lexicographically ordering Z ⊕ Z ⊕ Z. The element (1, 0, 0) is excluded by the convex `-subgroups generated by (0, 1, 0) and

(0, 0, 1), both of which are prime, as are all convex subgroups of an o-group.

However, the latter is a proper subgroup of the former. 71

Lemma 5.16. (RCA0). Let G be an abelian `-group, with G = {g0 = e, g1,...}.

Then there is a uniform sequence of infinite trees hT1,T2,...i such that ∀m ≥

1, f ∈ [Tm] ↔ f is the characteristic function of a prime subgroup not containing gm.

<ω Proof. For each σ ∈ 2 we write gi ∈ σ ↔ σ(i) = 1 and abuse this notation by

<ω writing things like |gi| ∈ σ to mean σ(k) = 1, where gk = |gi|. For each σ ∈ 2 , define

Sσ = {k < |σ| : gk ∈ σ or

∃i < |σ|(gi ∈ σ &(gi · gk = e or |gk| ≤ |gi|)) or

∃i, j < |σ|(gi, gj ∈ σ &(gk = gi · gj or gk = gi ∧ gj))}.

Note that τ ⊆ σ → Sτ ⊆ Sσ. We define the relations R,Q :

R(σ) := ∀i, j[(i < j < |σ| & gi ∧ gj = e) → (gi ∈ σ or gj ∈ σ)].

Q(σ) := ∀k < |σ|(gk ∈ Sσ → k ∈ σ).

Note that τ ⊆ σ → (R(σ) → R(τ)& Q(σ) → Q(τ)).We say σ is m-acceptable if and only if R(σ)& Q(σ)& |gm| 6∈ σ. Then τ ⊆ σ & σ is m-acceptable →

<ω τ is m-acceptable. For m ≥ 1, we define Tm = {σ ∈ 2 : σ is m-acceptable}.

Claim. If f ∈ [Tm] then f is the characteristic function of a prime subgroup not containing gm. 72

Let f ∈ [Tm]. Then f(|gm|) = 0, since f  n is m-acceptable for all n. Since

Q(f  n) for each n, f codes a convex `-subgroup and since R(f  n) for each n,

this convex `-subgroup is prime.

If I can prove in RCA0 that Tm is infinite, WKL0 will prove that there is a

path, and hence a prime subgroup not containing gm.

Lemma 5.17. (RCA0) Every tree Tx, x ≥ 1, is infinite. Formally, ∀n(Tx has a

node of length n).

Proof. Suppose not. Fix x ≥ 1 and let n be N-least such that Tx has no node of

length n. If n is less than the index of |gx|, then we are done, since the string

n 1 satisfies Q, R and is x-acceptable, hence belongs to Tx. So assume n ≥ the

index of |gx|. Let Y = {hi, ji : i < j < n & gi ∧ gj = e}. We will construct a

node σ of length n that is x-acceptable, so belongs to Tx. To satisfy R, σ must contain at least one element from every pair in Y . Intuitively, it is easy to find a convex `-subgroup excluding a nonidentity element – one may choose the identity subgroup or consider a principal polar. The difficulty arises when we try to also require it to be prime. Thus, our first priority is to choose one element from each pair in Y . The next priority is to make sure that Q is satisfied.

Claim. (RCA0) If Y contains m pairs, there is a way to choose one element from each pair so that our choices do not generate |gx|, in the sense of satisfying Q.

Suppose the m pairs of Y are written as hai,0, ai,1i, with 0 ≤ i ≤ m − 1. A 73

binary string of length m can then represent a choice of one element from each

m 0 pair. Given τ ∈ 2 , we form στ with |στ | = n by bounded Σ1 comprehension:

define

Sστ = {k : k < n & gk ∈ CL({ai,τ(i) : i < m)}}

and then let στ (k) = 1 ⇐⇒ k ∈ Sστ .

m n To prove the claim, we need to show that there is a τ ∈ 2 such that στ ∈ 2

Q|τ|−1 is x-acceptable. For τ with |τ| ≤ m, let gτ = i=0 ai,τ(i). Then the meet of all ^ possible products with one factor from the first k pairs in Y is gτ . First we τ∈2k ^ prove by induction that, for k ≤ m, gτ = e. Case k = 1. We have one pair τ∈2k a0,0, a0,1 such that a0,0 ∧ a0,1 = e. In this case, the only possibilies for gτ are a0,0 and a0,1, which clearly meet to the identity.

Suppose we have proved the result for k < m, and consider the case k + 1.

By the distributive law,

^ ^ ^ [( gτ ) · ak,0] = (gτ · ak,0) = ( gτ ). τ∈2k τ∈2k τ∈2k+1:τ(k)=0

Similarly, ^ ^ [( gτ ) · ak,1] = ( gτ ). τ∈2k τ∈2k+1:τ(k)=1 Thus, ^ ^ ^ gτ = [( gτ ) · ak,0] ∧ [( gτ ) · ak,1]. τ∈2k+1 τ∈2k τ∈2k ^ By distribution again, this equals ( gτ ) · (ak,0 ∧ ak,1). The first factor is equal τ∈2k to e by induction, and the second is equal to e by definition of Y . This proves 74

^ ^ that gτ = e. By induction, gτ = e. τ∈2k+1 τ∈2m p m−1 p Since G is abelian, it follows that (gτ ) = Πi=0 (ai,τ(i)) . By repeated appli-

^ p cation of Corollary 2.10, we have (gτ ) = e. τ∈2m Suppose, for a contradiction, that every way of choosing one element from

m each pair in Y forced the generation of |gx|, that is, for all τ ∈ 2 , |gx| ∈ Sστ .

m p By Lemmas 4.4 and 4.5, there is a p ≥ 0 such that ∀τ ∈ 2 (|gx| ≤ (gτ ) ). Thus

V p |gx| ≤ τ∈2m (gτ ) = e, a contradiction since gx 6= e. Therefore, there is at least one τ which does not generate |gx|. This completes the proof of the claim. To

m finish the proof of the lemma, fix any τ ∈ 2 such that |gx| 6∈ Sστ . Clearly στ is x-acceptable, and |στ | = n, a contradiction. Therefore, each tree is infinite.

So, we have a uniform sequence of infinite trees such that a path through

Tm codes a prime convex `-subgroup not containing gm. What we really want, however, is a set uniformly coding a sequence of excluding prime subgroups. We

∗ accomplish this by coding all the trees Tm into a single tree T such that a path

∗ through T uniformly computes a path through each Tm.

Theorem 5.18. (WKL0) Let e = g0, g1,... be an enumeration of the `-group

G. Then there is a set K such that ∀x ≥ 1, Kx is a prime subgroup of G not containing gx.

Lemma 5.19. (RCA0) If hT1,...i is a sequence of infinite trees, then there is an

∗ 0 ∗ infinite tree T and a Σ0 function φ(f, i) s.t.∀f ∈ [T ](φ(f, i) is a path in Ti). 75

We define, by induction, the tree T ∗ and a labeling function l : T ∗ → N×2<ω.

The root node λ of T ∗ is labeled l(λ) = h1, λi. Induction step: If σ ∈ T ∗ and

∗ l(σ) = hi, τi, then σ ∗ j ∈ T ↔ τ ∗ j ∈ Ti. Furthermore, we define

  hn + 1, λi if i = 1 l(σ ∗ j) =  hi − 1, τˆ ∗ ki if i > 1

where n = max{j :(∃σˆ ( σ)π1(l(ˆσ)) = j} andτ, ˆ k have the property that ifσ ˆ is

the longest proper substring of σ such that π1(l(ˆσ)) = i − 1, then l(ˆσ) = hi − 1, τˆi,

andσ ˆ ∗ k ⊆ σ. Here, π1 denotes the projection function on the first component.

(See Figure 5.1.) To prove that T ∗ is infinite, we describe a method of producing

nodes longer than a specified height. Let m > 1 be fixed. Since each Ti is infinite,

there exist σi ∈ Ti for i = 1, . . . , m such that |σi| = m. If we run through the

∗ construction substituting σi for Ti, we will be able to produce a node in T of

length greater than m.

Given {σi : 1 ≤ i ≤ m}, let f be the function which on input hσ, l(σ)i

returns the pair hσ ∗ j, l(σ ∗ j)i as defined in the construction above, replacing Ti

with σi, or more precisely, the tree consisting of all strings contained in σi.

Suppose f k(h1, λi) = hµ, hi, τii. Then f(hµ, hi, τii) has two components.

The first component is found by looking in σi for the string τ. We extend µ

by the next bit of σi after τ, resulting in µ ∗ σi(|τ| + 1) The second component 76

has two cases. If i = 1, then the second component is hn + 1, λi, where n =

max{j :(∃σˆ ( µ)π1(l(ˆσ)) = j}. If i > 1, then we look for the longest proper substringσ ˆ ( µ whose label has first component i − 1. Because we are working with the non-branching tree generated from σi instead of the whole of Ti, we know that l(ˆσ) = hi − 1, σi−1  ji for some j. In this case, the second component is hi − 1, σi−1  j + 1i.

This construction “visits” T1, then T2,T1, then T3,T2,T1, etc. So, just given

(m−1)(m) σi ∈ Ti of length m we will be able to define at least 2 iterates of f. Since

each iteration produces a longer node of T ∗, this is enough to produce a node

longer than m bits.

∗ Claim. If f is a path in T , then f computes a path in Ti for each i.

Proof. By primitive recursion, we can define a function g(i, j) such that g(i, j) is

∗ th ∗ the level in T coding the j level of Ti. For any f ∈ [T ], l(f  g(i, j)) has the

th form hi, τi, where |τ| = j. Thus, f(g(i, j)) is the j bit of the path in Ti coded

by f, and the function hi(j) := f(g(i, j)) gives a path through Ti.

∗ Now, RCA0 proves that T exists and that paths through Ti correspond to

∗ prime subgroups of G not containing gi. WKL0 proves that T has a path, and we

have seen that such a path can uniformly compute paths through each Ti. Thus,

we have proved Theorem 5.18. 77

Initial Subtree of T ∗

4, λ " #

010110

1, 00 1, 01 " " ## " " ##

01011 01111

2, 1 2, 1 " " ## " " ##

0101 T1 0111

3, λ 3, λ " # " #

T2 010 011

1, 0 1, 1 " " ## " " ##

01 11

2, λ T 2, λ " # 1 " #

0 1

λ, 1, λ " #

Information Used

T1 T2 T3 T4

Fig. 5.1: Initial Subtree of T ∗ Chapter 6

Holland’s Embedding Theorem

The central object of study in this chapter is the following theorem:

Holland’s Embedding Theorem 2. Let G be a lattice ordered group. Then there is an embedding of G into the group of order-preserving permutations of some linear order.

At face value, this theorem purports both the existence of a linear order and the group of order-preserving permutations of that linear order. For this to be non-trivial, the linear order must of course be infinite. Moreover, the per- mutation group is then a third-order object which we cannot define directly in

Z2. We carefully sidestep this obstacle by restating the theorem. Specifically, using the purported linear order L, we show that there is a uniform way of obtaining, for each g ∈ G, a specific function fg : L → L which is in fact

an order-preserving permutation, and verify all the desired properties of the

embedding without explicitly forming the range of the embedding as a set.

78 79

Holland’s Embedding Theorem 3. If G is an `-group, then there exists a

linear order L and a function f(g, l): G × L → L such that for each g ∈ G, the

function f(g, l), abbreviated fg, is an order preserving bijection on L, and satisfies

the properties

• ∀l ∈ L(fe(l) = l).

• [∀l ∈ L(fg(l) = fh(l))] ↔ g = h.

• ∀g, h ∈ G, l ∈ L(fg∧h(l) = min{fg(l), fh(l)}).

• ∀g, h ∈ G, l ∈ L(fg∨h(l) = max{fg(l), fh(l)}).

• ∀g, h ∈ G, l ∈ L(fgh(l) = fh(fg(l))}.

The first two properties above indicate, respectively, that the group identity induces the identity permutation, and that each element induces a distinct permu- tation. The last three ensure that the embedding respects the `-group structure of G – that is, that if one forms the meet, join, or group product of two elements, then one gets the same permutation as though one had taken the meet, join, or composition of the permutations induced by those elements individually.

6.1 Summary of Original Proof of Holland’s Theorem

1. Fix a sequence of values {Vg}, one for each g 6= e. 80

2. Since values are prime, the induced order on each collection of right cosets

G/Vg is linear.

3. The collections of cosets G/Vg may be ordered lexicographically to form a

linear order L.

4. Each element of G naturally induces an order-preserving permutation of L

by right multiplication.

5. The `-group structure of G naturally embeds into the `-group structure of

order-preserving permutations of L.

6. Proof is relatively constructive and can be done in RCA0 except for obtain-

ing a sequence of values Vg which, as we saw in Chapter 5, is equivalent to

ACA0.

There are two qualities of values that are used in the proof, namely, that they are prime and that each element is excluded from some value – that is, there is no use of values’ maximality except for proving that values are prime. This observation motivated the definition of excluding primes in Chapter 5 and led to a couple of questions, namely: “Is the existence a sequence of excluding primes equivalent to something weaker than ACA0?” and “Can Holland’s Theorem be proved using excluding primes instead of values?”

The answer to the first question is “Yes, but so far only for abelian `-groups”, as was shown in Chapter 5. We now address the second question. 81

6.2 Proof of Holland’s Theorem using excluding primes

Let G be an `-group. For g 6= e, let P (g) denote a prime subgroup not containing g. Suppose we are given a set K = {hx, yi : y ∈ P (x)} coding a sequence of excluding primes. (If G is abelian, then by Theorem 5.18, K can be formed in

WKL0.) By Theorem 3.7, RCA0 is sufficient to establish the induced orders on the sets of right cosets G/P (g) for each nonidentity g. Thus, given K we can form

K¯ = {hx, yi : y is the N-least element of a right coset in G/P (x)}.

Since P (g) is prime for each g, each column of K¯ (representing cosets

G/P (g)) is linearly ordered under the induced order

¯ Thus, we may totally order K: ha, bi < hx, yi iff a

Holland’s Theorem.

Lemma 6.1. (RCA0) If P is a prime subgroup of an `-group G, then an element g ∈ G induces an order-preserving permutation fg of G/P , defined by fg(P x) =

P xg.

Proof. Since P is prime, G/P is totally ordered under the induced order. Let the

−1 induced order be represented by . Since fg(P yg ) = P y, it is clear that the map is surjective. Suppose fg(P x) = fg(P y). Then P xg = P yg. It follows that

∃p ∈ P (x = py), so P x = P y, and we have shown fg is injective. Now, we show it is order-preserving. Suppose P x  P y. By the definition of the induced order, 82

this means ∃p ∈ P (x ≤ py). Then xg ≤ pyg, so P xg  P yg.

Lemma 6.2. (RCA0) Given an `-group G and sequence of excluding primes K,

there exists a function f(g, hx, yi): G × K¯ → K¯ so that for all g ∈ G, f(g, k) is

an order-preserving permutation of K¯ .

Proof. We define f(g, hx, yi) using right multiplication: f(g, hx, yi) = hx, zi,

¯ ∼ where z is the least coset representative of yg in Kx = G/P (x). By Lemma

¯ 6.1, this mapping is a bijection of each column Kx which preserves each induced

¯ order

Lemma 6.3. (RCA0) The function f(g, k) respects the `-group structure of G,

and the only element which induces the identity permutation of K¯ is the identity

of G.

Proof. We need to establish the following:

1. ∀g, h ∈ G(fg ◦ fh = fhg).

2. ∀g ∈ G(fg = idK¯ ↔ g = e).

3. ∀g, h ∈ G(fg∧h = fg ∧ fh = min{fg, fh}).

4. ∀g, h ∈ G(fg∨h = fg ∨ fh = max{fg, fh}).

(1) Follows directly from the definition of f(g, k) by right multiplication.

¯ (2) It is clear that fe is the identity on K. On the other hand, suppose g 6= e. 83

¯ Then we have a column in K corresponding to G/P (g). Since g 6∈ P (g), fg per-

¯ ¯ mutes Kg nontrivially, so cannot be the identity on K.

(3) By Theorem 3.8, for any convex subgroup C, C(g ∧ h) = Cg ∧ Ch. In the case that C is prime, the induced order is linear so the meet is the mini-

¯ ∗ ∗ mum. Let k = hx, yi ∈ K. fg∧h(hx, yi) = hx, y(g ∧ h) i = hx, (yg ∧ yh) i =

∗ ∗ hx, minG/P (x){yg, yh} i = minK¯ {fg(k), fh(k)}. The asterisk reminds us that

technically these equalities use the least coset representative, e.g., yg∗ is the N- least coset representative of P (x)yg.

(4) Similar to (3).

Note: the last two criteria feature a consequence of the standard lattice order on the group of automorphisms of a linear order as mentioned in Example

1.8.

While we do not explicitly form the set of automorphisms of the linear

order K¯ , we produce a function f(g, k) which, practically speaking, provides an

`-embedding of G into Aut(K¯ ). We have:

Theorem 6.4. (RCA0 + Existence of a sequence of excluding primes) ` Holland’s

Theorem.

By Theorem 5.18, we then have:

Corollary 6.5. WKL0 ` Holland’s Embedding Theorem for abelian `-groups.

This Corollary is certainly a successful result – the standard assumption of 84

a sequence of values requires ACA0 even for abelian `-groups and, in that con-

text, the strictly weaker assumption of a sequence of excluding primes suffices for

the proof of Holland’s Theorem. Without having a reversal, however, this raises

another question: “Might Holland’s Theorem be proven by still weaker assump-

tions?”

6.3 A Reversal

Theorem 6.6. (RCA0) Let G be an `-group. The following are equivalent:

1. Holland’s Theorem

2. The existence of a sequence of prime subgroups P (g) s.t. ∀g 6= e(g 6∈ P (g)).

Proof. We have just seen that 2 → 1. We now show that 1 → 2. If G is an `-group

then by Holland’s Theorem there is a linear order L and a function f(g, l): L →

L satisfying all the properties mentioned in the revised statement of Holland’s

Embedding Theorem for Z2. Suppose we are given such G, L, f. Define Fl = {g ∈

G : f(g, l) = l}, the set of g which “fix” l. It is easy to check that for each l, Fl is a prime subgroup of G.

• e ∈ Fl because fe = idL.

−1 • x ∈ Fl ↔ x ∈ Fl.

• x, y ∈ Fl → xy ∈ Fl, since fxy(l) = fy(fx(l)) = fy(l) = l. 85

• x ∈ Fl → x∨e ∈ Fl, since fx∨e(l) = max{fx(l), fe(l)} = l (here we are using

criterion 2.22).

• Suppose e ≤ x ≤ g and g ∈ Fl. Then fe(l) ≤ fx(l) ≤ fg(l), so l ≤ fx(l) ≤ l

and x ∈ Fl. Hence Fl is a convex `-subgroup.

• Suppose a ∧ b = e. Then fa∧b(l) = fe(l) = l. But we also have fa∧b(l) =

min{fa(l), fb(l)}, so either a ∈ Fl or b ∈ Fl, and therefore Fl is a prime

subgroup.

Furthermore, for each g 6= e, fg is nontrivial, so there is an l so that g 6∈ Fl.

Let l be the N-least such that f(g, l) 6= l. Then g 6∈ Fl.

This is an effective way of finding l s.t. g 6∈ Fl. Since Fl has a quantifier-free definition, given L we can form in RCA0 a set K = {hx, yi} such that Kx codes a prime subgroup not containing gx for x ≥ 1. Bibliography

[1] Marlow Anderson and Todd Feil. Lattice-Ordered Groups: an introduction. D. Reidel Publishing Company, Dordrecht, Holland, 1988.

[2] R.G. Downey and Stuart A. Kurtz. “Recursion Theory and Ordered Groups”. Annals of Pure and Applied Logic, 32:137–151, 1986.

[3] K. Hatzikiriakou and S.G. Simpson. “WKL0 and Orderings of Countable Abelian Groups”. Contemporary Mathematics, 106:177–180.

[4] V.M. Kopytov and N.Ya. Medvedev. The Theory of Lattice-Ordered Groups, pages 1–55. Kluwer, Dordrecht, The Netherlands, 1994.

[5] Stephen G. Simpson. Subsystems of Second Order Arithmetic. Springer-Verlag, Berlin, Germany, 1999.

[6] D.R. Solomon. “Reverse Mathematics and Fully Ordered Groups”. Notre Dame Journal of Formal Logic, 39(2):157–189, 1998.

[7] D.R. Solomon and Rod Downey. “Reverse Mathematics, Archimedean Classes, and Hahn’s Theorem”. In Stephen Simpson, editor, Reverse Mathematics 2001, pages 147,163. AK Peters, 2005.

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