Primitive-recursive certificates for GP′′=1G_P^{\prime\prime}=1 and effective enumeration via Laurent relations

17.124

Problem

Can the ordinary finite presentations defining metabelian groups be recursively enumerated?

M={⟨x1,…,xn∣r1,…,rm⟩:G′′=1}∈?Σ10.\mathcal M=\{\langle x_1,\ldots,x_n\mid r_1,\ldots,r_m\rangle:G''=1\} \quad\stackrel{?}{\in}\quad\Sigma^0_1.

A positive answer requires finite checkable witnesses for G′′=1G''=1. It does not require a terminating decision procedure on presentations of nonmetabelian groups.

The enumeration theorem

Let Fn=F(x1,…,xn)F_n=F(x_1,\ldots,x_n) and let RR be a finite list of words in FnF_n. We consider ordinary presentations

P=⟨x1,…,xn∣R⟩,G(P)=Fn/⟨ ⁣⟨R⟩ ⁣⟩,M={P:G(P)′′=1}.P=\langle x_1,\ldots,x_n\mid R\rangle,\qquad G(P)=F_n/\langle\!\langle R\rangle\!\rangle,\qquad \mathcal M=\{P:G(P)''=1\}.

The empty generating set is allowed. Encodings with a letter outside {x1,…,xn}\{x_1,\ldots,x_n\} are rejected.

Theorem 1. There is a primitive-recursive predicate VV on natural-number encodings of presentations and finite witnesses such that

P∈M⟺∃c∈N: V(P,c)=1.(2) \tag{2} P\in\mathcal M \quad\Longleftrightarrow\quad \exists c\in\mathbb N:\ V(P,c)=1.

In particular, M\mathcal M is recursively enumerable.

This answers the enumeration question in Kourovka Problem 17.124 (5) and the enumeration and witness questions of Shpilrain (7, Problems 4 and 5). The ordinary-presentation recognition question also occurs in (2, Problem 42, arXiv version), with attribution to Miller (6).

The covering presentations come from Bieri and Strebel (3, §3.5, equations (3.1)–(3.4), Lemma 3.5 and §3.6). We make their finite data effectively recognizable and obtain ordinary finite presentations PRP_R satisfying

PR′′=1,G′′=1, G finitely presented ⟹ ∃PR↠G.(3) \tag{3} P_R''=1,\qquad G''=1,\ G\text{ finitely presented} \ \Longrightarrow\ \exists P_R\twoheadrightarrow G.

A finite witness for the epimorphism consists of identities in a free group, each a product of conjugates of defining relators. These two ingredients give (2).

The algorithms of Baumslag, Cannonito and Robinson (1, p. 630 and Theorems 2.1, 3.1) take finite presentations in the metabelian variety. Here the input is Fn/⟨ ⁣⟨R⟩ ⁣⟩F_n/\langle\!\langle R\rangle\!\rangle in the variety of all groups. The geometric and module-theoretic ingredients remain classical; the construction makes their finite data sufficient for ordinary-presentation enumeration.

For abelian-by-cyclic groups, recursive enumerability of ordinary finite presentations is already recorded in the proof of (4, Lemma 11.4). Theorem 1 treats arbitrary metabelian groups. Its certificate gives explicit finite Laurent data and normal-closure derivations witnessing metabelianity.

Conventions and classical inputs

We use right conjugation and right commutators:

aw=w−1aw,[a,b]=a−1b−1ab.a^w=w^{-1}aw,\qquad [a,b]=a^{-1}b^{-1}ab.

A group GG is metabelian when its derived subgroup G′G' is abelian, equivalently when [[a,b],[c,d]]=1[[a,b],[c,d]]=1 for all a,b,c,d∈Ga,b,c,d\in G. Given letters t1,…,tkt_1,\ldots,t_k, put

q(v)=t1v1⋯tkvk(v∈Zk).q(v)=t_1^{v_1}\cdots t_k^{v_k}\qquad(v\in\mathbb Z^k).

We retain the literal inverse

q(v)−1=tk−vk⋯t1−v1.q(v)^{-1}=t_k^{-v_k}\cdots t_1^{-v_1}.

Unsubscripted norms are Euclidean.

Lemma 2. Let k≥1k\ge1 and let A\mathcal A be a finite alphabet, with a designated letter aija_{ij} for each 1≤i<j≤k1\le i<j\le k. For ρ>0\rho>0, let HρH_\rho have generators t1,…,tkt_1,\ldots,t_k and A\mathcal A, and relators

[ti,tj]=aij,[a,bq(v)]=1(a,b∈A, v∈Zk, ∥v∥<ρ).[t_i,t_j]=a_{ij},\qquad [a,b^{q(v)}]=1 \quad(a,b\in\mathcal A,\ v\in\mathbb Z^k,\ \lVert v\rVert<\rho).

Suppose ρ>2k\rho>2k and u,v∈Zku,v\in\mathbb Z^k satisfy

∥u∥≤ρ2k,∥v∥<ρ+12k,∥u+v∥<ρ.(8) \tag{8} \lVert u\rVert\le \frac{\rho}{2k},\qquad \lVert v\rVert<\rho+\frac1{2k},\qquad \lVert u+v\rVert<\rho.

Then, for all a,b∈Aa,b\in\mathcal A, the following hold in HρH_\rho:

[a,bq(u)q(v)]=1,[a,bq(u)−1q(v)−1]=1.[a,b^{q(u)q(v)}]=1,\qquad [a,b^{q(u)^{-1}q(v)^{-1}}]=1.

If all ordered commutation relators are imposed, with no radius restriction, the normal closure of A\mathcal A is abelian.

These are the ordered and inverse-ordered cases of (3, Lemma 3.4b); the normal-closure assertion is (3, Corollary 3.3). The collection argument (3, Lemma 3.2) is valid in every quotient of HρH_\rho.

For the second input, let AA be an abelian normal subgroup of a group EE with E/A≅ZkE/A\cong\mathbb Z^k. Conjugation gives AA a right module structure over Z[t1±1,…,tk±1]\mathbb Z[t_1^{\pm 1},\ldots,t_k^{\pm 1}], independent of the chosen lifts of the quotient generators. Let A∗A^* denote the inverse-action module, in which tit_i acts as ti−1t_i^{-1} acts on AA.

Lemma 3. Let k≥1k\ge1. Suppose EE is finitely presented and metabelian, E/A≅ZkE/A\cong\mathbb Z^k, and AA is finitely generated as a Z[t1±1,…,tk±1]\mathbb Z[t_1^{\pm 1},\ldots,t_k^{\pm 1}]-module. There is a finite family of nonzero Laurent polynomials

λ=∑u∈Sλcλ,utu(cλ,u∈Z∖{0}),\lambda=\sum_{u\in S_\lambda} c_{\lambda,u}t^u \quad(c_{\lambda,u}\in\mathbb Z\setminus\{0\}),

indexed by a signed family Λ=Λ+⊔Λ−\Lambda=\Lambda_+\sqcup\Lambda_-, such that

A(λ−1)=0(λ∈Λ+),A∗(λ−1)=0(λ∈Λ−),∀v∈Rk∖{0} ∃λ∈Λ:min⁡u∈Sλv⋅u>0.\begin{gather*} A(\lambda-1)=0\quad(\lambda\in\Lambda_+),\qquad A^*(\lambda-1)=0\quad(\lambda\in\Lambda_-),\\[3pt] \forall v\in\mathbb R^k\setminus\{0\}\ \exists\lambda\in\Lambda: \min_{u\in S_\lambda}v\cdot u>0. \end{gather*}

Apply the necessity direction of (3, Theorem A) and the centralizer criterion (3, Lemma 2.6) to AA and A∗A^*. For A=0A=0, use the monomials ti±1t_i^{\pm1}; for A≠0A\ne0, an identity-acting polynomial is nonzero. The minus sign changes the module action; the displayed support remains SλS_\lambda.

Rational separation and the commutation radius

Fix k≥1k\ge1 and a finite family (Sλ)λ∈Λ(S_\lambda)_{\lambda\in\Lambda} of nonempty subsets of Zk\mathbb Z^k. Consider

∀v≠0 ∃λ ∀u∈Sλ:v⋅u>0.(12) \tag{12} \forall v\ne0\ \exists\lambda\ \forall u\in S_\lambda: v\cdot u>0.

Only rational arithmetic is needed to test this condition.

Lemma 4. Condition (12) is decidable. If it holds, there is an integer r≥0r\ge0 such that, for ε=2−r\varepsilon=2^{-r},

∀v∈Rk, ∥v∥∞=1 ⟹ ∃λ ∀u∈Sλ: v⋅u>ε.(13) \tag{13} \forall v\in\mathbb R^k,\ \lVert v\rVert_\infty=1 \ \Longrightarrow\ \exists\lambda\ \forall u\in S_\lambda:\ v\cdot u>\varepsilon.

Whether a proposed rr satisfies (13) is decidable by a primitive-recursive computation on the finite integer data.

Proof. Put Σ={v∈Rk:∥v∥∞=1}\Sigma=\{v\in\mathbb R^k:\lVert v\rVert_\infty=1\}. Its faces are

Fj,s={v∈[−1,1]k:vj=s},Σ=⋃j=1k ⋃s∈{−1,1}Fj,s.F_{j,s}=\{v\in[-1,1]^k:v_j=s\},\qquad \Sigma=\bigcup_{j=1}^{k}\ \bigcup_{s\in\{-1,1\}}F_{j,s}.

Failure of (13) on Fj,sF_{j,s} is equivalent to the existence of (uλ)λ∈∏λSλ(u_\lambda)_\lambda\in\prod_\lambda S_\lambda and vv satisfying

v⋅uλ≤ε(λ∈Λ),−1≤vi≤1,vj=s.(15) \tag{15} v\cdot u_\lambda\le\varepsilon\quad(\lambda\in\Lambda), \qquad -1\le v_i\le1,\qquad v_j=s.

Fourier–Motzkin elimination is the following rational equivalence. For αix+βi⋅y≤γi\alpha_i x+\beta_i\cdot y\le\gamma_i, put I±={i:±αi>0}I_\pm=\{i:\pm\alpha_i>0\} and I0={i:αi=0}I_0=\{i:\alpha_i=0\}. Then

∃x ∀i: αix+βi⋅y≤γi⟺{βi⋅y≤γi(i∈I0),γn−βn⋅yαn≤γp−βp⋅yαp(n∈I−, p∈I+).\exists x\ \forall i:\ \alpha_i x+\beta_i\cdot y\le\gamma_i \quad\Longleftrightarrow\quad \begin{cases} \beta_i\cdot y\le\gamma_i &(i\in I_0),\\[4pt] \dfrac{\gamma_n-\beta_n\cdot y}{\alpha_n} \le\dfrac{\gamma_p-\beta_p\cdot y}{\alpha_p} &(n\in I_-,\ p\in I_+). \end{cases}

After kk eliminations, only rational inequalities remain. At each stage, rational values of the remaining coordinates admit a rational choice of the eliminated coordinate. Thus rational and real feasibility agree. Applying the same test with ε=0\varepsilon=0 decides (12).

Under (12), Λ≠∅\Lambda\ne\varnothing and compactness gives

0<μ:=min⁡v∈Σmax⁡λmin⁡u∈Sλv⋅u,2−r<μ ⟹ (13).0<\mu:=\min_{v\in\Sigma}\max_\lambda \min_{u\in S_\lambda}v\cdot u, \qquad 2^{-r}<\mu\ \Longrightarrow\ \text{(13)}.

Hence some r∈Nr\in\mathbb N succeeds. For a supplied rr, the kk elimination steps, finite choices of supports and exact rational comparisons are primitive recursive. ◻

For data passing this margin test put

C=2−rk,D=1+∑λ∑u∈Sλ∥u∥1,R=1+2k(1+D+D2k 2r).(18) \tag{18} C=\frac{2^{-r}}{k},\qquad D=1+\sum_\lambda\sum_{u\in S_\lambda}\lVert u\rVert_1,\qquad R=1+2k\bigl(1+D+D^2k\,2^r\bigr).

Thus DD is an integer, 0<C≤10<C\le1, and R=1+2k(1+D+D2/C)R=1+2k(1+D+D^2/C). In particular,

R>2k,D<R2k,R>max⁡{C,D2/C}.(19) \tag{19} R>2k,\qquad D<\frac{R}{2k},\qquad R>\max\{C,D^2/C\}.

By homogeneity and ∥v∥∞≥1/k\lVert v\rVert_\infty\ge1/k for ∥v∥=1\lVert v\rVert=1, (13) implies that for each Euclidean unit vector vv some support satisfies v⋅u>Cv\cdot u>C throughout.

Lemma 5. Put δ=min⁡{C/4,1/(4k)}\delta=\min\{C/4,1/(4k)\}. If ρ≥R\rho\ge R and ρ≤∥v∥<ρ+δ\rho\le\lVert v\rVert<\rho+\delta, there is a support SλS_\lambda such that

∥v+u∥<ρ(u∈Sλ).\lVert v+u\rVert<\rho \qquad(u\in S_\lambda).

Proof. Put s=∥v∥s=\lVert v\rVert. The margin in direction −v/∥v∥-v/\lVert v\rVert gives λ\lambda with v⋅u≤−Csv\cdot u\le-Cs for all u∈Sλu\in S_\lambda. For each such uu, ∥u∥<D\lVert u\rVert<D, and

∥v+u∥2=s2+2v⋅u+∥u∥2<s2−2Cs+D2<(ρ+δ)2−2C(ρ+δ)+D2=ρ2−2(C−δ)ρ+δ2−2Cδ+D2≤ρ2−32Cρ+D2<ρ2.\begin{align*} \lVert v+u\rVert^2 &=s^2+2v\cdot u+\lVert u\rVert^2\\ &<s^2-2Cs+D^2\\ &<(\rho+\delta)^2-2C(\rho+\delta)+D^2\\ &=\rho^2-2(C-\delta)\rho+\delta^2-2C\delta+D^2\\ &\le\rho^2-\tfrac32 C\rho+D^2 <\rho^2. \end{align*}

Indeed s≥ρ≥R>Cs\ge\rho\ge R>C, δ≤C/4\delta\le C/4 and ρ>D2/C\rho>D^2/C. The same λ\lambda works for every u∈Sλu\in S_\lambda. ◻

Metabelian presentations from Laurent relations

A cover datum consists of k≥1k\ge1, a finite nonempty alphabet A\mathcal A with designated letters aija_{ij}, a finite family of signed nonzero Laurent polynomials, and an integer rr passing the margin test. Write Λ+\Lambda_+ and Λ−\Lambda_- for the two signed families, and fix an order on every support. Using the radius (18), define PRP_R on the alphabet {t1,…,tk}∪A\{t_1,\ldots,t_k\}\cup\mathcal A by

[ti,tj]=aij(1≤i<j≤k),[a,bq(v)]=1(a,b∈A, ∥v∥<R),a=∏u∈Sλ(acλ,u)q(u)(a∈A, λ∈Λ+),a=∏u∈Sλ(acλ,u)q(u)−1(a∈A, λ∈Λ−).\begin{alignat*} {2} &[t_i,t_j]=a_{ij} &\qquad &(1\le i<j\le k), \tag{22a}\\[3pt] &[a,b^{q(v)}]=1 &\qquad &(a,b\in\mathcal A,\ \lVert v\rVert<R), \tag{22b}\\[3pt] &a=\prod_{u\in S_\lambda}(a^{c_{\lambda,u}})^{q(u)} &\qquad &(a\in\mathcal A,\ \lambda\in\Lambda_+), \tag{22c}\\[3pt] &a=\prod_{u\in S_\lambda}(a^{c_{\lambda,u}})^{q(u)^{-1}} &\qquad &(a\in\mathcal A,\ \lambda\in\Lambda_-). \tag{22d} \end{alignat*}

Products follow the chosen support order; the minus case uses the literal inverse q(u)−1q(u)^{-1}. The relator set is finite since

BR={v∈Zk:∥v∥<R}={v∈[−R,R]k∩Zk:∑ivi2<R2}.B_R=\{v\in\mathbb Z^k:\lVert v\rVert<R\} =\left\{v\in[-R,R]^k\cap\mathbb Z^k:\sum_i v_i^2<R^2\right\}.

Proposition 6. Every group PRP_R constructed from an accepted cover datum is metabelian.

Proof. In a group satisfying (22a), (22c) and (22d), write

C(ρ):[a,bq(v)]=1(a,b∈A, v∈Zk, ∥v∥<ρ).\mathcal C(\rho):\quad [a,b^{q(v)}]=1\quad (a,b\in\mathcal A,\ v\in\mathbb Z^k,\ \lVert v\rVert<\rho).

We prove C(ρ)⇒C(ρ+δ)\mathcal C(\rho)\Rightarrow\mathcal C(\rho+\delta) for ρ≥R\rho\ge R.

Only vectors vv with ρ≤∥v∥<ρ+δ\rho\le\lVert v\rVert<\rho+\delta need consideration. Choose λ\lambda by Lemma 5. Then, for every u∈Sλu\in S_\lambda,

∥u∥<D<ρ/(2k),∥v∥<ρ+1/(2k),∥u+v∥<ρ.(25) \tag{25} \lVert u\rVert<D<\rho/(2k),\qquad \lVert v\rVert<\rho+1/(2k),\qquad \lVert u+v\rVert<\rho.

If λ\lambda has sign ++, Lemma 2 and (25) give

[a,bq(u)q(v)]=1(u∈Sλ).[a,b^{q(u)q(v)}]=1\qquad(u\in S_\lambda).

Writing CH(a)C_H(a) for the centralizer of aa in the ambient group, relation (22c) therefore yields

bq(v)=∏u∈Sλ(bcλ,u)q(u)q(v)∈CH(a).b^{q(v)} =\prod_{u\in S_\lambda}(b^{c_{\lambda,u}})^{q(u)q(v)} \in C_H(a).

If λ\lambda has sign −-, the inverse-word assertion of Lemma 2, with the alphabet letters interchanged, gives

[b,aq(u)−1q(v)−1]=1(u∈Sλ).[b,a^{q(u)^{-1}q(v)^{-1}}]=1\qquad(u\in S_\lambda).

Now (22d) yields

aq(v)−1=∏u∈Sλ(acλ,u)q(u)−1q(v)−1∈CH(b).a^{q(v)^{-1}} =\prod_{u\in S_\lambda}(a^{c_{\lambda,u}})^{q(u)^{-1}q(v)^{-1}} \in C_H(b).

Conjugating this commutation identity by q(v)q(v) gives [a,bq(v)]=1[a,b^{q(v)}]=1. Thus both signs prove the propagation step.

In PRP_R, relation (22b) supplies C(R)\mathcal C(R), so

C(R+nδ)(n∈N),⋃n≥0{v∈Zk:∥v∥<R+nδ}=Zk.\mathcal C(R+n\delta)\quad(n\in\mathbb N),\qquad \bigcup_{n\ge0}\{v\in\mathbb Z^k:\lVert v\rVert<R+n\delta\}=\mathbb Z^k.

All ordered commutators vanish. Lemma 2 and (22a) imply, for AR=⟨ ⁣⟨A⟩ ⁣⟩A_R=\langle\!\langle \mathcal A\rangle\!\rangle,

AR′=1,(PR/AR)′=1,PR′≤AR,PR′′=1.□A_R'=1,\qquad (P_R/A_R)'=1,\qquad P_R'\le A_R,\qquad P_R''=1.\quad\Box

 ◻

Cofinality of the covers

Lemma 7. If 1→K→E→G→11\to K\to E\to G\to1 is a central extension, GG is finitely presented, and KK is free abelian of finite rank, then EE is finitely presented.

Proof. Let G=⟨x1,…,xm∣r1,…,rℓ⟩G=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_\ell\rangle. Choose lifts yi∈Ey_i\in E and a basis z1,…,zsz_1,\ldots,z_s of KK; write

rj(y)=∏h=1szhbjh,bjh∈Z.r_j(y)=\prod_{h=1}^{s}z_h^{b_{jh}},\qquad b_{jh}\in\mathbb Z.

For alphabets Y={y1,…,ym}Y=\{y_1,\ldots,y_m\} and Z={z1,…,zs}Z=\{z_1,\ldots,z_s\}, set

R0={[zh,zh′]:1≤h<h′≤s}∪{[zh,yi]:1≤h≤s, 1≤i≤m},R1={rj(y)(∏h=1szhbjh)−1:1≤j≤ℓ}.\begin{align*} \mathcal R_0 &=\{[z_h,z_{h'}]:1\le h<h'\le s\} \cup\{[z_h,y_i]:1\le h\le s,\ 1\le i\le m\},\\[3pt] \mathcal R_1 &=\left\{r_j(y)\left(\prod_{h=1}^{s}z_h^{b_{jh}}\right)^{-1} :1\le j\le\ell\right\}. \end{align*}

The lifts induce an epimorphism

φ:E~:=⟨Y⊔Z∣R0∪R1⟩↠E.\varphi:\widetilde E:= \langle Y\sqcup Z\mid\mathcal R_0\cup\mathcal R_1\rangle \twoheadrightarrow E.

If w∈ker⁡φw\in\ker\varphi, its image in GG is trivial. Express its YY-word as a product of conjugates of rj±1r_j^{\pm1}. The relations R1\mathcal R_1 and centrality relations R0\mathcal R_0 give

w=∏h=1szhnhin E~,1=φ(w)=∏h=1szhnhin K.w=\prod_{h=1}^{s}z_h^{n_h}\quad\text{in }\widetilde E, \qquad 1=\varphi(w)=\prod_{h=1}^{s}z_h^{n_h}\quad\text{in }K.

Independence of the basis implies n1=⋯=ns=0n_1=\cdots=n_s=0. Hence ker⁡φ=1\ker\varphi=1 and E~≅E\widetilde E\cong E. ◻

Proposition 8. Every finitely presented metabelian group GG is an epimorphic image of some certified cover PRP_R.

Proof. Put A=G′A=G' and Q=G/AQ=G/A. Choose a surjection π:Zk↠Q\pi:\mathbb Z^k\twoheadrightarrow Q with k≥1k\ge1 and form the pullback

E={(g,z)∈G×Zk:gA=π(z)}.E=\{(g,z)\in G\times\mathbb Z^k:gA=\pi(z)\}.

Both projections are surjective. Their kernels give exact sequences

1⟶A⟶E⟶Zk⟶1,1⟶ker⁡π⟶E⟶G⟶1.(37) \tag{37} 1\longrightarrow A\longrightarrow E\longrightarrow\mathbb Z^k\longrightarrow1, \qquad 1\longrightarrow\ker\pi\longrightarrow E\longrightarrow G\longrightarrow1.

The second kernel consists of (1,z)(1,z) with z∈ker⁡πz\in\ker\pi, and is central and free abelian of finite rank. Lemma 7 makes EE finitely presented. As a subgroup of the metabelian group G×ZkG\times\mathbb Z^k, it is metabelian.

For a finite generating set g1,…,gmg_1,\ldots,g_m of GG,

G′=⟨ ⁣⟨[gi,gj]:i<j⟩ ⁣⟩,A=∑i<j[gi,gj] ZQ.G'=\langle\!\langle [g_i,g_j]:i<j\rangle\!\rangle,\qquad A=\sum_{i<j}[g_i,g_j]\,\mathbb ZQ.

The second equality uses additive module notation for the abelian group AA. Pullback along π\pi preserves these module generators.

Choose lifts ti∈Et_i\in E of the standard basis of Zk\mathbb Z^k. Take a finite module-generating alphabet A\mathcal A for AA, including letters representing every [ti,tj][t_i,t_j]. A letter representing the identity may be added to ensure that the alphabet is nonempty. Lemma 3 supplies signed Laurent identities on this module. Lemma 4 supplies a finite margin certificate.

The selected elements satisfy (22a) by construction, (22b) because A′=1A'=1, and (22c)–(22d) by the module identities. They induce ψ:PR→E\psi:P_R\to E. If BB is the image of A\mathcal A, then

A=⟨bq(v):b∈B, v∈Zk⟩≤im⁡ψ,E=A⟨t1,…,tk⟩=im⁡ψ.A=\langle b^{q(v)}:b\in B,\ v\in\mathbb Z^k\rangle \le\operatorname{im}\psi, \qquad E=A\langle t_1,\ldots,t_k\rangle=\operatorname{im}\psi.

Thus ψ\psi is surjective, and PR↠E↠GP_R\twoheadrightarrow E\twoheadrightarrow G. ◻

Epimorphisms and normal closures

For P=⟨X∣R⟩P=\langle X\mid R\rangle and w∈F(X)w\in F(X),

w∈⟨ ⁣⟨R⟩ ⁣⟩⟺w=∏ν=1mhν−1rjνϵνhν in F(X),ϵν∈{−1,1}.(40) \tag{40} w\in\langle\!\langle R\rangle\!\rangle \quad\Longleftrightarrow\quad w=\prod_{\nu=1}^{m}h_\nu^{-1}r_{j_\nu}^{\epsilon_\nu}h_\nu \text{ in }F(X), \qquad \epsilon_\nu\in\{-1,1\}.

Here the right side means existence of m≥0m\ge0, hν∈F(X)h_\nu\in F(X), rjν∈Rr_{j_\nu}\in R and the indicated signs. A specified identity is checked by free reduction.

Lemma 9. Let H=⟨y1,…,yℓ∣S⟩H=\langle y_1,\ldots,y_\ell\mid S\rangle and G=⟨x1,…,xn∣R⟩G=\langle x_1,\ldots,x_n\mid R\rangle. Then

∃ f:H↠G⟺∃u1,…,uℓ∈F(X), v1,…,vn∈F(Y):s(u1,…,uℓ)∈⟨ ⁣⟨R⟩ ⁣⟩(s∈S),vj(u1,…,uℓ)xj−1∈⟨ ⁣⟨R⟩ ⁣⟩(1≤j≤n).\exists\,f:H\twoheadrightarrow G \quad\Longleftrightarrow\quad \begin{gathered} \exists u_1,\ldots,u_\ell\in F(X),\ v_1,\ldots,v_n\in F(Y):\\ s(u_1,\ldots,u_\ell)\in\langle\!\langle R\rangle\!\rangle\quad(s\in S),\\ v_j(u_1,\ldots,u_\ell)x_j^{-1}\in\langle\!\langle R\rangle\!\rangle\quad(1\le j\le n). \end{gathered}

Together with products (40) for these finitely many normal-closure memberships, the words ui,vju_i,v_j have a primitive-recursive checking predicate.

Proof. The first family of identities induces

f:H⟶G,yi⟼ui⟨ ⁣⟨R⟩ ⁣⟩.f:H\longrightarrow G,\qquad y_i\longmapsto u_i\langle\!\langle R\rangle\!\rangle.

The second gives f(vj)=xj⟨ ⁣⟨R⟩ ⁣⟩f(v_j)=x_j\langle\!\langle R\rangle\!\rangle, hence im⁡f=G\operatorname{im}f=G. Conversely, choose words representing the images and preimages of the generators under an epimorphism. Its defining identities lie in ⟨ ⁣⟨R⟩ ⁣⟩\langle\!\langle R\rangle\!\rangle and therefore admit (40). Substitution and free reduction of finite words, and the finite integer and index checks in (40), are primitive recursive. ◻

Proof of Theorem 1. Write A(d,r)\mathsf A(d,r) for the margin condition (13) on finite Laurent data dd, and Pd,rP_{d,r} for the presentation (22a)–(22d) at the radius (18). Let E(H,P;η)\mathsf E(H,P;\eta) check the words and products in Lemma 9. For a decoded witness c=(d,r,η)c=(d,r,\eta), set

V(P,c)=1⟺WF(P) ∧(n(P)=0 ∨ [A(d,r)∧E(Pd,r,P;η)]),(43) \tag{43} \begin{aligned} V(P,c)=1\quad\Longleftrightarrow\quad& \mathsf{WF}(P)\ \land\\ &\bigl(n(P)=0\ \lor\ [\mathsf A(d,r)\land\mathsf E(P_{d,r},P;\eta)]\bigr), \end{aligned}

where WF\mathsf{WF} checks the generating alphabet and n(P)n(P) is its size. All finite syntax checks are included in A\mathsf A and E\mathsf E. The radius is an integer expression in the data, and

{v∈Zk:∥v∥<R}={v∈[−R,R]k∩Zk:∑ivi2<R2}.\{v\in\mathbb Z^k:\lVert v\rVert<R\} = \left\{v\in[-R,R]^k\cap\mathbb Z^k:\sum_i v_i^2<R^2\right\}.

Consequently, constructing Pd,rP_{d,r} uses a bounded finite list. Lemma 4 and Lemma 9 show that VV is primitive recursive.

For n(P)>0n(P)>0, soundness is the implication

V(P,c)=1 ⟹ Pd,r′′=1,Pd,r↠G(P) ⟹ G(P)′′=1,V(P,c)=1 \ \Longrightarrow\ P_{d,r}''=1,\quad P_{d,r}\twoheadrightarrow G(P) \ \Longrightarrow\ G(P)''=1,

by Proposition 6. For completeness, Proposition 8 and Lemma 4 give

G(P)′′=1 ⟹ ∃d,r: A(d,r),Pd,r↠G(P).G(P)''=1 \ \Longrightarrow\ \exists d,r:\ \mathsf A(d,r),\quad P_{d,r}\twoheadrightarrow G(P).

Lemma 9 supplies η\eta. If n(P)=0n(P)=0 and PP is well formed, G(P)=1G(P)=1, as required by the first branch of (43). This proves (2). Finally, projection of the primitive-recursive relation {(P,c):V(P,c)=1}\{(P,c):V(P,c)=1\} gives the asserted recursive enumeration. ◻

References

Preprint · Lean (GitHub)

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