A positive parameter-free formula defining {c1,c−1}\{c_1,c_{-1}\} in the residually finite integer Heisenberg group

21.106

Problem

Is every parameter-free first-order group formula with one free variable concise in residually finite groups?

Gφ={g∈G:G⊨φ(g)},∣Gφ∣<∞⟹?∣⟨Gφ⟩∣<∞.G_\varphi=\{g\in G:G\models\varphi(g)\},\qquad |G_\varphi|<\infty\quad\stackrel{?}{\Longrightarrow}\quad |\langle G_\varphi\rangle|<\infty.

This concerns formulas with quantifiers, not only group words. The counterexample uses a positive formula in a torsion-free nilpotent group of class two.

Setup and result

Let φ(z)\varphi(z) be a parameter-free first-order formula in the group language {1,⋅,−1}\{1,\cdot,{}^{-1}\}. For a group GG, write

Gφ={g∈G:G⊨φ(g)},φ(G)=⟨Gφ⟩.G_\varphi=\{g\in G:G\models\varphi(g)\}, \qquad \varphi(G)=\langle G_\varphi\rangle.

The formula φ\varphi is concise in a class C\mathcal C of groups if GφG_\varphi finite implies φ(G)\varphi(G) finite for every G∈CG\in\mathcal C. Conte and Petschick (2) study this extension of conciseness from group words to first-order formulae. Problem 21.106 of the Kourovka Notebook (4) asks whether every formula with one free variable is concise in the class of residually finite groups.

Throughout, the commutator convention is [g,h]=g−1h−1gh[g,h]=g^{-1}h^{-1}gh. Consider the formula

Φ(z):=∃a ∃b ∀x ∀y ∃r ∃s(z=[a,b] ∧ [r,a]=1 ∧ [s,b]=1∧ [r,b]=[x,y] ∧ [a,s]=[x,y]).(2) \tag{2} \begin{aligned} \Phi(z):={}&\exists a\,\exists b\,\forall x\,\forall y\,\exists r\,\exists s\\ &\bigl(z=[a,b]\ \land\ [r,a]=1\ \land\ [s,b]=1\\ &\qquad{}\land\ [r,b]=[x,y]\ \land\ [a,s]=[x,y]\bigr). \end{aligned}

Each commutator abbreviates a term in the group language. Thus zz is the only free variable, and no group elements occur as parameters. The formula is positive, since its quantifier-free part is a conjunction of equations.

Theorem 1. Let H=H(Z)H=\mathrm{H}(\mathbb{Z}) be the integer Heisenberg group, with multiplication

(a,b,c)(d,e,f)=(a+d,b+e,c+f+ae).(a,b,c)(d,e,f)=(a+d,b+e,c+f+ae).

Then

HΦ={(0,0,1),(0,0,−1)},Φ(H)=Z(H)={(0,0,n):n∈Z}≅(Z,+).\begin{gathered} H_\Phi=\{(0,0,1),(0,0,-1)\},\\ \Phi(H)=Z(H)=\{(0,0,n):n\in\mathbb{Z}\}\cong(\mathbb{Z},+). \end{gathered}

Moreover, HH is residually finite, generated by two elements, torsion-free, and nilpotent of class exactly two. In particular, Φ\Phi is not concise in the class of residually finite groups.

Conte and Petschick prove that every existential formula is concise in torsion-free groups of nilpotency class two (2, Theorem 1.1). The universal quantifiers in (2) allow a counterexample in this same class. Further positive results for other classes of groups are obtained by Ciobanu and Conte (1). The question about group words in Problem 21.105 of (4) is distinct from the question answered here.

A formula defining units

Let RR be a commutative ring with identity. We realize the Heisenberg group H(R)\mathrm{H}(R) as the set R3R^3 with the multiplication in Theorem 1. These coordinates correspond to the upper unitriangular matrices

(a,b,c)⟷(1ac01b001).(a,b,c)\longleftrightarrow \begin{pmatrix}1&a&c\\0&1&b\\0&0&1\end{pmatrix}.

The identity is (0,0,0)(0,0,0), and

(a,b,c)−1=(−a,−b,−c+ab).(a,b,c)^{-1}=(-a,-b,-c+ab).

For t∈Rt\in R, put ct=(0,0,t)c_t=(0,0,t). For g=(g1,g2,g3)g=(g_1,g_2,g_3) and h=(h1,h2,h3)h=(h_1,h_2,h_3), write

Δ(g,h)=g1h2−g2h1.\Delta(g,h)=g_1h_2-g_2h_1.

Direct multiplication gives

[g,h]=cΔ(g,h),ctcu=ct+u.(8) \tag{8} [g,h]=c_{\Delta(g,h)},\qquad c_tc_u=c_{t+u}.

Every ctc_t is central. We use R×R^\times for the group of units of RR, identifying a unit with its underlying ring element.

Proposition 2. For every commutative ring RR with identity,

H(R)Φ={cu:u∈R×}.\mathrm{H}(R)_\Phi=\{c_u:u\in R^\times\}.

Proof. Suppose that Φ(z)\Phi(z) holds, and fix witnesses a,ba,b for the first two existential quantifiers. Apply the universal quantifiers to x=(1,0,0)x=(1,0,0) and y=(0,1,0)y=(0,1,0). Since [x,y]=c1[x,y]=c_1, the resulting witnesses r,sr,s satisfy

z=cΔ(a,b),Δ(r,a)=Δ(s,b)=0,Δ(r,b)=Δ(a,s)=1.\begin{gather*} z=c_{\Delta(a,b)},\qquad \Delta(r,a)=\Delta(s,b)=0,\\ \Delta(r,b)=\Delta(a,s)=1. \end{gather*}

The determinant identity

Δ(a,b)Δ(r,s)=Δ(r,a)Δ(s,b)+Δ(r,b)Δ(a,s)(11) \tag{11} \Delta(a,b)\Delta(r,s) =\Delta(r,a)\Delta(s,b)+\Delta(r,b)\Delta(a,s)

follows by expansion over RR. Substitution yields Δ(a,b)Δ(r,s)=1\Delta(a,b)\Delta(r,s)=1, so Δ(a,b)\Delta(a,b) is a unit.

Conversely, let u∈R×u\in R^\times, and choose v∈Rv\in R with uv=1uv=1. Set a=(1,0,0)a=(1,0,0) and b=(0,u,0)b=(0,u,0). For arbitrary x,y∈H(R)x,y\in\mathrm{H}(R), put t=Δ(x,y)t=\Delta(x,y) and take

r=(vt,0,0),s=(0,t,0).r=(vt,0,0),\qquad s=(0,t,0).

Equation (8) gives

[a,b]=cu,[r,a]=[s,b]=1,[r,b]=cvtu=ct=[a,s].[a,b]=c_u,\qquad [r,a]=[s,b]=1,\qquad [r,b]=c_{vtu}=c_t=[a,s].

Since [x,y]=ct[x,y]=c_t, all five equations in (2) hold with z=cuz=c_u. The elements a,ba,b depend only on uu, whereas r,sr,s are chosen after x,yx,y, as required. ◻

The integer counterexample

Proof of Theorem 1. The units of Z\mathbb{Z} are 11 and −1-1. Proposition 2 therefore gives HΦ={c1,c−1}H_\Phi=\{c_1,c_{-1}\}. These two elements are distinct. By (8), c1n=cnc_1^n=c_n for every n∈Zn\in\mathbb{Z}, so

Φ(H)=⟨c1,c−1⟩={cn:n∈Z}.\Phi(H)=\langle c_1,c_{-1}\rangle=\{c_n:n\in\mathbb{Z}\}.

The map n↦cnn\mapsto c_n is an injective homomorphism from (Z,+)(\mathbb{Z},+) onto this subgroup. In particular, the subgroup is infinite.

Put X=(1,0,0)X=(1,0,0) and Y=(0,1,0)Y=(0,1,0). If g=(a,b,c)g=(a,b,c) is central, then (8) gives [g,X]=c−b=1[g,X]=c_{-b}=1 and [g,Y]=ca=1[g,Y]=c_a=1. Thus a=b=0a=b=0, proving Z(H)={cn:n∈Z}Z(H)=\{c_n:n\in\mathbb{Z}\}.

We next verify residual finiteness. Recall that a group is residually finite if every nonidentity element has nonidentity image under some homomorphism to a finite group. For each positive integer mm, coordinate reduction defines a surjective homomorphism

πm:H⟶H(Z/mZ).\pi_m:H\longrightarrow\mathrm{H}(\mathbb{Z}/m\mathbb{Z}).

The homomorphism property follows from the multiplication formula, and the target has m3m^3 elements. If g≠1g\ne1, choose a nonzero coordinate tt of gg and take m=∣t∣+1m=|t|+1. Then 0<∣t∣<m0<|t|<m, so m∤tm\nmid t and πm(g)≠1\pi_m(g)\ne1.

Finally, [X,Y]=c1[X,Y]=c_1 and, for every (a,b,c)∈H(a,b,c)\in H,

(a,b,c)=XaYbcc−ab.(a,b,c)=X^aY^b c_{c-ab}.

Hence X,YX,Y generate HH. All commutators are central by (8), and [X,Y]≠1[X,Y]\ne1, so HH has nilpotency class exactly two. The projection (a,b,c)↦(a,b)(a,b,c)\mapsto(a,b) is a homomorphism to (Z2,+)(\mathbb{Z}^2,+). If an element of HH has finite order, its first two coordinates therefore vanish. Such an element is ctc_t, and ctk=cktc_t^k=c_{kt} for every positive integer kk; hence it has finite order only when t=0t=0. This proves torsion-freeness and completes the proof. ◻

References

Preprint · Lean (GitHub)

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