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Cone chains, torsion-freeness and the integral detector

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Coning off every sheet of the Cayley graph gives a two-dimensional complex whose cellular chains, by the cycle coordinates, form an exact sequence. Two consequences follow. A subgroup of prime order would act freely on this complex, which is impossible for an acyclic complex of dimension two; so the graphical group \(\Gamma\) is torsion-free. And the same exact sequence makes restriction \(H^2(B\Gamma;\mathbb Z)\to H^2(B\mathsf H;\mathbb Z)\) onto. An explicit integral \(2\)-cocycle on the Heisenberg group evaluates to \(-1\) on the torus spanned by \((z,x)\); extended to \(\Gamma\), it is the class that detects the torus in the main theorem [OpenAI-BCI, Sections 7.4-7.7, Lemmas 7.6-7.10, Proposition 7.11].

The setting of Cycle coordinates and disjoint domains is in force: the formal right Cayley graph \(\mathcal C\), the family \(\mathcal Q\) of sheets, and the isomorphisms \(\Phi_R\) of Lemma 1.1 there. Domains play no role here. We use Lemmas 3.1 and 3.2 of Torsion-freeness (the periodic resolution over a cyclic group of prime order), and two standard facts of algebraic topology, not proved in the programme:

(BG) (the bar construction; [Hatcher, Example 1B.7, Theorem 1B.8]). For a group \(\Lambda\), the \(\Delta\)-complex \(E\Lambda\) whose \(n\)-simplices are the ordered \((n+1)\)-tuples \((g_0,\ldots,g_n)\) of elements of \(\Lambda\), with faces given by deleting entries, is contractible, and \(\Lambda\) acts freely on it by left multiplication; the quotient \(B\Lambda\) is a \(K(\Lambda,1)\) CW complex, whose simplices are written \([g_1|\cdots|g_n]\) for the orbit of \((1,g_1,g_1g_2,\ldots,g_1\cdots g_n)\). The homotopy type of a CW complex \(K(\Lambda,1)\) is determined by \(\Lambda\), and a homomorphism \(f\) induces the cellular map \(Bf\) sending \([g_1|\cdots|g_n]\) to \([f(g_1)|\cdots|f(g_n)]\).

(CC) (cellular cohomology; [Hatcher, Theorem 3.5]). For a CW complex, singular cohomology with coefficients in \(G\) is the cohomology of the cellular cochain complex, which is \(\operatorname{Hom}(-,G)\) of the cellular chain complex, naturally for cellular maps.

1. The cone complex and exact chains

Starting from \(\mathcal C\), add a vertex \(c_Q\) for every sheet \(Q\), an oriented edge \(\sigma_{Q,u}\) (a spoke) from \(c_Q\) to every vertex \(u\) of \(Q\), and, for every positive edge \(e:u\to v\) of \(Q\), a triangle \(\tau_{Q,e}\) with cellular boundary \[ \partial\tau_{Q,e}=e+\sigma_{Q,u}-\sigma_{Q,v} .\tag{1.1} \] Call the resulting two-dimensional complex \(\mathcal Z\). Different sheets give different spokes and triangles, also where their images in \(\mathcal C\) agree; the complex is defined by its cells, with finite chains only.

Lemma 1.1 (exact cone chains; [OpenAI-BCI, Lemma 7.6]). For every commutative ring \(R\), the augmented cellular chain complex \[ \begin{gathered} 0\to C_2(\mathcal Z;R)\xrightarrow{\partial_2}{}\\ C_1(\mathcal Z;R)\xrightarrow{\partial_1}C_0(\mathcal Z;R)\\ \xrightarrow{\varepsilon}R\to0 \end{gathered}\tag{1.2} \] is exact.

Proof. For an edge chain \(b\) of a sheet \(Q\), let \(c_Q*b\) be the same combination of its triangles, and for a vertex chain \(t\) of \(Q\) let \(\sigma_Q(t)\) be the corresponding spoke chain. By (1.1), \(\partial(c_Q*b)=b-\sigma_Q(\partial b)\).

At \(C_2\). A finite \(2\)-chain is \(\sum_Qc_Q*b_Q\) for finitely many edge chains \(b_Q\). If its boundary vanishes, its components on the distinct spoke sets give \(\partial b_Q=0\) for every \(Q\), and its component on \(\mathcal C\) gives \(\sum_Qb_Q=0\). Injectivity of \(\Phi_R\) gives all \(b_Q=0\).

At \(C_1\). Let \(b+\sum_Q\sigma_Q(t_Q)\) be a finite \(1\)-cycle, with \(b\) on \(\mathcal C\). Its coefficient at \(c_Q\) is the augmentation of \(t_Q\), so \(t_Q\) has total coefficient zero; as \(Q\) is connected, \(t_Q=\partial a_Q\) for a finite edge chain \(a_Q\) of \(Q\) (join the vertices of its support to one of them by paths). Adding \(\sum_Q\partial(c_Q*a_Q)\) removes all spokes, and leaves a \(1\)-cycle on \(\mathcal C\), which is \(\sum_Qz_Q\) with sheet cycles \(z_Q\), by surjectivity of \(\Phi_R\); and \(z_Q=\partial(c_Q*z_Q)\). So the original cycle is a boundary.

At \(C_0\). \(\mathcal C\) is connected and every cone vertex is joined to it, so a vertex chain of total coefficient zero is a boundary, and \(\varepsilon\) is onto. \(\square\)

Left translation acts cellularly on \(\mathcal Z\), preserving all orientations. Put \(A=\mathbb Z\Gamma\) and \(B=\mathbb Z\mathsf H\). Then \[ \begin{gathered} C_1(\mathcal Z;\mathbb Z),\ C_2(\mathcal Z;\mathbb Z)\\ \text{are free }A\text{-modules},\\ C_0(\mathcal Z;\mathbb Z)\cong A\oplus A\otimes_B\mathbb Z . \end{gathered}\tag{1.3} \] Indeed an element stabilizing a formal edge fixes its initial vertex; one stabilizing a spoke fixes its endpoint in \(\mathcal C\); one stabilizing a triangle fixes its positive base edge. So these stabilizers are trivial. The group vertices form one free orbit, and the cone vertices the orbit \(\Gamma/\mathsf H\), since the stabilizer of a sheet is a conjugate of \(\mathsf H\) (Proposition 5.1 of The private completion and the graphical group). Neither torsion-freeness nor equivariance of any domain choice is used.

2. Torsion-freeness

Lemma 2.1 (comparison of free resolutions; [OpenAI-BCI, Lemma 7.7]). Let \(P_\bullet\to M\) and \(Q_\bullet\to N\) be exact augmented complexes of free modules over a ring \(A\). Every homomorphism \(M\to N\) lifts to a chain map \(P_\bullet\to Q_\bullet\), and any two lifts are chain homotopic. In particular two free resolutions of one module are chain homotopy equivalent, also after tensoring with a right module or applying \(\operatorname{Hom}_A(-,L)\).

Proof. On a basis of \(P_0\), choose images lifting the prescribed images under the augmentations. Having defined the map up to degree \(n-1\), the desired image of the boundary of a basis vector of \(P_n\) is a cycle of \(Q_{n-1}\), hence a boundary by exactness; choose a preimage and extend linearly. For two lifts, let \(f\) be their difference, with zero augmentation. Inductively, \(f_n-h_{n-1}\partial_n\) takes values in the cycles of \(Q_n\) (apply \(\partial\) and use the chain-map and previous homotopy identities), so it is \(\partial_{n+1}h_n\) on a basis; then \(f_n=\partial_{n+1}h_n+h_{n-1}\partial_n\), with \(h_{-1}=0\) and the zero augmentation at \(n=0\). Lifts of the identity in both directions give the homotopy equivalence; tensor and \(\operatorname{Hom}\) preserve the homotopy identities. No finite bases are needed. \(\square\)

Lemma 2.2 (no element of finite order; [OpenAI-BCI, Lemma 7.8]). The group \(\Gamma\) is torsion-free.

Proof. \(\mathsf H\) is torsion-free (Corollary 1.2 of The Heisenberg group and its laws). Suppose \(P\le\Gamma\) has prime order \(p\). The stabilizer of a cone vertex is a conjugate of \(\mathsf H\), so it meets \(P\) trivially; the other stabilizers are trivial by (1.3). So \(P\) acts freely on the cells of every dimension (on cells, not merely without fixed points of edges: no element can reverse a positive edge, since orientations are preserved). Choosing orbit representatives, the chains of \(\mathcal Z\) with coefficients \(\mathbb F_p\) are free \(\mathbb F_p[P]\)-modules, and Lemma 1.1 with \(R=\mathbb F_p\) makes them a free resolution of the trivial module \(\mathbb F_p\) of length two. Tensoring with \(\mathbb F_p\) gives zero homology above degree two, whereas by Lemmas 3.1 and 3.2 of Torsion-freeness this homology is \(\mathbb F_p\) in every degree. Hence \(\Gamma\) has no subgroup of prime order, and an element of finite order \(n>1\) would have a power of prime order. \(\square\)

3. The bar model and a Heisenberg cocycle

For a group \(\Lambda\), let \(\mathcal B_n(\Lambda)\) be the free abelian group on the tuples \((g_0,\ldots,g_n)\in\Lambda^{n+1}\), with the diagonal left action, the alternating deletion boundary, and augmentation \(1\) on every tuple of length one. It is free over \(\mathbb Z\Lambda\), with the orbit representatives with \(g_0=1\), and prepending \(1\), \(s(g_0,\ldots,g_n)=(1,g_0,\ldots,g_n)\), is a contracting homotopy of abelian groups: \(\partial s+s\partial=\mathrm{id}\), including the augmentation. So \(\mathcal B_\bullet(\Lambda)\) is a free resolution of the trivial module \(\mathbb Z\). It is the cellular chain complex of \(E\Lambda\) in (BG), so by (BG) and (CC) the cohomology of \[ \operatorname{Hom}_{\mathbb Z\Lambda}(\mathcal B_\bullet(\Lambda),\mathbb Z) \] is \(H^*(B\Lambda;\mathbb Z)\), and for a subgroup inclusion, restriction of these cochains to tuples of the subgroup is the pull-back along the induced map of classifying spaces. Write \([g_1|\cdots|g_n]\) for \((1,g_1,g_1g_2,\ldots,g_1\cdots g_n)\); then \[ \begin{gathered} \partial[g|h]=g[h]-[gh]+[g],\\ \partial[g|h|k]=g[h|k]-[gh|k]\\ +[g|hk]-[g|h], \end{gathered} \] and with trivial coefficients these are the formulas of inhomogeneous group cochains.

In the matrix coordinates \((\alpha,\beta,\gamma)\) of Lemma 1.1 of the lesson on the Heisenberg group, \(x^ny^lz^c=(n,l,c+nl)\) and \[ \begin{aligned} &(\alpha,\beta,\gamma)(\alpha',\beta',\gamma')\\ &\quad=(\alpha+\alpha',\,\beta+\beta',\\ &\qquad\quad\gamma+\gamma'+\alpha\beta') . \end{aligned}\tag{3.1} \]

Lemma 3.1 (a primitive Heisenberg detector; [OpenAI-BCI, Lemma 7.9]). The integer function \[ \begin{gathered} \omega\bigl((\alpha,\beta,\gamma),(\alpha',\beta',\gamma')\bigr)\\ =\alpha\gamma'+\binom\alpha2\beta' \end{gathered}\tag{3.2} \] is a normalized group \(2\)-cocycle on \(\mathsf H\). Its class \(c_{\mathsf H}\in H^2(B\mathsf H;\mathbb Z)\) evaluates to \(-1\) on the torus with ordered generators \((z,x)\).

Proof. \(\binom\alpha2=\alpha(\alpha-1)/2\) is an integer for all integers \(\alpha\), and \(\omega\) vanishes when either argument is the identity. For \(g,h,k\) with coordinates \((\alpha,\beta,\gamma),(\alpha',\beta',\gamma'),(\alpha'',\beta'',\gamma'')\), the cocycle identity \[ \omega(g,h)+\omega(gh,k)=\omega(h,k)+\omega(g,hk) \] reads, by (3.1), \[ \begin{aligned} &\alpha\gamma'+\tbinom\alpha2\beta'+(\alpha+\alpha')\gamma''\\ &\qquad+\tbinom{\alpha+\alpha'}2\beta''\\ &\quad=\alpha'\gamma''+\tbinom{\alpha'}2\beta''\\ &\qquad+\alpha(\gamma'+\gamma''+\alpha'\beta'')\\ &\qquad+\tbinom\alpha2(\beta'+\beta''), \end{aligned} \] which holds by \(\binom{\alpha+\alpha'}2=\binom\alpha2+\binom{\alpha'}2+\alpha\alpha'\). Equivalently, \((t,g)(t',h)=(t+t'+\omega(g,h),gh)\) is an associative multiplication on \(\mathbb Z\times\mathsf H\), a central extension by \(\mathbb Z\).

The subgroup generated by \(z=(0,0,1)\) and \(x=(1,0,0)\) is free abelian of rank two, with \(z^ux^v=(v,0,u)\). Subdivide the square of the torus with vertices \(1,z,zx,x\) by the diagonal from \(1\) to \(zx\); its two oriented triangles are \((1,z,zx)-(1,x,xz)\), that is, the bar chain \[ [z|x]-[x|z] . \] Sending the grid vertex \((u,v)\) to \(z^ux^v\) defines an equivariant map from the triangulated universal cover of the torus into \(E\Lambda\), \(\Lambda=\langle z,x\rangle\), which induces the identity on fundamental groups; so this chain represents the image of the fundamental class \([T^2]\). The evaluation is \[ \omega(z,x)-\omega(x,z)=0-1=-1 , \] since \(z\) has \(\alpha=0\), and \(x=(1,0,0)\), \(z=(0,0,1)\) give \(\omega(x,z)=1\). \(\square\)

4. Degree-two classes extend

Lemma 4.1 (integral restriction is onto in degree two; [OpenAI-BCI, Lemma 7.10]). The inclusion \(\mathsf H\le\Gamma\) induces a surjection \[ \begin{gathered} H^2(B\Gamma;\mathbb Z)\\ \longrightarrow H^2(B\mathsf H;\mathbb Z) . \end{gathered}\tag{4.1} \]

Proof. Put \(M=\mathbb Z\), \(C_0=C_0(\mathcal Z;\mathbb Z)\cong A\oplus A\otimes_BM\) and \(K=\ker(C_0\to M)\). By Lemma 1.1 and (1.3), \(K\) has a free resolution of length one, \[ \begin{gathered} 0\to P_1\xrightarrow{d_P}P_0\to K\to0,\\ P_1=C_2(\mathcal Z;\mathbb Z),\quad P_0=C_1(\mathcal Z;\mathbb Z) . \end{gathered} \] As a right \(B\)-module, \(A\) is free (coset representatives), so \(I_\bullet=A\otimes_B\mathcal B_\bullet(\mathsf H)\) is a free resolution of \(A\otimes_BM\). Resolve \(C_0\) by \(Q_0=A\oplus I_0\) and \(Q_n=I_n\) for \(n\ge1\). Lift \(K\to C_0\) to a chain map \(f:P_\bullet\to Q_\bullet\): choose \(f_0\) on a basis lifting the inclusion; the elements \(f_0d_P(P_1)\) have zero augmentation in \(C_0\), so exactness gives \(f_1:P_1\to Q_1\) with \(d_Qf_1=f_0d_P\).

Define a free augmented complex \(T_\bullet\to M\) by \(T_0=Q_0\) and \(T_n=Q_n\oplus P_{n-1}\) for \(n\ge1\) (with \(P_j=0\) for \(j\ge2\)), with boundary \(d_T(q,p)=(d_Qq+f_{n-1}p,\,-d_Pp)\) for \(n\ge2\), \(d_T(q,p)=d_Qq+f_0p\) for \(n=1\), and augmentation \(Q_0\to C_0\to M\). The identities defining \(f\) give \(d_T^2=0\). It is exact. In degree two, a cycle \((q,p)\) has \(d_Pp=0\), so \(p=0\), and \(q\) is a boundary in \(Q_\bullet\); the same holds in higher degrees. In degree one, if \(d_Qq+f_0p=0\), then the image of \(p\) in \(K\) has zero image in \(C_0\), so \(p=d_Pp_1\); then \(q+f_1p_1\) is a \(Q\)-cycle, equal to \(d_Qq_2\), and \((q,p)=d_T(q_2,-p_1)\). In degree zero, an element with zero image in \(M\) has image in \(K\); subtract a suitable \(f_0p\) and lift the rest through \(d_Q\). So \(T_\bullet\) is a free resolution of \(M\).

The inclusion \(Q_\bullet\to T_\bullet\) lifts \(C_0\to M\). A degree-two cocycle \(\phi:Q_2\to M\) extends to \(\psi(q,p)=\phi(q)\) on \(T_2=Q_2\oplus P_1\), still a cocycle since \(T_3=Q_3\) and \(d_T\) on \(T_3\) has no \(P_1\) component. So \[ \begin{gathered} \operatorname{Ext}^2_A(M,M)\\ \longrightarrow\operatorname{Ext}^2_A(C_0,M) \end{gathered}\tag{4.2} \] is onto. The isomorphism \[ \begin{gathered} \operatorname{Hom}_A(A\otimes_B\mathcal B_n(\mathsf H),M)\\ \to\operatorname{Hom}_B(\mathcal B_n(\mathsf H),M),\\ \phi\mapsto\bigl(b\mapsto\phi(1\otimes b)\bigr), \end{gathered} \] with inverse \(\eta\mapsto(a\otimes b\mapsto a\eta(b))\), commutes with the boundaries, and the extra free summand \(A\) in degree zero contributes no cohomology in degree two; so the target of (4.2) is \(H^2(B\mathsf H;\mathbb Z)\).

It remains to identify (4.2) with restriction. Choose a comparison map \(T_\bullet\to\mathcal B_\bullet(\Gamma)\) lifting the identity of \(M\) (Lemma 2.1). The composite \(I_\bullet\to Q_\bullet\to T_\bullet\to\mathcal B_\bullet(\Gamma)\) lifts the augmentation \(A\otimes_BM\to M\), and so does \[ \theta\bigl(a\otimes(h_0,\ldots,h_n)\bigr)=a\,(h_0,\ldots,h_n), \] the tuple read in \(\Gamma\). By Lemma 2.1 the two lifts are homotopic. After the isomorphism above, precomposition with \(\theta\) restricts a \(\Gamma\)-cochain to tuples of \(\mathsf H\), which is the pull-back along \(B\mathsf H\to B\Gamma\). So (4.2) is (4.1). \(\square\)

5. The detector

Proposition 5.1 (torsion-freeness and the integral detector; [OpenAI-BCI, Proposition 7.11]). The graphical group \(\Gamma\) is torsion-free. The homomorphism \(i_\Gamma:\mathbb Z^2\to\Gamma\) sending the ordered basis to \((z,x)\) is injective, and there is \(c_\Gamma\in H^2(B\Gamma;\mathbb Z)\) with \[ \bigl\langle(Bi_\Gamma)^*c_\Gamma,[T^2]\bigr\rangle=-1 .\tag{5.1} \]

Proof. Lemma 2.2 gives torsion-freeness. The embedding \(\mathsf H\le\Gamma\) of Proposition 5.1 of the lesson on the graphical group and \(z^ux^v=(v,0,u)\) give injectivity of \(i_\Gamma\). Choose \(c_\Gamma\) mapping to \(c_{\mathsf H}\) under the surjection of Lemma 4.1; that map is restriction, so Lemma 3.1 gives (5.1). \(\square\)

6. Exercises

Exercise 6.1. Verify the cocycle identity of Lemma 3.1 directly for \(g=h=k=x\).

Exercise 6.2. In the central extension defined by \(\omega\), with multiplication \((t,g)(t',h)=(t+t'+\omega(g,h),gh)\), show that \(X=(0,x)\) and \(Z=(0,z)\) satisfy \([X,Z]=T\) with \(T=(1,1)\), and that \(Y=(0,y)\) commutes with \(Z\).

Exercise 6.3. Why does the proof of Lemma 2.2 need the free action on cells of every dimension, and not only on vertices?

Exercise 6.4. Show that \(H^2(B\mathbb Z^2;\mathbb Z)\cong\mathbb Z\), generated by a class taking the value \(\pm1\) on \([T^2]\), using the bar cochain \(\omega_0((u,v),(u',v'))=uv'\) on \(\mathbb Z^2\).

7. Solutions

6.1. With \(x=(1,0,0)\): \(x^2=(2,0,0)\), \(\omega(x,x)=1\cdot0+\binom12\cdot0=0\), and \(\omega(x^2,x)=0\); both sides vanish because \(\beta=\gamma=0\) throughout.

6.2. In matrix coordinates \(x=(1,0,0)\), \(y=(0,1,0)\), \(z=(0,0,1)\). Then \(\omega(x,z)=1\cdot1=1\) and \(\omega(z,x)=0\), so \(XZ=(1,xz)\) and \(ZX=(0,zx)=(0,xz)\). Hence \(XZ=T\cdot ZX\) with \(T=(1,1)\), which is central because \(\omega\) vanishes when an argument is the identity; that is, \([X,Z]=XZ(ZX)^{-1}=T\). Also \(\omega(y,z)=0\cdot1=0\) and \(\omega(z,y)=0\), so \(YZ=(0,yz)=(0,zy)=ZY\).

6.3. Freeness of a \(\mathbb F_p[P]\)-module of chains requires that \(P\) permute the cells of that dimension without fixed cells; a fixed edge or triangle would give a non-free summand \(\mathbb F_p\) in \(C_1\) or \(C_2\), and the chain complex would not be a free resolution.

6.4. For \(\Lambda=\mathbb Z^2\) the cocycle \(\omega_0\) is the bilinear form \(uv'\); on \([z|x]-[x|z]\) with \(z=(1,0)\), \(x=(0,1)\) it gives \(1\cdot1-0\cdot0=1\). Bilinear forms are cocycles, and \(H^2(B\mathbb Z^2;\mathbb Z)=H^2(T^2;\mathbb Z)\cong\mathbb Z\) by (BG), since the torus is a \(K(\mathbb Z^2,1)\); the class of \(\omega_0\) evaluates to \(1\) on the fundamental class, so it generates.

References