# The counterexample

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

This lesson gives the group and the complex that answer the finite D(2) problem in the negative. A group \(P\) with five generators and four relations has a perfect normal subgroup \(K\) that is generated, as a normal subgroup, by two elements. The plus construction for \(K\) gives a complex \(X\) of dimension three with the condition D(2). A character \(\rho\) of the quotient \(G=P/K\) has the value \(-1\) on an element of order two, and the second homology of \(X\) twisted by \(\rho\) vanishes. By the tower obstruction, no finite complex of dimension two has these two properties together.

- The group \(P\), the elements \(t_1,t_2\), and the proof that \(K\) is perfect (Lemma 1.1).
- The character \(\rho\) (Section 2).
- The matrix of Fox derivatives and \(H_2(B;\rho)=0\) (Lemma 3.1).
- The theorem (Theorem 4.1).

The lesson follows [OpenAI-W, Section 4]. It uses presentation complexes and Fox derivatives from [Wall's D(2) problem and twisted homology](walls-d2-problem-and-twisted-homology.md), the corollary on elements of finite order from [The tower obstruction](the-tower-obstruction.md), and [The finite plus construction](the-finite-plus-construction.md).

## 1. The group and a perfect subgroup

Let \(P\) be the group with the generators \(x_1,a_1,x_2,a_2,s\) and the relations
\[
\begin{gathered}
x_1a_1x_1^{-1}=a_1^4,\qquad x_2a_2x_2^{-1}=a_2^3,\\
x_1=a_2^{13},\qquad sx_2s^{-1}=a_1^5 .
\end{gathered}
\]
Let \(B\) be its presentation complex: one vertex, five loops, four cells of dimension two. Put
\[
t_1=x_1^2,\qquad t_2=x_2^3,
\]
let \(K\) be the normal closure of \(t_1,t_2\) in \(P\), and \(G=P/K\).

The homomorphism from the free group to \(\mathbb Z\) with \(s\mapsto1\) and \(x_1,a_1,x_2,a_2\mapsto0\) respects the four relations and is \(0\) on \(t_1,t_2\). So it defines a surjective homomorphism \(G\to\mathbb Z\), and \(G\) is infinite.

We write \([g,h]=ghg^{-1}h^{-1}\).

**Lemma 1.1** ([OpenAI-W]). The group \(K\) is perfect.

**Proof.** *Two identities.* From \(xax^{-1}=a^n\) follows \(x^kax^{-k}=a^{n^k}\), by induction on \(k\). So
\[
t_1a_1t_1^{-1}=a_1^{16},\qquad t_2a_2t_2^{-1}=a_2^{27}.
\]
The square of the third relation is \(t_1=a_2^{26}\), and the cube of the fourth is \(st_2s^{-1}=a_1^{15}\). Hence
\[
\begin{gathered}
t_1=a_2^{26}=a_2^{27}a_2^{-1}=[t_2,a_2],\\
st_2s^{-1}=a_1^{15}=a_1^{16}a_1^{-1}=[t_1,a_1].
\end{gathered}
\]

*The module.* Write \(K_{\mathrm{ab}}\) additively, and \([k]\) for the class of \(k\in K\). The group \(P\) acts on \(K_{\mathrm{ab}}\) by \(g\cdot[k]=[gkg^{-1}]\), and \(K\) acts trivially; so \(K_{\mathrm{ab}}\) is a left module over \(\mathbb Z[G]\) (Exercise 4.2 of the lesson on the plus construction). For \(t\in K\) and \(a\in P\),
\[
[\,[t,a]\,]=[t]+[at^{-1}a^{-1}]=(1-a)\cdot[t].
\]
Let \(u_1=[t_1]\) and \(u_2=[t_2]\). The two identities give
\[
\begin{gathered}
u_1=(1-a_2)\,u_2,\\
s\,u_2=(1-a_1)\,u_1 .
\end{gathered}
\]

*The first equation.* The element \(a_2\) commutes with \(t_1=a_2^{26}\), so \(a_2u_1=u_1\). And \(a_2^{26}=t_1\) lies in \(K\), so it acts trivially on \(K_{\mathrm{ab}}\). Apply \(1+a_2+\dots+a_2^{25}\) to the first equation:
\[
26\,u_1=(1-a_2^{26})\,u_2=0 .
\]

*The second equation.* The element \(a_1\) commutes with \(st_2s^{-1}=a_1^{15}\), so \(a_1\) fixes \(su_2\). And \(a_1^{15}=st_2s^{-1}\) lies in \(K\) and acts trivially. Apply \(1+a_1+\dots+a_1^{14}\) to the second equation:
\[
15\,s\,u_2=(1-a_1^{15})\,u_1=0 .
\]
The action of \(s\) is invertible, so \(15u_2=0\), and by the first equation \(15u_1=0\).

*The end.* From \(26u_1=0\) and \(15u_1=0\) and \(26\cdot11-15\cdot19=1\) we get \(u_1=0\). Then \(su_2=(1-a_1)u_1=0\), so \(u_2=0\). The module \(K_{\mathrm{ab}}\) is generated by \(u_1\) and \(u_2\), because every element of \(K\) is a product of conjugates of \(t_1^{\pm1}\) and \(t_2^{\pm1}\). So \(K_{\mathrm{ab}}=0\). \(\blacksquare\)

The exponents are chosen for this proof: \(4^2-1=15=3\cdot5\) and \(3^3-1=26=2\cdot13\) have no common factor. The third and fourth relation make each of \(t_1\), \(t_2\) a commutator that involves the other one.

## 2. A character

Let \(\zeta\) be a complex number with \(\zeta^2+\zeta+1=0\), so that \(\zeta^3=1\) and \(\zeta\ne1\). Put
\[
\begin{gathered}
\rho(x_1)=-1,\quad\rho(a_1)=\zeta,\quad\rho(x_2)=\zeta^2,\\
\rho(a_2)=-1,\quad\rho(s)=1 .
\end{gathered}
\]
The values lie in a commutative group, so the four relations hold for them if
\[
\begin{gathered}
\zeta=\zeta^4,\qquad-1=(-1)^3,\\
-1=(-1)^{13},\qquad\zeta^2=\zeta^5 ,
\end{gathered}
\]
and all four are true. So \(\rho\) is a character of \(P\). Moreover \(\rho(t_1)=(-1)^2=1\) and \(\rho(t_2)=\zeta^6=1\), so \(\rho\) is trivial on \(K\) and defines a character of \(G\), again called \(\rho\).

**Lemma 2.1.** The class of \(x_1\) in \(G\) has the order two, and \(\rho(x_1)=-1\).

**Proof.** \(x_1^2=t_1\in K\), so the square of the class is trivial. The class is not trivial, because \(\rho\) has the value \(-1\) on it. \(\blacksquare\)

## 3. The twisted second homology

Write the relations as the words
\[
\begin{gathered}
r_1=x_1a_1x_1^{-1}a_1^{-4},\\
r_2=x_2a_2x_2^{-1}a_2^{-3},\\
r_3=x_1a_2^{-13},\\
r_4=sx_2s^{-1}a_1^{-5}.
\end{gathered}
\]

**Lemma 3.1** ([OpenAI-W]). The matrix of the Fox derivatives of \(r_1,\dots,r_4\) by \(x_1,a_1,x_2,a_2,s\), evaluated at \(\rho\), is the matrix \(D\) with the four rows

- for \(r_1\): \((1-\zeta,\ -2,\ 0,\ 0,\ 0)\),
- for \(r_2\): \((0,\ 0,\ 2,\ \zeta^2-1,\ 0)\),
- for \(r_3\): \((1,\ 0,\ 0,\ -1,\ 0)\),
- for \(r_4\): \((0,\ -(1+\zeta),\ 1,\ 0,\ 1-\zeta^2)\).

Its rows are linearly independent. So \(H_2(B;\rho)=0\).

**Proof.** *Relators of the form \(xax^{-1}a^{-n}\).* By Exercise 6.2 of the first lesson the derivatives by \(x\) and \(a\) are \(1-xax^{-1}\) and \(x-xax^{-1}\sum_{l=1}^na^{-l}\). At a character they have the values
\[
1-\rho(a),\qquad\rho(x)-\sum_{l=0}^{n-1}\rho(a)^{-l}.
\]
For \(r_1\): \(1-\zeta\), and \(-1-(1+\zeta^2+\zeta+1)=-2\). For \(r_2\): \(1-(-1)=2\), and \(\zeta^2-(1-1+1)=\zeta^2-1\).

*The relator \(r_3\).* The derivative by \(x_1\) is \(1\). The derivative by \(a_2\) is \(-x_1\sum_{l=1}^{13}a_2^{-l}\), with the value \(-(-1)\sum_{l=1}^{13}(-1)^l=-1\).

*The relator \(r_4\).* The derivative by \(s\) is \(1-sx_2s^{-1}\), with the value \(1-\zeta^2\). The derivative by \(x_2\) is \(s\), with the value \(1\). The derivative by \(a_1\) is \(-sx_2s^{-1}\sum_{l=1}^5a_1^{-l}\). Since \(\zeta^{-1}=\zeta^2\), the sum \(\sum_{l=1}^5\zeta^{-l}\) is \(\zeta^2+\zeta+1+\zeta^2+\zeta=-1\), and the value is \(\zeta^2=-(1+\zeta)\).

*The rank.* Take the columns of \(x_1,a_1,x_2,s\):
\[
\det\begin{pmatrix}
1-\zeta&-2&0&0\\
0&0&2&0\\
1&0&0&0\\
0&-(1+\zeta)&1&1-\zeta^2
\end{pmatrix}.
\]
Expanding along the third row and then along the last column gives
\[
\begin{aligned}
&(1-\zeta^2)\cdot\det\begin{pmatrix}-2&0\\0&2\end{pmatrix}\\
&\qquad=-4\,(1-\zeta^2)\ne0 .
\end{aligned}
\]
So the four rows are linearly independent, and \(H_2(B;\rho)=0\) by Lemma 5.1 of the first lesson. \(\blacksquare\)

## 4. The theorem

**Theorem 4.1** ([OpenAI-W]). Assume the homotopy invariance in TOP5. There is a finite connected CW complex \(X\) of dimension three, with one vertex, five edges, six cells of dimension two and two cells of dimension three, such that
\[
\begin{gathered}
H_i(\widetilde X;\mathbb Z)=0\ \text{ for }i>2,\\
H^3(X;M)=0
\end{gathered}
\]
for every left module \(M\) over \(\mathbb Z[\pi_1(X)]\), and such that \(X\) is not homotopy equivalent to any finite CW complex of dimension at most two.

**Proof.** *The complex.* By Lemma 1.1 the proposition on the plus construction applies to \(B\), \(t_1,t_2\) and \(K\). It gives a complex \(X\supseteq B\) with two more cells of dimension two and two cells of dimension three, with \(\pi_1(X)=G\), with the condition D(2), and with
\[
H_2(X;\rho)\cong H_2(B;\rho)=0
\]
by property 3 of that proposition and Lemma 3.1.

*No model of dimension two.* Suppose that \(h:Y\to X\) is a homotopy equivalence from a finite CW complex \(Y\) of dimension at most two. Then \(Y\) is connected, and \(\theta=h_*:\pi_1(Y)\to G\) is an isomorphism. By TOP5,
\[
H_2(Y;\rho\theta)\cong H_2(X;\rho)=0 .
\]
Let \(g\in\pi_1(Y)\) be the element with \(\theta(g)\) the class of \(x_1\). By Lemma 2.1 it has the order two, and \(\rho\theta(g)=-1\). This contradicts Corollary 3.1 of the lesson on the tower. \(\blacksquare\)

This is Theorem 1.1 of the first lesson.

**The roles of the two complexes.** The character \(\rho\) has vanishing second homology on \(B\), a complex of dimension two. This does not contradict the tower obstruction: the fundamental group of \(B\) is \(P\), and the obstruction says that \(\rho\) is \(1\) on every element of finite order of \(P\). Since \(\rho(x_1)=-1\), it shows that \(x_1\) has infinite order in \(P\). In \(G\) the element \(x_1\) has the order two. A complex of dimension two with the fundamental group \(G\) is, for example, \(B\) with the two cells for \(t_1,t_2\). For it the matrix \(D\) gets two more rows, the Fox derivatives of \(x_1^2\) and \(x_2^3\) at \(\rho\). They are \(1+\rho(x_1)=0\) and \(1+\zeta^2+\zeta^4=0\) in their columns and \(0\) elsewhere. So the second homology twisted by \(\rho\) has the dimension two there, as the obstruction demands. The two cells of dimension three of \(X\) remove exactly these two classes. A complex of dimension two cannot do this.

**What is proved and what is assumed.** The lessons prove the obstruction, the properties of the plus construction, the perfectness of \(K\) and the computation of the matrix. They use the standard theorems TOP1 to TOP6 of the first lesson; all of them are proved in the programme except the homotopy invariance in TOP5, which is a named hypothesis. The computation of \(D\) and of its minors was repeated by a program that differentiates the four words.

**What the theorem does not say.** It concerns finite complexes; the group \(G\) is infinite. For finite fundamental groups the problem is not touched by this example. The complex \(X\) is not given by explicit attaching maps for its cells of dimension three: they come from the theorem of Hurewicz.

## 5. Exercises

**5.1.** Check the four relations for \(\rho\), and compute \(\rho(a_1^{15})\) and \(\rho(a_2^{26})\).

**5.2.** Show that in \(G\) the classes satisfy \(x_1^2=1\), \(a_2^{26}=1\), \(x_2^3=1\) and \(a_1^{15}=1\).

**5.3.** Compute the minor of \(D\) for the columns \(a_1,x_2,a_2,s\).

**5.4.** Let \(\rho'\) be the character with \(\rho'(a_1)=1\) and the other values as for \(\rho\). Is it a character of \(G\)? Compute the first row of the matrix for \(\rho'\).

**5.5.** Compute \(\chi(X)\), and derive from \(H_0(X;\rho)=0\), \(H_2(X;\rho)=0\) and \(H_3(X;\rho)=0\) the dimension of \(H_1(X;\rho)\).

## 6. Solutions

**5.1.** The relations are checked in Section 2. \(\rho(a_1^{15})=\zeta^{15}=1\) and \(\rho(a_2^{26})=(-1)^{26}=1\); both elements lie in \(K\).

**5.2.** \(x_1^2=t_1\) and \(x_2^3=t_2\) lie in \(K\). By the proof of Lemma 1.1, \(a_2^{26}=t_1\) and \(a_1^{15}=st_2s^{-1}\) lie in \(K\).

**5.3.** The submatrix has the rows \((-2,0,0,0)\), \((0,2,\zeta^2-1,0)\), \((0,0,-1,0)\), \((-(1+\zeta),1,0,1-\zeta^2)\). Expanding along the first row and then along the last column gives \(-2\,(1-\zeta^2)\) times the determinant of the matrix with the rows \((2,\zeta^2-1)\) and \((0,-1)\), which is \(-2\). So the minor is \(4\,(1-\zeta^2)\). It is not zero either.

**5.4.** The relations: \(1=1^4\), and \(\rho'(x_2)=\zeta^2\) must be \(\rho'(a_1)^5=1\), which is false. So \(\rho'\) is not a character of \(P\). (With \(\rho'(x_2)=1\) as well it is one; then the first row is \((1-1,\ -1-4,\ 0,0,0)\), that is \((0,-5,0,0,0)\).)

**5.5.** \(\chi(X)=1-5+6-2=0\). The alternating sum of the dimensions of the twisted homology spaces is \(\chi(X)\), as in Lemma 4.3 of the first lesson. So \(-\dim H_1(X;\rho)=0\): the twisted homology of \(X\) vanishes in all degrees. (\(H_3(X;\rho)=0\) holds because \(\partial_3\) stays injective after tensoring, by the equation \(q\partial_3=\mathrm{id}\) of the plus construction.)

## References

- [OpenAI-W] OpenAI, *A Counterexample to Wall's D(2) Problem*, OpenAI Math Release preprint, 6 October 2026, Section 4. https://github.com/openai/math/tree/main/preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026