# The groups and the complexes

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

This lesson defines the complexes \(X_m\) of the theorem stated in [Coboundary expansion](coboundary-expansion.md) and proves their elementary properties. They are skeletons of coset complexes, in the sense of [Coset complexes](coset-complexes.md), of finite groups of matrices with entries in the ring \(\mathbb F_q[t]/(t^m)\).

- The subgroups are generated by elementary matrices attached to the edges of a cycle. They are described exactly by the pattern of their entries (Proposition 1.2); their orders and intersections follow (Corollary 1.3).
- The group \(U\) of the polynomial matrices of determinant \(1\) that are upper unitriangular at \(t=0\) is generated by these elementary matrices (Theorem 2.1). The proof is elementary: commutators, and the Euclidean algorithm for polynomials.
- The complexes \(T_m\) and their skeletons \(X_m\) are finite, pure and connected, every vertex of \(X_m\) lies in a number of faces that is bounded independently of \(m\), and the number of vertices tends to infinity (Theorem 3.2).
- The theorem on coboundary expansion follows from two further statements about the complexes \(T_m\), which are not proved in this course (Theorem 4.1).

The groups and complexes are those of [OpenAI-C, Sections 2 and 3]. The lesson uses matrices over a commutative ring, determinants, and division with remainder for polynomials over a field (core courses [Linear Algebra](https://kokunoyumeto.github.io/program-matematika-indonesia/en/#course-B40) and [Abstract Algebra I](https://kokunoyumeto.github.io/program-matematika-indonesia/en/#course-C30)).

## 1. Elementary matrices on a cycle

Let \(q\) be a prime number, \(k=\mathbb F_q\) the field with \(q\) elements, \(R=k[t]\) the polynomial ring, and \(N\ge3\) an integer. For \(a\ne b\) in \(\{1,\dots,N\}\) and \(f\in R\) let \(e_{ab}(f)\) be the matrix with \(1\) on the diagonal, \(f\) at the place \((a,b)\), and \(0\) elsewhere. Then \(e_{ab}(f)\,e_{ab}(g)=e_{ab}(f+g)\), and for distinct \(a,b,c\) the commutator of \(e_{ab}(f)\) and \(e_{bc}(g)\) is
\[
\begin{aligned}
&e_{ab}(f)\,e_{bc}(g)\,e_{ab}(f)^{-1}e_{bc}(g)^{-1}\\
&\quad=e_{ac}(fg).
\end{aligned}
\] Multiplying a matrix from the left by \(e_{ab}(f)\) adds \(f\) times its row \(b\) to its row \(a\); multiplying from the right adds \(f\) times its column \(a\) to its column \(b\).

Consider the directed cycle with the vertices \(1,\dots,N\) and the edges \(i\colon i\to i+1\) for \(1\le i<N\) and \(0\colon N\to1\). Let \(S=\{0,1,\dots,N-1\}\) be the set of its edges, and attach to each edge a matrix: to the edge \(i\) with \(1\le i<N\) the matrix \(x_i\), and to the edge \(0\) the matrix \(x_0\), where
\[
x_i=e_{i,i+1}(1),\qquad x_0=e_{N,1}(t).
\]
For a proper subset \(J\) of \(S\) let \(U_J\) be the group generated by the \(x_j\) with \(j\in J\); \(U_\emptyset=\{1\}\). For \(i\in S\) let
\[
H_i=U_{S\setminus\{i\}} .
\]

For \(a\ne b\) there is exactly one path from \(a\) to \(b\) along the cycle in the direction of the edges that visits no vertex twice. For a proper subset \(J\subsetneq S\) define a subspace \(V_{ab}(J)\) of \(R\) over \(k\): it is \(k\) if all edges of this path are in \(J\) and the edge \(0\) is not among them; it is \(kt\), the multiples of \(t\) by elements of \(k\), if all its edges are in \(J\) and the edge \(0\) is among them; and it is \(0\) otherwise. Let \(P_J\) be the set of the \(N\times N\) matrices \(g\) with entries in \(R\) such that \(g_{aa}=1\) for all \(a\) and \(g_{ab}\in V_{ab}(J)\) for all \(a\ne b\).

**Lemma 1.1.** Let \(J\subsetneq S\) and let \(a,b,c\) be vertices with \(a\ne b\), \(b\ne c\), such that the paths from \(a\) to \(b\) and from \(b\) to \(c\) have all their edges in \(J\). Then \(c\ne a\), the path from \(a\) to \(c\) is the first path followed by the second, and \(V_{ab}(J)\,V_{bc}(J)\subseteq V_{ac}(J)\).

**Proof.** Follow the first path and then the second. If this walk visited a vertex twice, it would go once around the whole cycle and use every edge; but \(J\) is a proper subset. So it is the path from \(a\) to \(c\), and \(c\ne a\). It uses the edge \(0\) at most once. So either none of the two paths uses the edge \(0\), and \(k\cdot k=k\); or exactly one does, and the product is \(kt\), which is \(V_{ac}(J)\). \(\blacksquare\)

**Proposition 1.2.** For every proper subset \(J\) of \(S\), \(P_J\) is a finite group of matrices of determinant \(1\), and \(U_J=P_J\).

**Proof.** *\(P_J\) is a group.* Choose an edge \(e\notin J\) and number the vertices in the order in which the cycle visits them, starting at the end of \(e\). A path with edges in \(J\) does not use \(e\), so it leads from a vertex to a later one. Hence for \(g\in P_J\) the matrix \(g-I\) is strictly upper triangular after this renumbering; so \(\det g=1\), and \((g-I)^N=0\). For \(g,h\in P_J\) and \(a\ne c\),
\[
(gh)_{ac}=g_{ac}+h_{ac}+\sum_{b\ne a,c}g_{ab}h_{bc},
\]
and by Lemma 1.1 every term lies in \(V_{ac}(J)\); the diagonal entries of \(gh\) are \(1\) because a product \(g_{ab}h_{ba}\) with both factors different from \(0\) is excluded by Lemma 1.1. So \(gh\in P_J\). In the same way all powers of \(g-I\) have their entries in the spaces \(V_{ab}(J)\), so \(g^{-1}=\sum_{r<N}(-1)^r(g-I)^r\) is in \(P_J\). The set \(P_J\) is finite because the spaces \(V_{ab}(J)\) are.

*\(U_J\subseteq P_J\).* The generators \(x_j\), \(j\in J\), are in \(P_J\): the path from \(j\) to \(j+1\) is the edge \(j\), and the path from \(N\) to \(1\) is the edge \(0\).

*The elementary matrices of \(P_J\) are in \(U_J\).* Let the path from \(a\) to \(b\) have its edges in \(J\), and \(f\in V_{ab}(J)\). We show \(e_{ab}(f)\in U_J\) by induction on the number of edges of the path. If it is one edge \(j\), then \(e_{ab}(f)\) is a power of \(x_j\), because \(f\) is \(c\) or \(ct\) with \(c\in k\), and \(c\) is a sum of ones. Otherwise let \(b'\) be a vertex of the path between \(a\) and \(b\). Then \(f=f_1f_2\) with \(f_1\in V_{ab'}(J)\) and \(f_2\in V_{b'b}(J)\): if \(f=c\) or \(f=ct\) with \(c\in k\), take \(f_1=c\) or \(ct\) and \(f_2=1\) or \(t\), according to which of the two parts of the path uses the edge \(0\). So \(e_{ab}(f)\) is the commutator of \(e_{ab'}(f_1)\) and \(e_{b'b}(f_2)\), which are in \(U_J\) by induction.

*\(P_J\subseteq U_J\).* Let \(g\in P_J\). Call the number of edges of the path from \(a\) to \(b\) the *length* of the place \((a,b)\). If \(g\ne I\), choose a place \((a,b)\) of smallest length with \(g_{ab}\ne0\), and replace \(g\) by \(g'=e_{ab}(-g_{ab})\,g\in P_J\). This adds \(-g_{ab}\) times the row \(b\) to the row \(a\). So \(g'_{ab}=0\), and the only other entries that change are at places \((a,c)\) with \(g_{bc}\ne0\); by Lemma 1.1 these places are longer than \((a,b)\). So the number of nonzero entries of smallest length decreases, or the smallest length increases. Lengths are at most \(N-1\), so after finitely many steps the matrix is \(I\). Hence \(g\) is a product of elementary matrices of \(P_J\). \(\blacksquare\)

**Corollary 1.3.**

1. \(|U_J|=q^{p(J)}\), where \(p(J)\) is the number of pairs \((a,b)\), \(a\ne b\), such that the path from \(a\) to \(b\) has all its edges in \(J\). In particular \(|H_i|=q^{\binom N2}\) and \(|U_{\{i\}}|=q\).
2. \(U_J\cap U_K=U_{J\cap K}\) for proper subsets \(J,K\). For a nonempty \(A\subseteq S\), \(\bigcap_{i\in A}H_i=U_{S\setminus A}\); in particular \(\bigcap_{i\in S}H_i=\{1\}\).
3. Every element of \(U_J\) has determinant \(1\), and its value at \(t=0\) is upper triangular with ones on the diagonal.
4. Let \(m\ge2\). Reduction of the entries modulo \(t^m\) is injective on the union of the groups \(U_J\). So it is injective on each \(U_J\), and the images satisfy \(\overline{U_J}\cap\overline{U_K}=\overline{U_{J\cap K}}\).

**Proof.** (1) The entries at the \(p(J)\) places with \(V_{ab}(J)\ne0\) can be chosen freely in a space with \(q\) elements. For \(J=S\setminus\{i\}\) the edges of \(J\) form one path through all \(N\) vertices, and every pair of vertices is joined by a part of it in exactly one direction. (2) \(V_{ab}(J)\cap V_{ab}(K)=V_{ab}(J\cap K)\), because the path from \(a\) to \(b\) does not depend on \(J\); and the intersection of the sets \(S\setminus\{i\}\), \(i\in A\), is \(S\setminus A\). (3) An entry in \(k\) belongs to a path without the edge \(0\), which leads from \(a\) to a larger \(b\); an entry in \(kt\) vanishes at \(0\). (4) All entries of all these matrices are polynomials of degree at most \(1\), and two different such polynomials are different modulo \(t^m\). If \(\overline g=\overline{g'}\) with \(g\in U_J\), \(g'\in U_K\), then \(g=g'\in U_J\cap U_K\). \(\blacksquare\)

## 2. The group \(U\) and its generators

Let \(\mathrm{SL}_N(R)\) be the group of the \(N\times N\) matrices over \(R\) with determinant \(1\). For such a matrix \(g\) let \(g(0)\) be the matrix of the values of its entries at \(t=0\). Let \(U\) be the set of the \(g\in\mathrm{SL}_N(R)\) for which \(g(0)\) is upper triangular with ones on the diagonal. It is a group, the inverse image of the group of upper unitriangular matrices under the homomorphism \(g\mapsto g(0)\). By Corollary 1.3(3) it contains all the groups \(U_J\).

**Theorem 2.1.** The group \(U\) is generated by \(x_0,x_1,\dots,x_{N-1}\).

**Proof.** Let \(U'\) be the group generated by the \(x_i\), and let \(\mathcal E\) be the set of the elementary matrices \(e_{ab}(f)\) with \(f\in R\) arbitrary if \(a<b\), and \(f\) divisible by \(t\) if \(a>b\). Then \(\mathcal E\subseteq U\).

*Step 1: \(\mathcal E\subseteq U'\).* Read the vertices modulo \(N\). For a vertex \(a\) and an integer \(\ell\ge1\) not divisible by \(N\), walk \(\ell\) steps from \(a\) along the cycle; let \(b\) be the end and \(j\) the number of times the edge \(0\) is used, and put
\[
E(a,\ell)=e_{ab}(t^j).
\]
Then \(b\ne a\), and \(j\ge1\) if \(a>b\). Conversely every \(e_{ab}(t^j)\) in \(\mathcal E\) is \(E(a,\ell)\) for \(\ell=b-a+jN\). If \(\ell=\ell_1+\ell_2\) with \(\ell_1,\ell_2\ge1\) and none of \(\ell_1,\ell_2,\ell\) divisible by \(N\), then the walk of \(\ell\) steps from \(a\) is a walk of \(\ell_1\) steps to a vertex \(c\) followed by a walk of \(\ell_2\) steps, the vertices \(a,c,b\) are distinct, and the numbers of uses of the edge \(0\) add up. So \(E(a,\ell)\) is the commutator of \(E(a,\ell_1)\) and \(E(c,\ell_2)\). We show \(E(a,\ell)\in U'\) by induction on \(\ell\). For \(\ell=1\) it is one of the \(x_i\). Let \(\ell\ge2\). If \(\ell-1\) is not divisible by \(N\), take \(\ell_1=1\). If \(\ell-1\) is divisible by \(N\), then \(\ell\ge N+1\ge4\); take \(\ell_1=2\), which is not divisible by \(N\) because \(N\ge3\), and \(\ell_2=\ell-2\ge2\), which is congruent to \(-1\) modulo \(N\). In both cases the induction applies. Finally, every \(f\) is a sum of terms \(ct^j\) with \(c\in k\), the matrix \(e_{ab}(ct^j)\) is a power of \(e_{ab}(t^j)\) because \(c\) is a sum of ones, and \(e_{ab}(f+g)=e_{ab}(f)e_{ab}(g)\). So \(\mathcal E\subseteq U'\).

*Step 2: \(U\) is generated by \(\mathcal E\).* We prove this for all \(N\ge1\), by induction on \(N\); for \(N=1\) the group is trivial. Let \(g\in U\) and let \(v_1,\dots,v_N\) be the entries of its first column. Then \(v_1(0)=1\) and \(v_a(0)=0\) for \(a>1\); write \(v_a=tu_a\) for \(a>1\). Multiplying \(g\) from the left by elements of \(\mathcal E\) we may perform the following operations, which keep these properties.

- For \(a>1\) and any \(f\in R\): add \(tf\) times the row \(1\) to the row \(a\). This replaces \(u_a\) by \(u_a+fv_1\).
- For \(b>1\) and any \(f\in R\): add \(f\) times the row \(b\) to the row \(1\). This replaces \(v_1\) by \(v_1+f\,tu_b\).

We reduce the first column to \((1,0,\dots,0)\) by induction on the degree \(D\) of \(v_1\). First replace every \(u_a\) by its remainder after division by \(v_1\); then every \(u_a\) is \(0\) or has degree less than \(D\). If all \(u_a\) are \(0\), the first column is \((v_1,0,\dots,0)\); expanding the determinant along it shows that \(v_1\) divides \(\det g=1\), so \(v_1\) is a constant, and \(v_1=1\) because \(v_1(0)=1\). Otherwise choose \(b\) with \(u_b\ne0\). The degree of \(tu_b\) is at most \(D\). Replace \(v_1\) by its remainder after division by \(tu_b\). The new \(v_1\) has degree less than \(D\), and its value at \(0\) is still \(1\). Now repeat.

So there is a product \(y\) of elements of \(\mathcal E\) such that \(yg\) has the first column \((1,0,\dots,0)\). Multiplying \(yg\) from the right by the matrices \(e_{1b}(f)\), \(b>1\), with \(f\) the negative of the entry at \((1,b)\), makes the first row \((1,0,\dots,0)\) and changes nothing else. The result has the form
\[
\begin{pmatrix}1&0\\0&h\end{pmatrix}
\]
with a matrix \(h\) of size \(N-1\), of determinant \(1\), whose value at \(0\) is upper triangular with ones on the diagonal. By induction \(h\) is a product of elementary matrices of the corresponding set for \(N-1\); placed in the rows and columns \(2,\dots,N\) these belong to \(\mathcal E\). So \(g\) is a product of elements of \(\mathcal E\) and their inverses. \(\blacksquare\)

## 3. The complexes

Let \(m\ge2\). Let \(\Gamma(m)\) be the group of the matrices in \(\mathrm{SL}_N(R)\) that are congruent to the identity matrix modulo \(t^m\). It is the kernel of the reduction of the entries modulo \(t^m\), so it is a normal subgroup of \(\mathrm{SL}_N(R)\), and it is contained in \(U\). Let
\[
G_m=U/\Gamma(m),
\]
which is the group of the reductions modulo \(t^m\) of the matrices in \(U\); it is finite, because \(R/(t^m)\) is finite. For a subgroup \(K\) of \(U\) let \(\overline K\) be its image in \(G_m\).

**Definition 3.1.** For \(m\ge2\), \(T_m\) is the coset complex of the group \(G_m\) with the subgroups \(\overline{H_i}\), \(i\in S\). For an integer \(d\ge1\) with \(d\le N-1\), \(X_m\) is the skeleton of dimension \(d\) of \(T_m\).

In the theorem on coboundary expansion, \(N=2d+1\), so that \(T_m\) has the dimension \(2d\). Let
\[
M=q^{\binom N2}.
\]

**Theorem 3.2.** Let \(m\ge2\).

1. \(T_m\) is a finite, pure and connected simplicial complex of dimension \(N-1\). Its chambers correspond to the elements of \(G_m\).
2. A face of \(T_m\) of type \(A\) lies in exactly \(|U_{S\setminus A}|\) chambers. Every vertex lies in exactly \(M\) chambers, every face of dimension \(N-2\) in exactly \(q\) chambers, and every face in at least \(1\) and at most \(M\) chambers.
3. The faces of \(T_m\) of a type \(A\) correspond to the orbits of \(\Gamma(m)\) on the faces of type \(A\) of the coset complex of \(U\) with the subgroups \(H_i\), and no element of \(\Gamma(m)\) other than \(1\) fixes a nonempty face of that complex.
4. \(X_m\) is finite, pure of dimension \(d\) and connected. Every vertex of \(X_m\) lies in at most \(M\binom{N-1}d\) faces of dimension \(d\).
5. \(X_m\) has \(N\,|G_m|/M\) vertices, and \(|G_m|\ge q^m\). So the number of vertices of \(X_m\) tends to infinity with \(m\), while the bound in (4) does not depend on \(m\).

**Proof.** By Corollary 1.3(2) and (4), for every nonempty \(A\subseteq S\)
\[
\bigcap_{i\in A}\overline{H_i}=\overline{U_{S\setminus A}},
\]
which is the image of \(\bigcap_{i\in A}H_i\): an element of the left side is the reduction of matrices \(g_i\in H_i\), which are all equal because the reduction is injective on the union of the groups. In particular the intersection of all \(\overline{H_i}\) is trivial.

(1) and (2): \(T_m\) is pure of dimension \(|S|-1=N-1\), its chambers correspond to the elements of \(G_m\), and a face of type \(A\) lies in \(|\overline{U_{S\setminus A}}|=|U_{S\setminus A}|\) chambers (Proposition 1.3 of the lesson on coset complexes). For a vertex, \(S\setminus A\) is the complement of one edge, and the number is \(M\); for a face of dimension \(N-2\), \(S\setminus A\) is one edge, and the number is \(q\) (Corollary 1.3(1)). In general \(U_{S\setminus A}\) is contained in some \(H_i\). By Theorem 2.1 the images of the \(x_j\) generate \(G_m\), and each \(x_j\) lies in the groups \(H_i\) with \(i\ne j\). So the subgroups \(\overline{H_i}\) generate \(G_m\), and \(T_m\) is connected by Proposition 2.1 of that lesson.

(3) This is Proposition 3.1 of the lesson on coset complexes for the normal subgroup \(\Gamma(m)\) of \(U\); its assumption was verified above, and \(\Gamma(m)\cap H_i=\{1\}\) because the reduction is injective on \(H_i\).

(4) The skeleton of a pure complex is pure, and it has the same vertices and edges, so it is connected. A vertex lies in \(M\) chambers, and a chamber has \(\binom{N-1}d\) faces of dimension \(d\) containing a given one of its vertices.

(5) There are \(|G_m|/|\overline{H_i}|=|G_m|/M\) vertices of each of the \(N\) types. The matrices \(e_{12}(f)\) with \(f\) of degree less than \(m\) are \(q^m\) elements of \(U\) with different reductions. \(\blacksquare\)

So the complexes \(X_m\) have the properties required in the theorem on coboundary expansion, except the expansion itself. The construction needs only that \(q\) is a prime; the two statements of the next section need more.

## 4. What remains

Let \(d\ge1\) and \(N=2d+1\), so \(T_m\) has the dimension \(n=2d\). Let \(Y_m\) be the skeleton of dimension \(n-1\) of \(T_m\). Consider the following two statements.

- **(E)** There is a number \(\eta>0\) such that for all \(m\ge2\), all \(i\) with \(0\le i<n-1\) and all cochains \(f\) of degree \(i\) of \(Y_m\): \(\|\delta_if\|\ge\eta\,\operatorname{dist}(f,Z^i(Y_m))\), with the norms of \(Y_m\).
- **(V)** For all \(m\ge2\) and all \(i\) with \(1\le i\le d-1\): \(H^i(T_m)=0\).

**Theorem 4.1.** If (E) and (V) hold, then every \(X_m\), \(m\ge2\), is an \(\varepsilon\)-coboundary expander with \(\varepsilon=\eta/M\). Together with Theorem 3.2 this gives the theorem on coboundary expansion for the dimension \(d\), with \(D=M\binom{2d}d\).

**Proof.** Let \(0\le i<d\). Then \(i+1\le d\le n-1\). The cochains of the degrees \(i\) and \(i+1\), the map \(\delta_i\), and the spaces \(Z^i\) and \(B^i\) are the same for \(X_m\), \(Y_m\) and \(T_m\). By (V), and for \(i=0\) because \(T_m\) is connected, \(Z^i=B^i\).

Every face of dimension \(n-1\) of \(T_m\) lies in exactly \(q\) chambers. So the weights of \(Y_m\) and of \(T_m\) agree in the degrees up to \(n-1\) (Exercise 7.4 of the lesson on coboundary expansion), and (E) says
\[
\begin{aligned}
&\|\delta_if\|_{T_m,i+1}\\
&\quad\ge\eta\,\operatorname{dist}_{T_m,i}(f,B^i).
\end{aligned}
\]
Every face of dimension \(d\) of \(T_m\) lies in at least \(1\) and at most \(M\) chambers. By Corollary 5.2 of that lesson, with \(s=d\), \(a=1\) and \(b=M\),
\[
\begin{aligned}
&\|\delta_if\|_{X_m,i+1}\\
&\quad\ge\frac\eta M\,\operatorname{dist}_{X_m,i}(f,B^i).
\end{aligned}
\]
\(\blacksquare\)

In [OpenAI-C] both statements are proved for \(d\ge3\) and for a prime \(q\) that is odd and large enough, depending on \(d\).

- Statement (E) is obtained from a theorem of I. Oppenheim and I. Valentiner-Branth on the cosystolic expansion of a class of coset complexes called Kac–Moody–Steinberg complexes [OpenAI-C, Section 5]. The properties of \(T_m\) that are needed to apply it are those of Corollary 1.3: the reduction is injective on the groups \(U_J\) and preserves their intersections.
- Statement (V) is proved in [OpenAI-C, Sections 3 and 4]. The infinite coset complex of \(U\) with the subgroups \(H_i\) is identified with a part of a building, and its homology with coefficients in \(\mathbb F_2\) is shown to vanish below the dimension \(n-1\). By Theorem 3.2(3) the complex \(T_m\) is its quotient by \(\Gamma(m)\), which acts freely on the faces; so the homology of \(T_m\) in low degrees is the homology of the group \(\Gamma(m)\). This is shown to vanish in the degrees \(1\) to \(d-1\) by results on the stability of the homology of general linear groups.

Neither the theorem on Kac–Moody–Steinberg complexes nor the theory of buildings and of the homology of groups used for (V) is available in this course. They are the two hypotheses on which the theorem on coboundary expansion rests here.

## 5. Exercises

**5.1.** For \(N=3\) write down the general element of \(H_0\), \(H_1\) and \(H_2\), and check \(|H_i|=q^3\).

**5.2.** For \(N=3\) express \(e_{13}(1)\), \(e_{21}(t)\) and \(e_{12}(t)\) through \(x_0,x_1,x_2\) by commutators.

**5.3.** Carry out the reduction of Step 2 in the proof of Theorem 2.1 for the first column \((1+t^2,\ t,\ 0)\) with \(N=3\).

**5.4.** Which steps of the proof of Theorem 2.1 use \(N\ge3\), and which hold for every \(N\)? Where does the proof use that \(q\) is a prime?

## 6. Solutions

**5.1.** \(H_0\) is generated by \(x_1,x_2\): the paths with edges in \(\{1,2\}\) are \(1\to2\), \(2\to3\) and \(1\to3\), all without the edge \(0\). So \(H_0\) consists of the upper triangular matrices with ones on the diagonal and entries of \(k\) above it. \(H_1\) is generated by \(x_2,x_0\): the paths are \(2\to3\), \(3\to1\) and \(2\to1\), the last two with the edge \(0\). So its elements have the entries \(g_{23}\in k\), \(g_{31}\in kt\), \(g_{21}\in kt\), and all other off-diagonal entries \(0\). \(H_2\) is generated by \(x_0,x_1\): the paths are \(3\to1\), \(1\to2\), \(3\to2\), so \(g_{31}\in kt\), \(g_{12}\in k\), \(g_{32}\in kt\). In each case three entries are free in a set with \(q\) elements.

**5.2.** \(e_{13}(1)\) is the commutator of \(x_1=e_{12}(1)\) and \(x_2=e_{23}(1)\). \(e_{21}(t)\) is the commutator of \(x_2=e_{23}(1)\) and \(x_0=e_{31}(t)\). For \(e_{12}(t)\) the walk from \(1\) to \(2\) has \(4\) steps; with \(\ell_1=2\) and \(\ell_2=2\) it is the commutator of \(e_{13}(1)\) and \(e_{32}(t)\), and \(e_{32}(t)\) is the commutator of \(x_0=e_{31}(t)\) and \(x_1=e_{12}(1)\).

**5.3.** Here \(v_1=1+t^2\), \(u_2=1\), \(u_3=0\), and \(D=2\). The \(u_a\) are already reduced. Take \(b=2\): \(tu_2=t\), and \(1+t^2=t\cdot t+1\), so adding \(-t\) times the row \(2\) to the row \(1\) gives \(v_1=1\). Now \(D=0\); the remainder of \(u_2=1\) after division by \(1\) is \(0\), which is obtained by adding \(-t\) times the row \(1\) to the row \(2\). The column is \((1,0,0)\).

**5.4.** Step 1 uses \(N\ge3\) twice: the commutator formula needs a third vertex \(c\) different from \(a\) and \(b\), and for the walks whose length is congruent to \(1\) modulo \(N\) it uses that \(2\) is not divisible by \(N\). Step 2 holds for every \(N\ge1\). So for \(N=2\) the group \(U\) is generated by the set \(\mathcal E\), but the argument does not show that it is generated by \(x_0\) and \(x_1\).

That \(q\) is a prime is used in Step 1 and in Proposition 1.2: an element \(c\) of \(k\) is a sum of ones, so \(e_{ab}(cf)\) is a power of \(e_{ab}(f)\). For a field \(k\) that is not a prime field, the powers of \(x_j\) give only the entries in the prime field, and the groups generated by the \(x_j\) are smaller.

## References

- [OpenAI-C] OpenAI, *Bounded-degree coboundary expanders in every dimension*, OpenAI Math Release preprint, 24 September 2026, Sections 2 to 5. https://github.com/openai/math/tree/main/preprints/Bounded-degree-coboundary-expanders-in-every-dimension-September-24-2026