Proof. Choose a binary basis of \(K_2\). The matrix of multiplication by \(\lambda\in K_2\) is a unital ring representation; it is injective because its value on \(1\in K_2\) recovers \(\lambda\). Apply this map coefficientwise to a group-ring sum. The coefficient matrices commute with the formal group symbols, so products are preserved. Comparing the coefficient matrix at each group element proves injectivity. Thus \(XY=I_d\) and \(I_d-YX\ne0\). Now \((YX)^2=Y(XY)X=YX\), \(X(I_d-YX)=X-(XY)X=0\) and \((I_d-YX)Y=Y-Y(XY)=0\), proving all the identities. ◻

Proof. Put \(C=I_d-YX\). Choose an entry \(C_{ij}\) and a group element \(h_0\) whose coefficient in that entry is nonzero. The coefficient is \(1\) because the field is \(\mathbb F_2\). Left multiplication by \(h_0^{-1}\) therefore gives \[ h_0^{-1}C_{ij}=1+\sum_{s\in S}s \tag{4}\] for a finite \(S\subseteq H_2\setminus\{1\}\), possibly empty.

To isolate the identity term in Equation (4), we seek a nonzero rational projection \(p\) with \(psp=0\) for every \(s\in S\). The cyclic coordinates provide complementary projections that enforce these vanishing products. Let \(z_h\) generate the cyclic coordinate labelled by \(h\) and put \[p_h=\frac{1+z_h+z_h^2}{3},\qquad p=p_1\prod_{s\in S}(1-p_s).\] The \(p_h\) are self-adjoint idempotents: averaging the nine products in \((1+z_h+z_h^2)^2\) gives three copies of each element of \(C_3\). They commute because distinct coordinates commute. Consequently \(p=p^*=p^2\), and the displayed product shows that \(3^{1+|S|}p\) is integral. This is the elementary commuting-coordinate method in the works cited above; compare (Dicks and Schick 2002, Lemmas 3.1 and 3.4).

Conjugation obeys \(s p_hs^{-1}=p_{sh}\). For \(s\in S\), the product \(sps^{-1}\) contains \(p_s\), while \(p\) contains \(1-p_s\). All these coordinate factors commute, so \[ psp=p(sps^{-1})s=0. \tag{5}\] This argument does not need disjointness of the two collections of coordinate labels. Inverse pairs or finite-order elements in \(S\) cause no change.

The original coordinates \(\{1\}\cup S\) are distinct. The coefficient of the identity in their product is therefore the product of the individual identity coefficients. Hence \[ \tau_1(p)=[1]p=\frac13\prod_{s\in S}\frac23=\theta>0. \tag{6}\] For empty \(S\) this means \(p=p_1\) and \(\theta=1/3\).

The map \(\mathbb Z[1/3]\to\mathbb F_2\) is defined because \(3\) is odd. Reduce the projection modulo two and combine Equations (4) and (5) to obtain \[\bar p h_0^{-1} C_{ij}\bar p=\bar p.\] Let \(f\) be the row of length \(d\) with entry \(\bar p h_0^{-1}\) at position \(i\) and zero elsewhere. Let \(w\) be the column with entry \(\bar p\) at position \(j\) and zero elsewhere. Thus \(fCw=\bar p\). Define \[P=\operatorname{diag}(I_d,p),\qquad u=\begin{pmatrix}X\\fC\end{pmatrix},\qquad v=\begin{pmatrix}Y&Cw\end{pmatrix}.\] Using Equation (3) in this order gives \[uv=\begin{pmatrix}XY&XCw\\fCY&fC^2w\end{pmatrix} =\operatorname{diag}(I_d,\bar p)=\bar P.\] The rational trace is \(d+\theta\) by Equation (6). It is the rational orthogonal projection, not its binary reduction, that has this strictly positive extra trace. ◻

Proof. Since \(D\) is odd, every coefficient of \(P\) has a well-defined reduction modulo \(q\). Set \(\mathcal R=(\mathbb Z/q\mathbb Z)[G]\) and choose arbitrary finite lifts \(u_0,v_0\) of \(u,v\) over \(\mathcal R\). In the corner ring \(e\operatorname{M}_m(\mathcal R)e\), whose identity is \(e\), put \[Z=eu_0v_0e,\qquad W=e-Z.\] Modulo two, \(Z\) is \(\bar Puv\bar P=\bar P^3=\bar P\). Every coefficient of \(W\) is therefore even. A product of \(k\) such coefficients is divisible by \(2^k\), giving \(W^k=0\). This is nilpotence in the even ideal of the corner ring; the corner ring itself need not be nilpotent.

Because \(Z=e-W\), the finite geometric sum \[Q=e+W+\cdots+W^{k-1}\] satisfies \(QZ=ZQ=e-W^k=e\). When \(k=1\), this sum is just \(e\). Thus \(Q\) is a corner inverse of \(Z\), not a claimed inverse in the entire matrix ring. Set \[u_q=Qeu_0,\qquad v_q=v_0e.\] Then \(u_qv_q=Qeu_0v_0e=QZ=e\). Only finite sums and products were used, so the resulting supports remain finite. ◻

Proof. The matrix \(P+q(I_m-P)=qI_m+(1-q)P\) is integral because \(D\mid(q-1)\). Its reduction modulo \(q\) is \(e=P_q\), which is also the reduction of \(UV\). Thus every coefficient of \(UV-P-q(I_m-P)\) is an integer divisible by \(q\); the bottom-right block of Equation (7) is integral.

For \(J\) and \(B\) defined above, the complementary projection identities give \(J^{-1}=-qP-(I_m-P)\). For \(x\in[0,1]\) put \[L_x=\begin{pmatrix}I_r&0\\xU/q&I_m\end{pmatrix},\qquad R_x=\begin{pmatrix}I_r&xV/q\\0&I_m\end{pmatrix}.\] Their inverses negate the off-diagonal blocks, and multiplication without interchanging any factors gives \[ M(x):=L_xBR_x= \begin{pmatrix}qI_r&xV\\xU&x^2UV/q+J\end{pmatrix}. \tag{8}\] In particular \(A=M(1)\), with the rational group-ring inverse \(R_1^{-1}B^{-1}L_1^{-1}\). For general real \(x\), both \(M(x)\) and its inverse lie in the real group ring and depend continuously differentiably on \(x\). No identities \(PU=U\), \(VP=V\), or \(V=U^*\) are required of the chosen integer lifts. ◻

Proof. Put \(N=M^*M\). Continuity of \(M\) and its inverse on the compact interval gives constants \(0<\delta<c\) with \(\delta I\le N(x)\le cI\) for all \(x\). Write \(W=I-N/c\), so \(\|W\|\le\rho=1-\delta/c<1\). Functional calculus gives the uniformly norm-convergent series \[\log N=(\log c)I-\sum_{j\ge1}\frac{W^j}{j}.\] Its derivative series is also uniformly norm-convergent, since \[\left\|\frac{d}{dx}\frac{W^j}{j}\right\| \le\sup_{y\in[0,1]}\|W'(y)\|\rho^{j-1}.\] In the derivative of \(W^j\), cyclicity makes the traces of its \(j\) terms equal; commutation of \(N\) and \(N'\) is not assumed. Using \(W'=-N'/c\) therefore gives \[\frac{d}{dx}\tau_n(\log N) =\frac1c\sum_{j\ge1}\tau_n(W^{j-1}N') =\tau_n(N^{-1}N').\] Since \(N'=M'^*M+M^*M'\), the second summand contributes \(\tau_n(M'M^{-1})\). Cyclicity and \(MN^{-1}=(M^*)^{-1}\) turn the first into \[\tau_n((M^*)^{-1}M'^*) =\overline{\tau_n(M'M^{-1})}.\] Their sum is the asserted real part. This is the tracial logarithmic derivative of (Fuglede and Kadison 1952, Lemma 2), written with the unnormalized trace used here. ◻

Proof. Write \(M=L_xBR_x\). Both shears have inverses obtained by negating their off-diagonal blocks. The ordered product rule gives \[M'M^{-1}=L_x'L_x^{-1} +(L_xB)(R_x'R_x^{-1})(L_xB)^{-1}.\] The matrices \(L_x'L_x^{-1}\) and \(R_x'R_x^{-1}\) have zero diagonal blocks, so each has trace zero. Cyclicity gives trace zero also for the conjugate \((L_xB)(R_x'R_x^{-1})(L_xB)^{-1}\). Lemma 7 now gives zero derivative. ◻

Proof. Lemma 6 has already established integrality and the finite-support rational inverse. Apply Lemma 8 to Equation (8). At \(x=0\) the matrix is \(B=\operatorname{diag}(qI_r,-q^{-1}P-(I_m-P))\). Orthogonality gives \[B^*B=\operatorname{diag}(q^2I_r,q^{-2}P+I_m-P),\qquad \log(B^*B)=2\log q\,\operatorname{diag}(I_r,-P).\] Taking the trace and using shear invariance proves Equation (9); half this trace is Equation (10). The regular representation of the rational group-ring inverse is bounded, so \[\|T_A^{-1}\|^{-2}I\le T_A^*T_A\le\|T_A\|^2I.\] Its logarithm is therefore bounded. In the probability measure convention from the introduction this also gives \[ \int_{(0,\infty)}\log t\,d\mu_A(t) =\frac{2(r-\tau_m(P))\log q}{r+m}. \tag{11}\] ◻

Proof of Theorem 1. Take \(K_2,H_2,a_2,b_2\) from Theorem 2. Apply Lemma 3 with \(d=[K_2:\mathbb F_2]\ge1\) and then Proposition 4. It gives the group \(G_D=(\bigoplus_{H_2}C_3)\rtimes H_2\) and a projection \(P\) of size \(d+1\) whose binary reduction factors through \(d\) columns. Its trace is \(d+\theta\), where \[\theta=\frac13\left(\frac23\right)^{|S|}>0, \qquad D=3^{1+|S|}.\] The odd integer \(D\) clears the coefficients of \(P\). Choose \(k\ge1\) with \(q=2^k\equiv1\pmod D\). Proposition 9, with \(r=d\) and \(m=d+1\), produces \(A\in\operatorname{M}_{2d+1}(\mathbb Z[G_D])\) with a rational group-ring inverse and \[\log\det_{\mathcal N(G_D)}(T_A)=-\theta\log q.\] Since \(q>1\) and \(\theta>0\), this proves \[ 0<\det_{\mathcal N(G_D)}(T_A)=q^{-\theta}<1. \tag{12}\] With \(n=2d+1\), Equation (11) reads \[ \int_{(0,\infty)}\log t\,d\mu_A(t) =-\frac{2\theta\log q}{2d+1}<0. \tag{13}\] It is finite because of the spectral bounds in Proposition 9. Finally, a finite generating set of \(H_2\) together with \(z_1\) generates \(G_D\), since \(z_h=hz_1h^{-1}\). Thus \(G_D\) is finitely generated. ◻