Proof. The algebra is complete for the \(\ell\)-adic topology. For every \(c\in\mathcal A\), the geometric series for \((1-\ell c)^{-1}\) converges, so \(\ell\mathcal A\subseteq J\). The finite algebra \(\mathcal A/\ell\mathcal A\) has nilpotent radical and semisimple quotient a product of finite fields. Consequently \(J^N\subseteq\ell\mathcal A\) for some \(N\), proving that the two topologies agree and establishing the first assertions.

For (1), choose a lift \(v\) of a square root in \(\mathcal A/J\). Both \(v\) and \(2v\) are units. If \(e=v^2-u\in J^k\), replace \(v\) by \(v-e/(2v)\). The new error is \(e^2/(4v^2)\in J^{2k}\). Iteration converges to a square root of \(u\).

In a finite field \(k\) of odd cardinality, the sets of squares and of \(\bar u\) minus the squares each have \((|k|+1)/2\) elements. They intersect, so \(\bar u=\bar a^2+\bar b^2\). For (2), make such a choice in each residue field and lift it to \(a_0,b_0\in\mathcal A\). Then \(c=a_0^2+b_0^2\) is a unit and \(u/c\in1+J\). By (1), write \(u/c=v^2\). Taking \(a=va_0\) and \(b=vb_0\) proves the claim. ◻

Proof. Set \(P=B_1^{-1}B_2\), so \(B_2(x,y)=B_1(x,Py)\). Unimodularity makes \(P\) an integral automorphism. Symmetry of \(B_2\) makes it self-adjoint for \(B_1\), and invariance of both forms gives \(PT=TP\).

Write the characteristic polynomial of \(P\) modulo \(\ell\) as \(\prod_i f_i^{e_i}\), with distinct monic irreducible \(f_i\). Lift this coprime factorization to \(\det(X-P)=\prod_i F_i\) over \(\mathbb Z_\ell\), where \(F_i\bmod\ell=f_i^{e_i}\). For completeness, the lift is obtained successively modulo \(\ell^n\). For two coprime monic residue factors \(f,g\) of degrees \(a,b\), the correction map \[(u,v)\longmapsto ug+fv,\qquad \deg u<a,\quad\deg v<b,\] is an isomorphism onto the polynomials of degree less than \(a+b\): its kernel is zero by coprimality and the dimensions agree. It therefore corrects the product by any prescribed error at the next power of \(\ell\), without changing monicity. Completeness gives the lift, and iteration handles any number of factors. The Chinese remainder idempotents for the \(F_i\), evaluated at \(P\), give \[L=\bigoplus_i L_i,\qquad F_i(P_i)=0, \qquad P_i=P|_{L_i}.\] These idempotents are polynomials in \(P\). They are self-adjoint and commute with \(T\), so the sum is orthogonal for \(B_1\) and \(B_2\) and each \(L_i\) is \(T\)-stable. Each restricted form is unimodular.

On \(L_i\otimes L_j\), let \(\mathcal A_{ij}\) be the algebra generated by \(P_i\otimes1\) and \(1\otimes P_j\). Cayley–Hamilton bounds the degrees in these two generators, so this is a finite commutative \(\mathbb Z_\ell\)-algebra of endomorphisms. Every element is self-adjoint for the restricted form \(B_1\otimes B_1\) and commutes with \(T\otimes T\). The unit \[U_{ij}=P_i\otimes P_j\] relates the two restricted tensor forms. Its inverse belongs to \(\mathcal A_{ij}\), since the inverses of the generators are polynomials in them with integral coefficients.

If \(i=j\), consider a residue field of \(\mathcal A_{ii}\). The relations \(F_i(P_i)=0\) imply that the two generator images \(x,y\) are roots of \(f_i\). They are nonzero because the generators are units. In a finite field, two roots of the same irreducible polynomial over \(\mathbb F_\ell\) are Frobenius conjugates. Hence \(y=x^{\ell^h}\) for some \(h\geq0\), and \[xy=x^{1+\ell^h} =\left(x^{(1+\ell^h)/2}\right)^2.\] Lemma 15 supplies \(a\in\mathcal A_{ii}\) with \(a^2=U_{ii}\). Because \(a\) is self-adjoint, \[(B_1\otimes B_1)(ax,ay)=(B_2\otimes B_2)(x,y) \qquad(x,y\in L_i\otimes L_i).\]

For \(i<j\), pair \(L_i\otimes L_j\) with \(L_j\otimes L_i\). The factor swap identifies their operators and both their forms. On the resulting two equal blocks, use Lemma 15 to choose \(a,b\in\mathcal A_{ij}\) with \(a^2+b^2=U_{ij}\) and set \[M=\begin{pmatrix}a&b\\-b&a\end{pmatrix}.\] Writing \(*\) for adjoint with respect to the first tensor form on the two blocks gives \[M^*M=\begin{pmatrix}U_{ij}&0\\0&U_{ij}\end{pmatrix}.\] This is invertible, and \(M\) commutes with the tensor operator. The diagonal-block maps and these paired-block maps combine to an integral equivariant isometry from the second tensor form to the first. Its inverse has the direction asserted in the proposition. ◻

Proof. The reduction of \(X^m-1\) modulo \(2\) is separable, since its derivative is \(X^{m-1}\) there. Lifting its distinct irreducible factors gives \[\mathcal O=\mathbb Z_2[X]/(X^m-1)=\prod_i\mathcal O_i,\] where each \(\mathcal O_i\) is the integer ring of an unramified extension of \(\mathbb Q_2\). The action \(X\mapsto T\) decomposes \(L=\bigoplus_i M_i\). Every \(M_i\) is free over \(\mathcal O_i\): it is finite and torsion-free over this discrete valuation ring.

Invariance of a form \(B\) says that the adjoint of \(X\) is \(X^{-1}\). Let \(\sigma\) denote this involution of \(\mathcal O\). If it exchanges the factors indexed by \(i\) and \(i'\), then \(M_i\) and \(M_{i'}\) are isotropic and pair perfectly with each other; they are orthogonal to all other factors. Such a pairing has a unique equivariant isometry class for the prescribed modules. Indeed, it identifies \(M_{i'}\) with the \(\mathbb Z_2\)-dual of \(M_i\), with its \(\mathcal O_i\)-action transported by \(\sigma\). An isomorphism on \(M_i\) extends uniquely by its inverse dual to an isometry of the pair.

It remains to treat a nonzero \(M_i\) whose factor is preserved by \(\sigma\). Write \(\mathcal O_i=\mathcal D\) and \(\mathcal D_0=\mathcal D^\sigma\). The involution on \(\mathcal D\) is nontrivial. Otherwise the image \(\zeta\) of \(X\) would satisfy \(\zeta=\zeta^{-1}\); its odd order would force \(\zeta=1\), contrary to the hypothesis. Thus \(\mathcal D/\mathcal D_0\) is an unramified quadratic extension.

We record explicitly how \(B\) becomes an integral hermitian form. The trace pairing of \(\mathcal D\) over \(\mathbb Z_2\) is perfect: modulo \(2\) it is the nondegenerate trace pairing of a finite separable field extension, and its determinant is therefore a unit. Hence there is a unique \(\mathcal D\)-valued form \(h\) such that \[ \operatorname{Tr}_{\mathcal D/\mathbb Z_2}\bigl(a h(x,y)\bigr) =B(ax,y)\qquad(a\in\mathcal D). \tag{13}\] The adjoint relation and symmetry of \(B\) give \[h(ax,by)=a\sigma(b)h(x,y),\qquad h(y,x)=\sigma(h(x,y)).\] Thus \(h\) is hermitian, with linear first variable. It is unimodular: after extending to fraction fields, the \(B\)-dual of \(M_i\) equals its \(h\)-dual. Indeed, if \(B(x,M_i)\subseteq\mathbb Z_2\), then for every \(a\in\mathcal D\) and \(y\in M_i\), \[B(ax,y)=B(x,\sigma(a)y)\in\mathbb Z_2.\] Testing all \(a\) in (13) and using the perfect trace pairing gives \(h(x,M_i)\subseteq\mathcal D\); the converse is immediate. This argument takes no quotient by \(2\).

Every unimodular hermitian form over \(\mathcal D/\mathcal D_0\) is determined by its rank. Here are the details in residue characteristic two. Let \(k/k_0\) be the quadratic residue extension and let \(\bar h\) be the reduced hermitian form. There is a vector of nonzero length. If all lengths vanished, then for all \(x,y\) and \(a\in k\), \[0=\bar h(x+ay,x+ay) =\operatorname{Tr}_{k/k_0} \bigl(\sigma(a)\bar h(x,y)\bigr).\] The finite-field trace pairing would force \(\bar h=0\), contradicting nonsingularity. A lift of a vector of nonzero length has unit length \(u\in\mathcal D_0^\times\). Its line splits off integrally, by the projection \(x\mapsto u^{-1}h(x,v)v\). Its orthogonal complement is again unimodular. Induction diagonalizes \(h\) with all diagonal entries in \(\mathcal D_0^\times\).

Every such unit is a norm from \(\mathcal D^\times\). The residue norm \(k^\times\to k_0^\times\) is surjective. After lifting a residue preimage, improve a norm congruence modulo \(2^n\) to one modulo \(2^{n+1}\) using \[N_{\mathcal D/\mathcal D_0}(1+2^nz) \equiv1+2^n\operatorname{Tr}_{\mathcal D/\mathcal D_0}(z) \pmod{2^{n+1}},\qquad n\geq1.\] The trace modulo \(2\) is surjective, so the needed \(z\) always exists. The corrections converge. Rescaling each diagonal vector therefore makes every diagonal entry \(1\). The ranks of the \(M_i\) depend only on the operator module, so this proves uniqueness on every fixed factor and completes the proof. ◻

Proof. The polar pairing of \(Q_h\) is \(\tfrac12\operatorname{Tr}_{E/F}h(x,y)\). If it annihilates \(x\), testing scalar multiples of \(y\) and using the nondegenerate trace pairing shows \(h(x,y)=0\) for every \(y\). Thus \(Q_h\) is nonsingular. In particular a nonzero hermitian space contains a vector of nonzero length, and such a vector splits off an orthogonal hermitian line.

The quadratic cancellation used below is the reflection argument underlying Witt cancellation (Witt 1937); we include the needed form of it. For a quadratic form \(Q\) with polar form \(b\) and a vector \(z\) of nonzero length, the reflection \[\rho_z(v)=v-\frac{2b(v,z)}{Q(z)}z\] is an isometry, as direct expansion shows. If \(Q(x)=Q(e)=a\ne0\), then \(Q(x-e)+Q(x+e)=4a\). If \(Q(x-e)\ne0\), the reflection in \(x-e\) sends \(x\) to \(e\). Otherwise reflection in \(x+e\) sends \(x\) to \(-e\), and reflection in \(e\) sends \(-e\) to \(e\). It follows that an isometry \(\langle a\rangle\perp V\cong\langle a\rangle\perp W\) can be adjusted to match the two displayed vectors, and then restricts to an isometry \(V\cong W\).

Now write \(E=F(\sqrt d)\). Choose \(x\) of hermitian length \(a\ne0\) for \(h_1\), and let \(y\) be its image under an isometry of the length quadratic forms. The lines \(Ex\) and \(Ey\) are hermitian isometric, and their length quadratic forms are both \[aN_{E/F}=\langle a,-ad\rangle.\] Split these hermitian lines off. Cancel the two nonzero quadratic lines \(\langle a\rangle\) and \(\langle-ad\rangle\) successively by the preceding reflection argument. The complementary hermitian forms again have isometric length forms. Induction on their \(E\)-dimension proves the lemma. ◻

Proof. Take a common odd exponent for the operators. The cyclotomic decomposition of each \(V_j\) is orthogonal: its idempotents are unchanged under \(U_j\mapsto U_j^{-1}\), which is the adjoint involution. Each equivariant local isometry respects these decompositions, and corresponding component dimensions agree.

The eigenvalue-\(1\) component carries an ordinary rational symmetric form. Its local isometries give a rational isometry by Theorem 7. Odd order excludes eigenvalue \(-1\). For any remaining cyclotomic component put \[E=\mathbb Q(\zeta_d),\qquad F=\mathbb Q(\zeta_d+\zeta_d^{-1}), \qquad d>1.\] The component is an \(E\)-vector space, with \(U_j\) acting by \(\zeta_d\). Complex conjugation \(\sigma\) is a nontrivial involution of \(E\). The same trace construction as (13), now over \(\mathbb Q\), gives a unique nonsingular \(E/F\)-hermitian form \(h_j\) satisfying \[\operatorname{Tr}_{E/\mathbb Q}\bigl(a h_j(x,y)\bigr) =\beta_j(ax,y)\qquad(a\in E).\]

An equivariant isometry over a rational completion is linear over the corresponding algebra \(E\otimes_\mathbb Q\mathbb Q_\ell\), or over \(E\otimes_\mathbb Q\mathbb R\), and preserves the trace identity. Nondegeneracy of the extended trace pairing therefore makes it an isometry of the extended hermitian forms. Decomposing \(F\otimes_\mathbb Q\mathbb Q_\ell\) into its completion factors supplies an isometry at every place of \(F\) above \(\ell\); decomposing \(F\otimes_\mathbb Q\mathbb R\) supplies every archimedean place. This also covers finite places where \(E/F\) splits, since the trace pairing of the resulting quadratic étale algebra is still nondegenerate.

Consequently the nonsingular quadratic length forms \(Q_{h_1}\) and \(Q_{h_2}\) are isometric over every completion of \(F\). Hasse–Minkowski over \(F\) gives a global quadratic isometry (Milne 2022, Theorem 25.64(c)). Lemma 18 upgrades its existence to an \(E\)-linear hermitian isometry. Such an isometry preserves the rational symmetric form and intertwines \(U_j\). Combining the cyclotomic components proves the claim. ◻

Proof. Use the integral operator-module isomorphism of Corollary 13 to transport the two forms to a common module with a common operator. Proposition 16 gives the required square isometry at each odd prime. Proposition 17 applies at \(2\), because the prepared operators have odd order and no eigenvalue \(1\); tensor its isometry with itself. Lemma 14 supplies the real isometry, which also tensors. Lemma 19 now gives \(R\). Finally, the matrices of \(R\) and \(R^{-1}\) in integral lattice bases have denominators supported at finitely many primes. At every other prime \(R\) itself is an integral lattice isometry, so choose \(D_\ell=R\) there. ◻

Proof. Complex conjugation preserves the argument-zero Milnor page and exchanges its positive and negative normal directions; let \(C_j\) be its action on middle homology. We check the sign directly in the linking definition of the Seifert form (Milnor 1968). The ambient sphere has dimension \(2s-1\), and the cycles have dimension \(s-1\). Complex conjugation has degree \((-1)^s=1\) on that sphere, and interchanging the two linking cycles has sign \((-1)^{s^2}=1\). If \(y^+\) and \(y^-\) denote the two push-offs, then \[\begin{aligned} S_j(C_jx,C_jy) &=\operatorname{lk}(x,y^-) =\operatorname{lk}(y^-,x)\\ &=\operatorname{lk}(y,x^+)=S_j(y,x). \end{aligned}\] The third equality moves both cycles along the page-normal flow. Thus \(C_j^{\mathsf t}S_jC_j=S_j^{\mathsf t}\). Taking the monodromy of this congruence gives \(C_j^{-1}T_jC_j=T_j^{-1}\), and then \[C_j^{\mathsf t}B_jC_j =S_j^{\mathsf t}T_j^{-r}=S_jT_j^{1-r}=B_j.\] The map on oriented cycles already includes conjugation’s effect on page orientation; the calculation uses the orientation of the ambient sphere for linking. ◻

Proof. Put \(a=R^{-1}D_\ell\) and \(C=C_1^{(2)}\). Multiplying the two maps in (14) gives the exact identity \[ \widehat R^{-1}\widehat D_\ell=a\otimes CaC^{-1}. \tag{15}\] In particular, no compatibility of \(R\) with the reversers is required.

Let \(W_\gamma\) be the \(\gamma\)-eigenspace of \(\tau_1\) over \(\overline{\mathbb Q_\ell}\), and set \[m_\gamma=\dim W_\gamma, \qquad t_\gamma=\det(a|_{W_\gamma}).\] Invariance and nondegeneracy of \(b_1\) give a perfect pairing between \(W_\gamma\) and \(W_{\gamma^{-1}}\). Since \(a\) is an isometry, \[ m_{\gamma^{-1}}=m_\gamma, \qquad t_{\gamma^{-1}}=t_\gamma^{-1}. \tag{16}\] For \(\gamma=1\), this says \(t_1^2=1\); no choice of its sign is needed. The reverser carries \(W_{\eta^{-1}}\) onto \(W_\eta\), so the determinant of \(CaC^{-1}\) on \(W_\eta\) is \(t_{\eta^{-1}}\).

For each \(\lambda\), its entire eigenspace in the tensor square is \[\bigoplus_{\gamma\eta=\lambda}W_\gamma\otimes W_\eta,\] where the sum ranges over all ordered pairs of occurring eigenvalues. By the determinant formula for a tensor product, the determinant of (15) on this space is \[\begin{align*} \prod_{\gamma\eta=\lambda} t_\gamma^{m_\eta}t_{\eta^{-1}}^{m_\gamma} &= \frac{\displaystyle\prod_{\gamma\eta=\lambda}t_\gamma^{m_\eta}} {\displaystyle\prod_{\gamma\eta=\lambda}t_\eta^{m_\gamma}} =1. \end{align*}\] The last equality exchanges \(\gamma\) and \(\eta\) in the denominator. It includes diagonal pairs and all coincidences among product eigenvalues. Finally, tensoring with an identity gives, on each new eigenspace, a product of powers of the already unit determinants. ◻

Proof. Before adjoining \(\mathcal V\), the operator order divides \(3^a\cdot5\cdot7\) for some \(a\). The eigenvalues of \(v\) are \[\zeta_{11}^b\zeta_{13}^c, \qquad 1\le b\le10,\quad 1\le c\le12.\] Their nontrivial \(11\)- and \(13\)-primary components cannot cancel against an eigenvalue of the earlier tensor factors. Thus every final order contains both primes and remains odd. In particular, neither \(1\) nor \(-1\) is a final eigenvalue.

We calculate the signatures supplied by \(u\). Let \[A=\begin{pmatrix}1&-1\\0&1\end{pmatrix}.\] In the convention of Theorem 3, a one-cube Seifert matrix is \(S_c=-A\) (Sakamoto 1974, Corollary 3). Its chosen monodromy is \[t=-A^{-1}A^{\mathsf t} =\begin{pmatrix}0&-1\\1&-1\end{pmatrix}.\] Here the minus sign is the one-variable convention. The form \[J=At^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\] is alternating and \(t\)-invariant. Iterating Theorem 3 gives \(S_u=(-1)^{15}(-A)^{\otimes6}=-A^{\otimes6}\). The sixfold monodromy is \(t^{\otimes6}\), and \(2\cdot2\equiv1\pmod3\), so its symmetric form is \[B_u=S_u(t^{\otimes6})^2=-J^{\otimes6}.\] We compute the signatures of \(J^{\otimes6}\) below; passing to \(B_u\) reverses every sign.

Set \(\omega=e^{2\pi i/3}\) and \(e_+=(1,-\omega)\), \(e_-=(1,-\omega^2)\). These are eigenvectors of \(t\) with eigenvalues \(\omega\) and \(\omega^2\), respectively, and \[J(e_+,\overline{e_+})=i\sqrt3, \qquad J(e_-,\overline{e_-})=-i\sqrt3, \qquad J(e_+,\overline{e_-})=0.\] In the resulting orthogonal tensor basis, a vector with \(k\) copies of \(e_+\) has eigenvalue \(\omega^{-k}\) and length \(27(-1)^{k+1}\); there are \(\binom6k\) such vectors. We obtain the following signatures, up to reversal of every sign: \[\begin{array}{c|c|c|c} \text{eigenvalue}&k&\text{signature}&\text{dimension}\\\hline 1&0,3,6&(20,2)&22\\ \omega&2,5&(6,15)&21\\ \omega^2&1,4&(6,15)&21 \end{array}\] Every eigenspace of this common factor is therefore indefinite.

For any real nondegenerate symmetric form \(b\) invariant under a finite-order operator, its complex bilinear extension pairs the \(\gamma\)-eigenspace perfectly with the \(\gamma^{-1}\)-eigenspace. Consequently \(b(x,\overline y)\) restricts to a nondegenerate hermitian form on each eigenspace. Distinct eigenspaces are orthogonal for this hermitian form.

Apply this observation to each of the three factors in (18). A final eigenspace is the orthogonal direct sum of the tensor products of factor eigenspaces whose eigenvalues have the prescribed product. Each nonzero summand contains one of the indefinite spaces in the table. Diagonalizing the other two hermitian forms shows that its tensor form has both signs: multiply any nonzero diagonal entry from the other factors by a positive and a negative entry from the six-cube factor. Every summand is thus indefinite and has dimension at least \(21\). The same conclusions hold for their orthogonal direct sum. ◻

Proof. Put \(V=\Lambda_1\otimes\mathbb Q\). Since \(\Theta_1\) has finite order, its minimal polynomial is a product of distinct cyclotomic polynomials. Decompose the rational operator space as \[V=\bigoplus_{d\in\mathcal D}V_d, \qquad E_d=\mathbb Q(\zeta_d),\quad F_d=\mathbb Q(\zeta_d+\zeta_d^{-1}), \quad n_d=\dim_{E_d}V_d.\] The action of \(\Theta_1\) gives \(V_d\) its \(E_d\)-vector space structure. Distinct \(V_d\) are orthogonal for \(\beta_1\). By Lemma 23, all \(d\) are odd and divisible by \(11\cdot13\), so \(E_d/F_d\) is quadratic with nontrivial conjugation. The adjoint of multiplication by \(a\in E_d\) is multiplication by \(\overline a\), because the adjoint of \(\Theta_1\) is \(\Theta_1^{-1}\).

For completeness, the nondegenerate trace pairing of \(E_d/\mathbb Q\) defines a unique \(E_d\)-valued form \(h_d\) by \[\mathop{\mathrm{Tr}}_{E_d/\mathbb Q}\bigl(a h_d(x,y)\bigr)=\beta_1(ax,y) \quad(a\in E_d).\] This form is linear in its first argument, conjugate-linear in its second, and satisfies \(h_d(y,x)=\overline{h_d(x,y)}\). These identities follow by testing against every \(a\) in the displayed trace equation. The same equation proves its nondegeneracy. It also shows that an \(E_d\)-linear map preserves \(\beta_1\) exactly when it preserves \(h_d\).

Let \(G\) be the algebraic \(\mathbb Q\)-group of equivariant self-isometries of \((V,\Theta_1,\beta_1)\) with determinant \(1\) on every eigenspace over an algebraic closure. These conditions are defined over \(\mathbb Q\). The preceding description identifies this algebraic group as \[ G=\prod_{d\in\mathcal D}\mathop{\mathrm{Res}}_{F_d/\mathbb Q}\mathop{\mathrm{SU}}(h_d). \tag{19}\] Indeed, commuting with \(\Theta_1\) is exactly \(E_d\)-linearity on each \(V_d\). Each embedding of \(E_d\) into an algebraic closure gives one \(\Theta_1\)-eigenspace, and the determinant on it is the corresponding image of the \(E_d\)-determinant. Requiring all these determinants to be \(1\) gives precisely \(\mathop{\mathrm{SU}}(h_d)\). In particular this requirement removes the unitary determinant tori, and the absence of eigenvalues \(\pm1\) leaves no orthogonal factor.

Every embedding \(F_d\hookrightarrow\mathbb R\) extends to a conjugate pair of embeddings of \(E_d\) into \(\mathbb C\). At this real place the trace identity writes the underlying real form as \(2\operatorname{Re}h_d\). After complexification, its pairing \((x,y)\mapsto\beta_1(x,\overline y)\) on either corresponding operator eigenspace is the associated complex hermitian form. Lemma 23 therefore makes it indefinite, and \(n_d\ge21\). The group \(\mathop{\mathrm{SU}}(h_d)\) is connected, simply connected, and absolutely almost simple over \(F_d\), since over an algebraic closure it becomes \(\mathrm{SL}_{n_d}\) (Milne 2022, Theorem 24.44, Remark 24.46, and Summary 24.66). At each real place it is a noncompact group \(\mathop{\mathrm{SU}}(p,q)\) with \(p,q>0\).

Theorem 8, applied to \(\mathop{\mathrm{SU}}(h_d)\) over \(F_d\) with all archimedean places omitted, now gives density of \(\mathop{\mathrm{SU}}(h_d)(F_d)\) in its finite adelic group (Platonov 1969, Theorem A), with (Platonov 1970); see also (Rapinchuk 2014, Theorem 2.3). Restriction of scalars and the finite product in (19) therefore give \[ \overline{G(\mathbb Q)}=G(\mathbb A_{\mathbb Q,f}). \tag{20}\] Here finite adeles mean the restricted product over all finite places. No isotropy over the number field itself is being assumed; the required noncompactness is supplied at the real places just checked.

We spell out how (20) gives an integral map. For each prime set \[a_\ell=\widehat R^{-1}\widehat D_\ell, \qquad \Lambda_{j,\ell}=\Lambda_j\otimes\mathbb Z_\ell.\] Lemma 22 gives \(a_\ell\in G(\mathbb Q_\ell)\), and \(a_\ell=1\) for all but finitely many \(\ell\). Thus \(a=(a_\ell)\) is a finite adelic point. Define \[K_\ell=\{g\in G(\mathbb Q_\ell):g\Lambda_{1,\ell} =\Lambda_{1,\ell}\}, \qquad K=\prod_\ell K_\ell.\] In a basis of the actual lattice \(\Lambda_1\), the subgroup \(K_\ell\) is \(G(\mathbb Q_\ell)\cap\mathop{\mathrm{GL}}_{\operatorname{rank}\Lambda_1}(\mathbb Z_\ell)\). It is compact and open in \(G(\mathbb Q_\ell)\). To identify their product with an adelic open subgroup, take the schematic closure of \(G\) in \(\mathop{\mathrm{GL}}(\Lambda_1)\) over \(\mathbb Z\). After inverting finitely many primes this is an integral model of the same affine group, and its \(\mathbb Z_\ell\)-points are exactly \(K_\ell\). Thus the \(K_\ell\) agree with the reference integral subgroups defining the restricted product at all but finitely many primes. Their product \(K\) is consequently compact and open in \(G(\mathbb A_{\mathbb Q,f})\). This definition uses the full lattice and requires no integral splitting of its cyclotomic components.

By (20), the nonempty open coset \(aK\) contains a point \(g\in G(\mathbb Q)\). For every prime we can write \(g=a_\ell k_\ell\) with \(k_\ell\in K_\ell\), and hence \[(\widehat Rg)(\Lambda_{1,\ell}) =\widehat R a_\ell(\Lambda_{1,\ell}) =\widehat D_\ell(\Lambda_{1,\ell}) =\Lambda_{2,\ell}.\] A full integral lattice satisfies \[\Lambda_j =\{x\in\Lambda_j\otimes\mathbb Q: x\in\Lambda_{j,\ell}\text{ for every prime }\ell\}:\] in a lattice basis this is the elementary identity \(\mathbb Q\cap\bigcap_\ell\mathbb Z_\ell=\mathbb Z\). Applying this characterization to \(\widehat Rg\) and its inverse shows that \(U=\widehat Rg\) carries \(\Lambda_1\) onto \(\Lambda_2\). It is an equivariant isometry because both factors are such isometries, which proves the proposition. ◻

Proof. The first nonzero-coordinate argument in Lemma 11 shows that each chain has no critical point anywhere in its affine space except the origin. The gradient equations of a sum in disjoint variables separate into the gradient equations of its summands. The pure powers have the same global property. Consequently the only critical point of each \(q_j\), and then of each \(f_j\), is the origin. The orders of \(q_1,q_2\) are two and three. Disjoint copies have no monomials that could cancel one another, and all added terms have degree at least three. The claimed orders follow. More precisely, the \(4E\) balancing squares in each copy of \(p_1\) are the only quadratic terms. Four copies give Hessian ranks \(16E\) for \(f_1\) and \(0\) for \(f_2\), so both ranks are even.

If an irreducible nonunit analytic germ \(a\) occurred with multiplicity at least two in \(f_j\), we could write \(f_j=a^2b\). Every first derivative of \(f_j\) would then vanish on \(V(a)\). A nonunit hypersurface germ in \(\mathbb C^N\) has dimension \(N-1>0\), contradicting the isolated critical point. Thus \(f_j\) is reduced as an analytic germ. The same argument for a repeated nonconstant polynomial factor would force the gradient to vanish on its positive-dimensional affine zero set. The global critical-locus statement just proved excludes this as well, so \(f_j\) is reduced as a polynomial. ◻

Proof. Every eigenvalue of \(T_j^{\mathrm f}\) has the form \[\xi\zeta_{11}^{a}\zeta_{13}^{b}, \qquad 1\leq a\leq10,\quad1\leq b\leq12,\] where the order of \(\xi\) divides \(3^A\cdot5\cdot7\) for some integer \(A\geq1\). The factors of orders \(11\) and \(13\) cannot cancel against each other or against \(\xi\). Thus every eigenvalue has order divisible by both \(11\) and \(13\).

Since \(T_j^{\mathrm f}\) is integral and has finite order, its characteristic polynomial is a product of cyclotomic polynomials \(\Phi_d\), with every occurring index \(d\) divisible by \(143\). For \(d>1\), the elementary identity \[\Phi_d(1)= \begin{cases} p,&d\text{ is a power of the prime }p,\\ 1,&d\text{ has at least two distinct prime factors} \end{cases}\] follows by induction from \(d=\prod_{e\mid d,\ e>1}\Phi_e(1)\): the prime-power divisors already contribute the full prime factorization of \(d\) when \(d\) is not a prime power. Hence \[ \det(1-T_j^{\mathrm f})=1. \tag{24}\] In particular, \(1-T_j^{\mathrm f}\) is an automorphism of the integral middle homology of \(F_j\).

The geometric monodromy acts as \((T_j^{\mathrm f})^{-1}\) in our convention. The identity \(1-(T_j^{\mathrm f})^{-1}=-(T_j^{\mathrm f})^{-1}(1-T_j^{\mathrm f})\) shows that its difference from the identity is also invertible over \(\mathbb Z\). The Wang exact sequence of the Milnor fibration now gives \[H_i(X_j;\mathbb Z)= \begin{cases} \mathbb Z,&i=0,1,\\ 0,&i\geq2. \end{cases}\] Indeed, the only possible additional groups are, equivalently, the kernel and cokernel of \(1-T_j^{\mathrm f}\) on \(H_n(F_j;\mathbb Z)\), and both vanish by (24). Alexander duality (Hatcher 2002, Corollary 3.45) therefore makes \(K_j\) an integral homology \((2n-1)\)-sphere. Its \((n-2)\)-connectivity makes it simply connected. The Hurewicz and Whitehead Theorems (Hatcher 2002, Theorem 4.32 and Corollary 4.33) show that it is a homotopy sphere. Since \(2n-1\geq5\), the high-dimensional Poincaré theorem included in Theorem 9 makes it homeomorphic to \(S^{2n-1}\).

Finally, the homotopy exact sequence of \(\operatorname{int}F_j\to X_j\to S^1\), together with the \((n-1)\)-connectivity of \(\operatorname{int}F_j\), gives \[\pi_1(X_j)\cong\mathbb Z, \qquad \pi_i(X_j)=0\quad(2\leq i\leq n-1).\] This is the simplicity condition for a \((2n-1)\)-knot. ◻

Proof. Theorem 1 supplies the ambient homeomorphism and the stated analytic properties. The calculation in the proof of Lemma 26 gives complex Hessian ranks \(16E\) for \(f_1\) and \(0\) for \(f_2\). Since \(E>0\) by Lemma 10, their coranks satisfy \[\operatorname{corank}_0(f_1)=N-16E<N=\operatorname{corank}_0(f_2).\] ◻

Proof. For holomorphic germs \(f,g:(\mathbb C^n,0)\to(\mathbb C,0)\) with \(n\ge2\) and isolated critical points at the origin, Saeki’s topological \(V\)-equivalence is precisely an ambient homeomorphism of their zero-set germs (Saeki 1989, Definition 3). Saeki’s Corollary 2 identifies this with topological right-left equivalence (Saeki 1989). King’s corollary (King 1978), recalled in (Saeki 1989, sec. 3, p. 28), then gives right equivalence up to conjugation. In the precise source-homeomorphism formulation recorded by Sampaio (Sampaio 2025, proof of Proposition 2.4), there is a homeomorphism germ \(\varphi:(\mathbb C^n,0)\to(\mathbb C^n,0)\) such that \[f=g\circ\varphi \quad\text{or}\quad f=(c_1\circ g\circ c_n)\circ\varphi,\] where \(c_k(z_1,\ldots,z_k)=(\overline z_1,\ldots,\overline z_k)\). Theorem 1 supplies these hypotheses for \(f=f_1\) and \(g=f_2\). Writing \(f_2(z)=\sum_\alpha a_\alpha z^\alpha\) with \(a_\alpha\in\mathbb R\), we have \[(c_1\circ f_2\circ c_N)(z) =\overline{f_2(\overline z)} =\sum_\alpha\overline{a_\alpha}z^\alpha =f_2(z).\] Consequently both alternatives give \(f_1=f_2\circ\varphi\). The multiplicities are \(2\) and \(3\) by Theorem 1, and these integers are unequal both in \(\mathbb Z\) and modulo two. ◻

Proof. Apply the quadratic case of Sampaio’s stabilization (Sampaio 2025, proof of Proposition 2.3). For the homeomorphism germ \(\varphi\) in Corollary 29, the product homeomorphism \(\widetilde\varphi(z,t)=(\varphi(z),t)\) satisfies \[\widetilde f_1=\widetilde f_2\circ\widetilde\varphi.\] It also carries the corresponding zero-set germs to one another. Moreover, \(\nabla\widetilde f_j(z,t)=(\nabla f_j(z),2t)\), so the critical point remains isolated on each side.

By the proof of Lemma 26, the quadratic part of \(f_1\) is \[Q=\sum_{\alpha=1}^{16E}u_\alpha^2\] in the independent balancing-square coordinates, with \(E>0\), whereas \(f_2\) has order three. Hence \[(\widetilde f_1)_*=Q+t^2,\qquad (\widetilde f_2)_*=t^2.\] Their global unit fibres are respectively \[\left\{v\in\mathbb C^{16E+1}: \sum_{\alpha=1}^{16E+1}v_\alpha^2=1\right\} \times\mathbb C^{N-16E}, \qquad \mathbb C^N\times\{-1,1\}.\] The quadric factor is connected. Indeed, writing \(v=x+iy\), its equation becomes \(|x|^2-|y|^2=1\) and \(x\cdot y=0\). The map \((u,y)\mapsto\sqrt{1+|y|^2}\,u+iy\), with \(u\in S^{16E}\) and \(y\perp u\), identifies this quadric with the tangent bundle of the connected sphere \(S^{16E}\). The first unit fibre is therefore connected, whereas the second has exactly two components. This gives the displayed zeroth homology groups and rules out homotopy equivalence. ◻