Ambiently homeomorphic isolated hypersurfaces of multiplicities two and three

Proof. Form the integer matrix with column b consisting of (v_p(A_b))_p and (v_p(R_b))_p, with separate rows for the two families of prime valuations. It is enough to find an integer kernel vector whose coordinate sum is nonzero.

Here is the finite certificate for this assertion. Retain only the columns for which each of

has no prime factor greater than 4450. Repeatedly delete any column that is the only column supported on some nonzero row; discard zero rows. The resulting matrix M and its augmentation \widetilde M by a final row of ones satisfy

and

Appendix A gives a complete program, using only integer arithmetic and elimination modulo the prime 1009, that verifies these counts and (3.2). Its sieve stores a prime divisor of every integer greater than one. Its six signed factor entries are precisely the two valuation columns above. In the elimination, a normalized pivot is stored for each leading row, so the number of stored pivots is the rank. Thus the certificate is a finite check of the displayed integer matrix, without numerical approximation.

Since M has 867 rows, (3.2) gives an 868-column square submatrix H of \widetilde M whose integer determinant is nonzero. The vector

is integral, satisfies the valuation equations on these columns, and has coordinate sum \det H. Extend z by zeros to all columns b\in\mathcal B, and change its overall sign if necessary. This gives e_b with E=|\det H|>0. Every omitted valuation row is zero on the retained columns, so all prime valuations of both products in (3.1) vanish. A positive rational number with this property is 1, proving the Lemma.□

Proof. Set c_1=c and successively perform the divisions

stopping at the first zero remainder. Each nonzero remainder is a positive integer strictly smaller than its predecessor, so this procedure terminates. Give x_i weight c_i/D. Then a_ic_i+c_{i+1}=D for each nonterminal term, and a_\ell c_\ell=D for the terminal term. All a_i\ge2. Consequently each nonterminal monomial has ordinary degree a_i+1\ge3. The terminal exponent divides the odd integer D and exceeds 2, so it too is at least 3.

To prove isolation, suppose a nonzero point is critical and let j be the first index with x_j\ne0. If j=\ell, the derivative in x_j is a_\ell x_\ell^{a_\ell-1}\ne0. Otherwise that derivative forces x_{j+1}=0, because the preceding coordinate is zero. The derivative in x_{j+1} is then x_j^{a_j}\ne0: its other term vanishes since a_{j+1}\ge2. Both cases contradict criticality. The same reasoning applies to the tail, which has the stated weights.□

Proof. Fix the integers in Lemma 3.1, and put D=3d. The four weights below are chosen for the triple-angle identity in the spectral comparison; their signed counts will give side 2 an excess of 4E variables, to be balanced by squares on side 1. Start with two lists of weights prescribed by the following signed counts; a positive entry means that many variables on side 2, and a negative entry means its absolute value on side 1:

All these weights are positive and less than 1/2. Their numerators with denominator D are 3b,b,d-b,d+b, respectively.

Realize each occurrence of a starting weight c/D by its own copy of the chain in Lemma 3.2. On the opposite side insert an independent copy of the tail of that chain, doing nothing there if the tail is empty. The added tail weights occur equally on both sides. All variables used for different copies are disjoint. Thus this operation realizes the signed weight counts exactly by sums of isolated chains, and all its monomials have degree at least 3.

The variable excess on side 2 is 4E. Add 4E independent squares to side 1, giving both sides the same number of variables. Every starting multiplicity in (3.4) is even, as is the number of copies of each tail, and 4E is even. Both variable counts are therefore even already. Finally add two independent pure cubes to each side, and call the resulting polynomials p_1,p_2. They have the same even number s_0 of variables. Sums in disjoint variables have critical loci equal to products of the individual critical loci, so both critical points are isolated. All coefficients can be taken to be 1. Squares occur only in p_1, and the common cubes occur in both polynomials, proving the asserted ordinary orders.

We next compare the spectra. In the convention of Theorem 2.3, write

All exponents lie in (2D)^{-1}\Z. Reduction modulo 2 puts their multiplicities in the group algebra of the finite cyclic group (2D)^{-1}\Z/2\Z. Its characters are \alpha\mapsto e^{\pi i k\alpha}, for integer k. By Fourier inversion, it suffices to prove that P_{p_1} and P_{p_2} take the same value at every such character. When an individual quotient in (3.5) is indeterminate, evaluate along t^u=\exp((\pi i k+\varepsilon)u) and let \varepsilon\to0.

Case 1: k=2m is even. For a weight w, its individual factor has value

The second value follows by taking the ratio of the two first derivatives in \varepsilon. All these values are nonzero, so common weights can be canceled.

The first column of (3.4) has exact reduced denominator d, and the other three have exact reduced denominator 3d, since 3\nmid b. Within either denominator group, therefore, all factors are exceptional in (3.6), or none are. The exceptional signed product for the first group is \prod_b A_b^{2e_b}=1. For the other group it is

The nonexceptional signs disappear because all signed counts are even. A square has exceptional value 1 and nonexceptional value -1; its total contribution is 1, since there are 4E squares. This proves equality in Case 1, including k=0.

Case 2: k is odd and 3\nmid k. Each factor is

All chain weights have a reduced denominator that is a nontrivial power of 3. None of their cotangents has a zero or pole in this case: either would imply that this denominator divides k. Squares also have finite nonzero factors. Hence common weights may again be canceled.

Put C(u)=\cot(\pi ku/2). The exact signed identity we need is

Indeed, with x=\pi kb/(6d) the usual triple-angle product gives

The shifts by k\pi/3 give the same three factors modulo \pi as the shifts by \pi/3, because 3\nmid k. Since k is odd,

Substitution proves (3.7). Its 2e_b-th power is exactly the assertion that the signed cotangent product in row b of (3.4) equals 1. Each square contributes C(1/2)=\cot(\pi k/4)\in\{1,-1\}, whose 4E-th power is 1. Finally the powers of i are the same because the two complete polynomials have the same number s_0 of variables. This proves equality in Case 2.

Case 3: k is odd and 3\mid k. A pure cube has spectrum t^{1/3}+t^{2/3}, whose value is -1+1=0. Before the final common cubes were inserted, both sides were already isolated sums of chains and squares, so their spectra were finite polynomials. Multiplication by a cube spectrum therefore makes both character values zero. This argument uses those finite polynomials, and does not cancel a zero against a pole of an individual weight quotient. It completes the Fourier comparison.

Weighted homogeneous monodromy has finite order and hence is semisimple. Theorem 2.4 now converts the equality of spectra modulo 2 in the common dimension into an isomorphism of real Seifert forms. The same equality gives equal eigenvalue multiplicities. All summands are chains, including the pure powers, so Theorem 2.5 gives an isomorphism of integral monodromy modules. These are separate isomorphisms; no compatibility between them is asserted here.

Finally, every nonsquare weight has denominator dividing D. Each square has spectrum t^{1/2}, so the 4E squares together contribute t^{2E}. All spectral exponents of either seed therefore belong to D^{-1}\Z. Their monodromy eigenvalues are D-th roots of unity. Semisimplicity then gives M_{p_j}^D=1 for j=1,2, as claimed.□

Proof. The new terms preserve isolation and ordinary orders. Formula (3.5) multiplies each seed spectrum by the same two pure-power spectra, preserving equality modulo 2. Theorems 2.4 and 2.5 consequently apply again. Each new eigenvalue has the form

Because D, 5, and 7 are pairwise coprime, its order divides 35D and is divisible by both 5 and 7. In particular it is not 1. Finite-order weighted homogeneous monodromy is semisimple, so the same bound holds for the order of the operators themselves.□

Proof. Suppress the subscript. The defining equality S^{\mathsf t}=ST implies

It also gives T^{\mathsf t}BT=B. Integrality and unimodularity follow from those of S and T. An isometry of S intertwines S^{-1}S^{\mathsf t}, and therefore preserves B as well. For tensor products, use the same r modulo a common odd exponent and the identity

The overall Seifert signs supplied by the integral Thom–Sebastiani theorem depend only on the factor dimensions [30]; they are thus the same on the two sides of every comparison here.□

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 \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

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

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 \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

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 \F_\ell are Frobenius conjugates. Hence y=x^{\ell^h} for some h\geq0, and

Lemma 4.2 supplies a\in\mathcal A_{ii} with a^2=U_{ii}. Because a is self-adjoint,

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 4.2 to choose a,b\in\mathcal A_{ij} with a^2+b^2=U_{ij} and set

Writing * for adjoint with respect to the first tensor form on the two blocks gives

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

where each \mathcal O_i is the integer ring of an unramified extension of \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 \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 \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

The adjoint relation and symmetry of B give

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\Z_2, then for every a\in\mathcal D and y\in M_i,

Testing all a in (4.2) 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,

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

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 [38]; 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

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

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 2.6. Odd order excludes eigenvalue -1. For any remaining cyclotomic component put

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 (4.2), now over \Q, gives a unique nonsingular E/F-hermitian form h_j satisfying

An equivariant isometry over a rational completion is linear over the corresponding algebra E\otimes_\Q\Q_\ell, or over E\otimes_\Q\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_\Q\Q_\ell into its completion factors supplies an isometry at every place of F above \ell; decomposing F\otimes_\Q\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 [16, Theorem 25.64(c)]. Lemma 4.5 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 3.4 to transport the two forms to a common module with a common operator. Proposition 4.3 gives the required square isometry at each odd prime. Proposition 4.4 applies at 2, because the prepared operators have odd order and no eigenvalue 1; tensor its isometry with itself. Lemma 4.1 supplies the real isometry, which also tensors. Lemma 4.6 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 [17]. 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

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

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 (5.1) gives the exact identity

In particular, no compatibility of R with the reversers is required.

Let W_\gamma be the \gamma-eigenspace of \tau_1 over \overline{\Q_\ell}, and set

Invariance and nondegeneracy of b_1 give a perfect pairing between W_\gamma and W_{\gamma^{-1}}. Since a is an isometry,

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

where the sum ranges over all ordered pairs of occurring eigenvalues. By the determinant formula for a tensor product, the determinant of (5.2) on this space is

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

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

In the convention of Theorem 2.2, a one-cube Seifert matrix is S_c=-A [30, Corollary 3]. Its chosen monodromy is

Here the minus sign is the one-variable convention. The form

is alternating and t-invariant. Iterating Theorem 2.2 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

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

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:

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 (5.5). 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\Q. Since \Theta_1 has finite order, its minimal polynomial is a product of distinct cyclotomic polynomials. Decompose the rational operator space as

The action of \Theta_1 gives V_d its E_d-vector space structure. Distinct V_d are orthogonal for \beta_1. By Lemma 5.3, 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/\Q defines a unique E_d-valued form h_d by

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 \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 \Q. The preceding description identifies this algebraic group as

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 \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\R extends to a conjugate pair of embeddings of E_d into \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 5.3 therefore makes it indefinite, and n_d\ge21. The group \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} [16, Theorem 24.44, Remark 24.46, and Summary 24.66]. At each real place it is a noncompact group \SU(p,q) with p,q>0.

Theorem 2.7, applied to \SU(h_d) over F_d with all archimedean places omitted, now gives density of \SU(h_d)(F_d) in its finite adelic group [25, Theorem A], with [26]; see also [27, Theorem 2.3]. Restriction of scalars and the finite product in (5.6) therefore give

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 (5.7) gives an integral map. For each prime set

Lemma 5.2 gives a_\ell\in G(\Q_\ell), and a_\ell=1 for all but finitely many \ell. Thus a=(a_\ell) is a finite adelic point. Define

In a basis of the actual lattice \Lambda_1, the subgroup K_\ell is G(\Q_\ell)\cap\GL_{\operatorname{rank}\Lambda_1}(\Z_\ell). It is compact and open in G(\Q_\ell). To identify their product with an adelic open subgroup, take the schematic closure of G in \GL(\Lambda_1) over \Z. After inverting finitely many primes this is an integral model of the same affine group, and its \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(\A_{\Q,f}). This definition uses the full lattice and requires no integral splitting of its cyclotomic components.

By (5.7), the nonempty open coset aK contains a point g\in G(\Q). For every prime we can write g=a_\ell k_\ell with k_\ell\in K_\ell, and hence

A full integral lattice satisfies

in a lattice basis this is the elementary identity \Q\cap\bigcap_\ell\Z_\ell=\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 3.2 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 \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

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

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

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 \Z. The Wang exact sequence of the Milnor fibration now gives

Indeed, the only possible additional groups are, equivalently, the kernel and cokernel of 1-T_j^{\mathrm f} on H_n(F_j;\Z), and both vanish by (6.4). Alexander duality [9, 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 [9, 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 2.8 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

This is the simplicity condition for a (2n-1)-knot.□

Proof. Theorem 1.1 supplies the ambient homeomorphism and the stated analytic properties. The calculation in the proof of Lemma 6.1 gives complex Hessian ranks 16E for f_1 and 0 for f_2. Since E>0 by Lemma 3.1, their coranks satisfy

□

Proof. For holomorphic germs f,g:(\C^n,0)\to(\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 [28, Definition 3]. Saeki’s Corollary 2 identifies this with topological right-left equivalence [28]. King’s corollary [12], recalled in [28, Section 3, p. 28], then gives right equivalence up to conjugation. In the precise source-homeomorphism formulation recorded by Sampaio [31, proof of Proposition 2.4], there is a homeomorphism germ \varphi:(\C^n,0)\to(\C^n,0) such that

where c_k(z_1,\ldots,z_k)=(\overline z_1,\ldots,\overline z_k). Theorem 1.1 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\R, we have

Consequently both alternatives give f_1=f_2\circ\varphi. The multiplicities are 2 and 3 by Theorem 1.1, and these integers are unequal both in \Z and modulo two.□

Proof. Apply the quadratic case of Sampaio’s stabilization [31, proof of Proposition 2.3]. For the homeomorphism germ \varphi in Corollary 6.4, the product homeomorphism \widetilde\varphi(z,t)=(\varphi(z),t) satisfies

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 6.1, the quadratic part of f_1 is

in the independent balancing-square coordinates, with E>0, whereas f_2 has order three. Hence

Their global unit fibres are respectively

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.□