The enhanced A-polynomial Newton polygon detects torus knots

The unknot and all positive nontrivial torus knots are thin by [F1]: their enhanced curves have respectively the unit equation and the two explicitly calculated monomial terms. These are actual proved formulas, rather than assumptions on an uncomputed torus character variety.

Suppose now that the knot exterior has the complete finite-volume hyperbolic metric. We first construct the necessary SL2 lift on this EXTERIOR; no lift of the holonomy of a closed filling is assumed. Alexander duality in [F3], in degrees one and two, gives $H_1(E_K;\mathbb{Z})=\mathbb{Z}$ and $H_2(E_K;\mathbb{Z})=0$, since the knot is a connected circle. The universal coefficient exact sequence therefore gives $H^2(E_K;\mathbb{Z}\mathrm{/}2)=0$. On a finite triangulation choose a spanning tree and the associated finitely many vertex base paths. A PSL2 representation labels each oriented edge by its based transport; the labels multiply to identity around every triangle. Lift each edge label to SL2, choosing the reverse lift as its inverse. A triangle product is a central sign. The four triangle signs around each tetrahedron sum to zero in $\mathbb{Z}\mathrm{/}2$: after conjugating to one vertex the intermediate edge factors cancel and the remaining signs commute. They are therefore a simplicial two-cocycle. Its cohomology class is zero, so it is the coboundary of a one-cochain. Multiply each edge lift by its corresponding sign. Every triangle product is now identity, and edge-path homotopy gives an actual SL2 representation lifting the specified PSL2 one. This finite argument uses the space cohomology directly and inserts no spin or closed-three-manifold parallelizability theorem.

We also need global image control before the SU2 filling operation, since the enhanced Definition alone omits rather than excludes higher-dimensional images. Take any irreducible component of the finite algebraic representation variety together with one common peripheral eigenline, and its actual generic matrices over its finitely generated function field $F$. Cover the eigenline projective coordinate by its two affine charts. If its peripheral image were two-dimensional, $M,L$ would be algebraically independent by [F5]; in particular a peripheral matrix generically has distinct eigenvalues. Its adjoint system is self-dual by trace. The explicit nonparabolic torus computation in [F5] gives boundary homology dimension two. Here is the rank computation over this arbitrary $F$, without assuming that its transcendence degree is one. Dualizing chains over the field, Lefschetz cap identifies restriction on $H^1$ with the pair connector on $H_2$, up to its degree sign, while restriction is transpose to inclusion on $H_1$. Exactness therefore gives $2=\dim \ker i_{\ast }+\mathrm{rank}{i}_{\ast }=2\mathrm{rank}{i}_{\ast }$; both image and kernel have dimension one. The extension of functionals needed to dualize is permitted by the standing full AC, exactly as in [F5]. For any $\mathbb{C}$-derivation $D:F\to F$, differentiating actual word matrices gives the cocycle $Z(g)=D\rho (g)\rho (g{)}^{-1}$. The finite word pairing with $u=\rho (\delta )-\mathrm{tr}\rho (\delta )I\mathrm{/}2$ annihilates every relator boundary and has peripheral values one common nonzero factor times $(DL\mathrm{/}L,DM\mathrm{/}M)$, as the explicit moving-eigenbasis calculation in [F5] shows. Hence ALL these pairs annihilate the SAME one-dimensional boundary kernel. But extend $M,L$ to a finite transcendence basis of $F\mathrm{/}\mathbb{C}$ and use the two coordinate derivations on that rational function field. Each extends over its finite algebraic extension to $F$: for a minimal polynomial $p(\alpha )=0$, set $D\alpha =-(Dp)(\alpha )\mathrm{/}{p}^{\mathrm{\prime }}(\alpha )$; characteristic zero ensures $p^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$, and the product/quotient rules verify the extension. Their resulting logarithmic pairs are independent, contradicting that common annihilator line. Thus no image component is two-dimensional. Components with generically reducible representations have $L=1$, because their diagonal characters factor through the meridian abelianization. The central-peripheral case has $M,L\in \{1,-1\}$ and is already zero-dimensional. This proves the dimension bound for the entire eigenvalue image.

We supply the classical satellite Alexander formula used in [L2], instead of assuming it from a satellite Definition. Write $X=V\setminus \mathrm{Int}\nu P$ for the pattern exterior and $E_K=X{\cup }_T{E}_C$ for the satellite exterior, with outer meridian ${\mu }_V$ identified with ${\mu }_C$ and preferred outer longitude with ${\lambda }_C$. Let $w\mathrm{\ne }0$ be the winding number, and let $\mathrm{\Lambda }=\mathbb{Z}[t^{\pm 1}]$ act on covers by the satellite's positive-meridian deck transformation. Total linking restricts to $w$ times companion linking on $E_C$, to $w$ on ${\mu }_V$ and zero on the outer longitude. The pulled-back cover of $T$ has $\mathrm{\mid }w\mathrm{\mid }$ cylinder components, so $H_1(\tilde{T})=\mathrm{\Lambda }\mathrm{/}(t^w-1)$ generated by a lift of ${\lambda }_C$; its $H_0$ is the same module. The cover of $E_C$ has $\mathrm{\mid }w\mathrm{\mid }$ components, with module $H_1=\mathrm{\Lambda }{\otimes }_{\mathbb{Z}[s^{\pm 1}]}{A}_C$, where $s$ acts by $t^w$. Its $H_0$ map from $T$ is an isomorphism, because the two linking images are the same subgroup $w\mathbb{Z}$. Its longitude class is zero in $H_1$: a Seifert surface for $C$ lifts, since linking vanishes on all its loops by normal push-off, and its boundary is that lifted longitude. Replace $C$ by the unknot to obtain the pattern closure $P(U)$. Its companion-side cover has $\mathrm{\mid }w\mathrm{\mid }$ contractible components, the same $H_0$ and zero $H_1$. In both lifted Mayer–Vietoris sequences the $H_0$ map is injective, since its companion-side coordinate is an isomorphism. The only $H_1(\tilde{T})$ image is consequently in the SAME pattern-side module. Quotienting that image yields, respectively, $A_K\cong A_{P(U)}\oplus \left(\mathrm{\Lambda }{\otimes }_{\mathbb{Z}[s^{\pm 1}]}{A}_C\right)$. All gluings and the chosen longitude bounds are in the actual peripheral marking; no companion character transfer is used in this module calculation.

We now close the classical knot-trichotomy operations, leaving precisely the actual hyperbolization input [L2]. For any smooth knot $J$, an essential smooth torus in its exterior gives the actual satellite construction in [F10], with nontrivial companion, noncore pattern and the correct preferred peripheral marking. If there is no essential torus, its exterior is embedded-atoroidal by definition. If it is Seifert-fibered, the other new supplier in [F10] recognizes the knot by an actual torus-knot ambient isotopy, including the unknot case. Otherwise $J$ is nontrivial: an unknot isotopy and the explicit tube control give a solid-torus exterior, which has a product Seifert fibration. The residual exterior is compact, connected, orientable and has one torus boundary by the actual JSJ proof1.1. It is irreducible and its boundary is incompressible by [F10]. Thus it has ALL the literal hypotheses of [L2], with embedded atoroidality and no assumed map-capture condition. With the metric conclusion treated as the external geometric premise described in [L2], this is the hyperbolic branch. Consequently a nonhyperbolic nontorus knot must have the displayed essential satellite torus, conditionally only on that precise hyperbolization theorem. The geometric and representation satellite operations have already been discharged by [F8, F9]; no abstract knot trichotomy is being imported in their place.

This proves the forward direction for every signed torus knot as well. Reflection carries its oriented meridian to the inverse positive meridian of the reflected knot: it preserves the carried knot orientation and reverses the ambient normal-plane orientation. It carries its preferred longitude to the preferred longitude, because the degree along the knot remains one and its exterior homology class remains zero; uniqueness in [F3] excludes any meridian shift. Composition of representations with this exterior diffeomorphism therefore identifies the peripheral loci by $(M,L)\mapsto (M^{-1},L)$, preserving irreducibility and all closure dimensions. Their reduced curve equations differ by that Laurent substitution and a unit. Reversing the knot orientation instead inverts both peripheral generators and gives $(M,L)\mapsto (M^{-1},L^{-1})$. These invertible integral changes of exponents preserve containment in an affine line. Finally swapping the two complex coordinates of the standard $S^3$ interchanges the two torus parameters and preserves ambient orientation; the radial torus parameter may be changed by a smooth radial isotopy. Thus, using the mirror and orientation conventions in [F1], every coprime torus knot reduces to the positive case, or to the unknot if one parameter has absolute value one. In the latter case the actual sine-cosine circle isotopy and ambient extension in the unit-slope supplier [F1] give the unknot, with its countable-choice hypothesis supplied by full AC. This is a proof of thinness for mirrors, not an assertion that the displayed positive formula applies unchanged to a negative knot.

The lift works for the local holonomy family, not merely its central point. Use the actual local slice representations constructed in [F2], not just the abstract quotient coordinate. The augmented-coordinate supplier cited in [F2] proves that three conditions transverse to the conjugation orbit give a holomorphic slice, with no large-conjugator identifications, and then gives its holomorphic family after the branched substitution $s=4{\cosh }^2(u\mathrm{/}2)$. The complete cusp group is nonelementary. Indeed, if the full discrete group fixed its cusp point, put that point at infinity. It contains a rank-two translation lattice. An element with dilation of modulus different from one would conjugate a nonzero translation to translations tending to identity, contrary to discreteness. The translations in the group form a discrete rank-two lattice containing the cusp lattice; conjugation preserves it. The linear parts therefore belong to the finite orthogonal symmetry group of this lattice, and the Euclidean quotient of each horizontal horosphere has the same finite positive area. Its hyperbolic volume is the area times ${\int }_0^{\mathrm{\infty }}{z}^{-3}dz=\mathrm{\infty }$, contradicting finite volume. Thus the holonomy has no common invariant line and its SL2 lift is irreducible. The same is true in a neighborhood: otherwise invariant lines for a convergent sequence of reducible tuples have a convergent subsequence in compact ${\mathbb{C}\mathbb{P}}^1$, giving an invariant line at the complete point. Now choose the local edge lifts holomorphically, with their values at the center equal to those of step 1.2. Each triangle product is a continuous central sign, hence stays identity. This yields the actual local family of irreducible SL2 representations. On the augmented curve its chosen common peripheral line has eigenvalues $M(u)={ϵ}_{\mu }{e}^{u\mathrm{/}2}$ and $L(u)={ϵ}_{\lambda }{e}^{v(u)\mathrm{/}2}$ for fixed signs, because their squares are the eigenvalue ratios in [F2].

There are therefore only finitely many exceptional eigenvalue pairs not lying on one-dimensional irreducible-character image curves. Indeed use the finite representation variety and its common-line incidence variety on the two affine eigenline charts; their coordinate rings are finite type and Noetherian. They have finitely many components by [F5]. For each one, step 1.3 makes its image closure either a curve or a point. The union of the curve images from the irreducible-character locus is precisely the enhanced curve of [F4]; images of remaining components of that locus contribute only finitely many points. Thus an actual irreducible SU2 representation either has its eigenvalue pair on the enhanced curve or in that finite exceptional set. This conclusion was not inferred merely from the wording that the Definition discards zero-dimensional components.

For $w\mathrm{\ne }0$, $\mathrm{\Lambda }$ is free of rank $\mathrm{\mid }w\mathrm{\mid }$ over $\mathbb{Z}[s^{\pm 1}]$ via $s=t^w$, with basis $1,t,\dots ,t^{\mathrm{\mid }w\mathrm{\mid }-1}$ after reversing the exponent convention if $w<0$. Extension of scalars is therefore exact, and [F6] supplies a square presentation of its companion summand by the companion Seifert matrix with $s$ replaced by $t^w$. The direct sum in step 1.4 has a block-diagonal square presentation. Taking its determinant gives ${\mathrm{\Delta }}_K(t)\doteq {\mathrm{\Delta }}_{P(U)}(t){\mathrm{\Delta }}_C(t^w)$. With symmetric normalization and value one at $t=1$, the Laurent unit ambiguity disappears. The winding-zero case gives the same polynomial identity with ${\mathrm{\Delta }}_C(1)=1$: its companion-side cover is now a disjoint union of copies of $E_C$, with $H_1=\mathrm{\Lambda }[{\mu }_C]$ and $H_0=\mathrm{\Lambda }$. The torus-side groups are $H_1={\mathrm{\Lambda }}^2$, $H_0=\mathrm{\Lambda }$, and the meridian map onto the companion $H_1$ is an isomorphism while the longitude maps to zero. The same maps hold for an unknot companion. Eliminate the companion meridian generator in both sequences; their identical remaining pattern quotient is $A_K\cong A_{P(U)}$. This proves the Alexander satellite formula also for $w=0$, without claiming that such a pattern can be an instanton L-space satellite.

This nonconstant peripheral arc lies in the enhanced curve of [F4]. To check the dimension condition rather than assume it, take the algebraic irreducible-character component through the complete irreducible point. The ordinary local quotient in [F2] is a smooth complex curve, so that component has dimension one; the local finite lift from PSL2 to SL2 does not change this dimension. Adjoining a peripheral eigenline away from $u=0$ is a finite choice of the two eigenlines, so its component is again one-dimensional. Its algebraic peripheral image closure has dimension at most one, and at least one because $M(u)$ is nonconstant. Hence this is an actual one-dimensional irreducible-character image component, not an isolated representation discarded by the enhanced Definition. The enhanced polynomial vanishes on the entire local arc.

We now complete the nonvertical thinness-to-SU2-averse filling operation. Suppose the thin polynomial has nonconstant support direction $(p,q)$ with $q\mathrm{\ne }0$, so its curve components are $M^p{L}^q=c_j$. Fact [F7] supplies finite multiplicative orders for ALL the constants $c_j$, using the entire thin segment. Let $N$ be divisible by all of them and by $\mathrm{\mid }q\mathrm{\mid }$. For every positive multiple $n$ of $N$, the slope $(np+1)\mathrm{/}(nq)$ is primitive: a common divisor of its coefficients divides $q$, while $np+1$ is congruent to one modulo every divisor of $q$. The slopes are distinct and tend to $p\mathrm{/}q$. If an irreducible SU2 representation extends over such a filling and lies on a curve component, its filling equation gives $1=M^{np+1}{L}^{nq}=M{c}_j^n=M$, impossible: a unitary determinant-one matrix with eigenvalue one is the identity, and a meridian normally generates the knot group, so its representation would be trivial. For any of the finite exceptional pairs from step 2.3 put $c=M^p{L}^q$. If $c$ is a root of unity, enlarge $N$ by its order; then its filling equation again forces $M=1$. If $c$ is not a root of unity, the equation $M{c}^n=1$ holds for at most one integer $n$, since two would give $c^{n-n^{\mathrm{\prime }}}=1$. Delete those finitely many integers. All remaining infinitely many primitive fillings have no irreducible SU2 representation. Every reducible unitary representation is abelian: an invariant line has invariant orthogonal complement, so all matrices are simultaneously diagonal. Hence these are actually SU2-abelian fillings. This proves the entire infinite-filling reduction on the stated nonvertical nonconstant branch. A vertical support direction $q=0$ produces no such sequence; ruling out that case for a nontrivial knot remains among the genuine representation/instanton operations in [L1], rather than being hidden by division by $q$.

The finite torus-pattern obstruction is also elementary once the stated coefficient restrictions in [L1] have been supplied. For the nonzero signed winding $w$ put $v=\mathrm{\mid }w\mathrm{\mid }>0$. Symmetric normalization gives ${\mathrm{\Delta }}_C(t^w)={\mathrm{\Delta }}_C(t^v)$, so the satellite product of step 2.4 uses this positive exponent. Write $a>b\ge 2$ coprime and $g=(a-1)(b-1)\mathrm{/}2$. The formula in [F6], expanded at infinity, is ${\mathrm{\Delta }}_{T_{a,b}}(t)=t^g(1-t^{-ab})(1-t^{-1})\mathrm{/}((1-t^{-a})(1-t^{-b}))$. Modulo terms of degree at most $g-a$, its expansion is ${\sum }_{i=0}^{\lfloor a\mathrm{/}b\rfloor }(t^{g-ib}-t^{g-ib-1})$. Thus, for $0\le v<a$, its coefficient $c_{g-v}$ is $1$ if $v=0(\mathrm{m}\mathrm{o}\mathrm{d}b)$, $-1$ if $v=1(\mathrm{m}\mathrm{o}\mathrm{d}b)$, and zero otherwise. Suppose the companion's normalized polynomial has leading terms $t^h-t^{h-1}+O(t^{h-2})$ and the satellite's nonzero coefficients alternate with absolute value one, exactly the two externally supplied instanton conclusions in [L1]. By step 2.4 its coefficient in degree $g+hv-v$ is $c_{g-v}-1$. If $v=1(\mathrm{m}\mathrm{o}\mathrm{d}b)$ this is $-2$, impossible. If $v=qb+r$ with $2\le r<b$, it is $-1$, while the preceding nonzero coefficient is the $-1$ in degree $g+hv-qb-1$: all intervening torus coefficients vanish and no lower companion term reaches those degrees. These consecutive negative coefficients contradict alternation. Consequently $b$ divides $v$ whenever $0<v<a$. This argument is a proved finite polynomial implication and uses the externally supplied instanton properties, without proving their gauge-theoretic foundations.

A nonzero Laurent polynomial with collinear integral support has the form $M^a{L}^{b}P(M^r{L}^s)$, where $(r,s)$ is a primitive nonzero integral vector and $P$ is a one-variable Laurent polynomial; subtract one support point and divide all support differences by their common primitive direction. If the support is a single point, it has no zero in $({\mathbb{C}}^{\ast }{)}^2$ and already cannot vanish on the arc in step 3.1. Otherwise $P$ has finitely many nonzero complex roots. A connected local arc on its zero set has $M^r{L}^s$ equal to one fixed root. Applying step 2.2 and differentiating at $u=0$ gives $r+s\tau =0$. Since $\mathrm{Im}\tau \mathrm{\ne }0$, this forces $s=0$ and then $r=0$, a contradiction. Thus the enhanced polynomial of a finite-volume hyperbolic knot is not thin. This proves the complete hyperbolic exclusion without treating a PSL2 filling representation as an already lifted SL2 representation.

In particular the torus-pattern configuration required by the minimal-satellite argument is incompatible with those coefficient restrictions. If $ab=k{v}^2$ for an integer $k\ge 1$, then $v<a$, since otherwise $ab\ge a^2>ab$. Step 3.3 makes $b\mid v$, whence $a=kb(v\mathrm{/}b{)}^2$ contradicts $\gcd (a,b)=1$ and $b\ge 2$. Mirroring a torus pattern does not change its symmetric Alexander polynomial, so the same contradiction applies after the prescribed mirror reduction. Also $v=1$ is already excluded as soon as both nontrivial pattern and companion have leading terms $t^g-t^{g-1}$ and $t^h-t^{h-1}$: their product has second leading coefficient $-2$. Thus the remaining satellite obstacles in [L2] are the actual geometric/representation transfers, nontriviality/nonzero-winding and genus reduction, not an unauthored Alexander product or finite torus-coefficient calculation.

We give the complete final assembly, retaining precisely its premises [L1, L2]. Suppose a nontrivial thin knot were not a torus knot. The unit/vertical exclusions required in [L1] make its support a nonconstant segment of finite slope; step 3.2 then supplies actual infinitely many SU2-abelian fillings. The SU2-averse and instanton conclusions required in [L1] make it or its mirror fibered with the alternating-unit and adjacent-top Alexander properties. Those three properties, genus and thinness are invariant under mirroring, so the later argument uses them for either choice. It is not hyperbolic by step 4.1. The actual classical cases of step 1.5, conditional only on [L2], consequently make it a satellite. Choose a thin satellite of least genus and an actual essential companion torus. Fact [F8] gives nonzero signed winding, nontrivial pattern closure, and $g(K)=g(P(U))+\mathrm{\mid }w\mathrm{\mid }g(C)$; its adjacent-top hypothesis is exactly the supplied instanton property, not inferred from thinness alone.

Fact [F9] now makes $P(U)$ and $C$ thin, of the original finite slope and that slope divided by $w^2$, respectively. Both are nontrivial, and their genera are positive; by [F8]'s formula each is strictly smaller than $g(K)$, even if $\mathrm{\mid }w\mathrm{\mid }=1$. Each is nonhyperbolic by step 4.1. By step 1.5 and the exact geometric input described in [L2] each is therefore torus or satellite, and least-genus choice excludes the satellite option. Thus they are torus knots. Their explicit signed torus formulas in [F1] and step 2.1 identify the two slopes. After a common mirror if necessary, and parameter swaps/reversals, write $P(U)=T_{a,b}$ and $C=T_{c,d}$ with $a>b\ge 2$, $c>d\ge 2$, $\gcd (a,b)=\gcd (c,d)=1$. Their slopes are $ab$ and $cd$, so $ab=cd{w}^2$. The common mirror preserves all polynomial and genus properties used here; no claim that both mirror choices are instanton L-space knots is needed.

The companion torus polynomial has the explicit leading terms $t^h-t^{h-1}$ by step 3.3, without any extra instanton assertion about that companion. The satellite itself has alternating unit coefficients from the exact input in [L1]. Put $v=\mathrm{\mid }w\mathrm{\mid }>0$. Symmetry gives ${\mathrm{\Delta }}_C(t^w)={\mathrm{\Delta }}_C(t^v)$ and the slope equation is $ab=cd{v}^2$, so step 4.2 applies with this POSITIVE $v$ and gives a contradiction. This completes the converse assembly with the precise externally supplied or explicitly conditional premises [L1, L2]. All classical geometry, signed peripheral representation transfer, exceptional-value filling control and Alexander coefficient operations used in that assembly are supplied locally or by [F1] through [F10]. This records the finite conditional local derivation; the full classification, including the residual input, follows independently from [F11].

By [F11], the original full iff holds for every smooth knot in the Statement, with the unknot included. The conventions comparison in [F11] identifies its thinness condition with the present one, so no component, constant case or mirror is omitted. The preceding local arguments are preserved; the full completion is source-qualified by the external classification.