Alexander polynomials from Seifert matrices

Put $n=2g$ and $M(t)=V-t{V}^T$. By [F1] this is the positive-push matrix in the Statement, and by [F4] $A_K$ is isomorphic to the cokernel of the square presentation matrix $M(t)$. The definition of $E_0$ uses the $n\times n$ minors; the only such minor of an $n\times n$ matrix is its determinant. Thus the presentation gives $E_0(\mathrm{coker}M)=(\det M)$, including $n=0$, where the sole empty minor is $1$. Presentation and isomorphism independence [F3] yields $E_0(A_K)=(\det M)$.

Evaluating at $t=1$ gives $\det M(1)=\det (V-V^T)=\det J_F=\pm 1$ by [F5]. Since $E_0(A_K)=(\det M)$ and ${\mathrm{\Delta }}_K$ is a generator of that principal ideal, ${\mathrm{\Delta }}_K=u\det M$ for a unit $u=\pm t^k$ by [F2]; every such unit evaluates to $\pm 1$ at $t=1$. Hence ${\mathrm{\Delta }}_K(1)=\pm 1$.

If the basis is changed by $P\in {\mathrm{G}\mathrm{L}}_n(\mathbb{Z})$, then $M^{\mathrm{\prime }}(t)=P^{T}M(t)P$ and $\det M^{\mathrm{\prime }}=(\det P{)}^2\det M=\det M$. Under an ambient isotopy, [F6] gives this same transported-matrix relation and [F7] preserves the Alexander ideal up to its unit generator. Thus the asserted ideal-generator relation is compatible with the stated choices.

Fix the arbitrary knot, surface, and basis in the Statement, put $n=2g$, and set $M=V-t{V}^T$. By [F2] the cokernel of $M$ is $A_K$, and [F3] gives $E_0(A_K)=(\det M)$. By [F4], ${\mathrm{\Delta }}_K$ also generates this ideal. The local lemma gives $(\det M)(1)=\pm 1$, so $\det M\mathrm{\ne }0$. If ${\mathrm{\Delta }}_K=a\det M$ and $\det M=b{\mathrm{\Delta }}_K$ for $a,b\in \mathrm{\Lambda }$, then $\det M=ba\det M$; since $\mathrm{\Lambda }$ is a domain and $\det M\mathrm{\ne }0$, $ba=1$. Thus $a$ is a unit, and [F4] identifies it as $\pm t^k$. Hence $\det M\doteq {\mathrm{\Delta }}_K$.

If the basis is changed by $P\in {\mathrm{G}\mathrm{L}}_n(\mathbb{Z})$, then $M^{\mathrm{\prime }}=P^{T}MP$ and $\det M^{\mathrm{\prime }}=(\det P{)}^2\det M=\det M$. Under an orientation-preserving ambient isotopy, [F5] gives the same transported-matrix relation and [F6] preserves ${\mathrm{\Delta }}_K$ up to a unit. Since the original $K,F$, and basis were arbitrary, the formula holds for every surface and basis in the Statement and is compatible with the oriented knot invariant.

By [A1] and [F8], the countable-choice hypotheses of the surface and genus suppliers hold. Fix any connected Seifert surface $F$ for $K$, of genus $g(F)$, and put $n=2g(F)$ and $p(t)=\det (V-t{V}^T)$. By [F1] and [F2], $p(1)=\det (V-V^T)=\pm 1$ and any generator ${\mathrm{\Delta }}_K$ is a unit associate of $p$. Thus ${\mathrm{\Delta }}_K(1)=\pm 1$.

Set $M(t)=V-t{V}^T$. Directly, $M(t{)}^T=V^T-tV=-tM(t^{-1})$. Taking determinants and using $n=2g(F)$ even gives $p(t)=\det M(t)=\det M(t{)}^T=(-t{)}^{n}p(t^{-1})=t^{n}p(t^{-1})$. Therefore $t^{-g(F)}p(t)$ is invariant under $t\mapsto t^{-1}$. Since ${\mathrm{\Delta }}_K$ is a unit associate of $p$, it is reciprocal up to a unit, and multiplying the centered representative by the sign of $p(1)$ makes its value at one equal to $1$.

Choose a surface $F$ attaining $g(F)=g(K)$ by [F4]; [A1] and [F8] discharge its countable-choice hypothesis. Set $n=2g(K)$, $M(t)=V-t{V}^T$, and $p(t)=\det M(t)$ for its matrix; [F3] gives the $n\times n$ size. Each entry of $M(t)$ is a polynomial of degree at most one with exponents in $\{0,1\}$. Every determinant term is a product of $n$ entries, so the exponents of $p$ lie between $0$ and $n$; hence $\mathrm{deg}p\le n=2g(K)$. By [F1], $p$ is a unit associate of ${\mathrm{\Delta }}_K$, and [F6] makes degree invariant under that shift, so $\mathrm{deg}{\mathrm{\Delta }}_K\le 2g(K)$. Applying the centering calculation of step 2.1 to this minimum-genus surface gives a symmetric representative supported in $[-g(K),g(K)]$.

Suppose two unit associates $q$ and $\tilde{q}=\epsilon t^{k}q$, with $\epsilon \in \{\pm 1\}$, both satisfy the exact equalities $q(t^{-1})=q(t)$ and $\tilde{q}(t^{-1})=\tilde{q}(t)$, and both have value $1$ at $t=1$. Their exact symmetry gives $\epsilon t^{-k}q(t)=\epsilon t^{k}q(t)$. Since $q\mathrm{\ne }0$ and the Laurent ring is a domain, $t^{2k}=1$; distinct Laurent monomials are unequal, so $k=0$. Evaluation at one then gives $\epsilon =1$. Thus the normalized symmetric representative is unique. Since ${\mathrm{\Delta }}_K(1)\mathrm{\ne }0$, the denominator $1-t$ has a simple zero at one that does not cancel, so $D_K$ is a proper fraction with a simple pole there.

The properties in the Statement are unchanged by a unit multiple: evaluation at one changes at most by sign, reciprocity is stated up to units, and exponent width is shift-invariant. By [F7], the Alexander polynomial and transported Seifert determinant class are unchanged up to units by orientation-preserving ambient isotopy. Hence the conclusions depend only on the oriented knot type.

By [A1] and [F3], the AC${}_{\omega }$ hypothesis of [F4] holds. Apply [F4] to choose connected Seifert surfaces $F$ for $K$ and $G$ for $J$. By [F5], choose ordered integral bases for $H_1(F;\mathbb{Z})$ and $H_1(G;\mathbb{Z})$.

Reduce each given trivial-arc ball to a small coordinate ball as follows. Write its oriented ball-arc parametrization as $e:D^3\to B$, with diameter on the first coordinate axis. On $B$ prescribe the inward field $X(e(x))=D{e}_x(-x)$. Extend it smoothly to $S^3$, tangent to the whole knot: near each arc endpoint take coordinates $(s,u,v)$ with the knot $u=v=0$ and the ball boundary $s=0$. The two normal components on the ball side vanish on the axis, so each is $ua+vb$ with smooth coefficients, by integrating normal derivatives. Extend these coefficients and the tangential component across $s=0$; away from the endpoints extend across the sphere in ordinary boundary charts. A finite partition of unity patches these extensions, all agreeing with $X$ on $B$, and a cutoff makes the field zero away from a neighborhood of $B$ while preserving tangency to the knot. This is the tangent-field extension construction in the proof of [F2], now applied at the boundary of the given ball. By [F12] its flow $H_t$ exists for every fixed finite time interval; uniqueness gives $H_t(e(x))=e(e^{-t}x)$ for $x\in D^3$. The flow preserves the oriented knot setwise, and for large finite $T$ its image of $B$ is an arbitrarily small coordinate ball in the interior chart $e$. Transport the boundary parametrization as $H_t\circ e$ throughout. The fixed gluing map in these parametrizations is unchanged, so the flows identify the cut-and-glued pairs and their boundary collars throughout the isotopy, just as in the representative-transport argument in the proof of [F2]. Thus the original connected sum is ambient isotopic to one using small coordinate balls.

By step 2.1 and [F2], it suffices to use standard round coordinate balls whose trivial-arc neighborhoods lie in boundary collars of $F$ and $G$; such collars exist by [F10]. Their complementary pieces are standard closed 3-balls. Remove a small boundary half-disk from each surface and join the exposed boundary arcs by the standard band in the gluing collar. The resulting embedded oriented surface $S$ has boundary $K\mathrm{\#}J$. By [F11], each summand surface is a thickening of a finite graph. Include a path from each attaching arc to its graph as a tree edge, then retract the band to its core interval. The two graphs are joined by one further bridge, so no cycle is added. Their old cycle bases therefore form a basis of $H_1(S;\mathbb{Z})$.

Use the union basis from step 3.1. Each basis cycle and its positive push-off can be kept in the interior of its own complementary 3-ball. By [F9], a cycle in one ball bounds an integral $2$-chain there; that chain is disjoint from every cycle in the other ball. Thus both cross-linking entries between the two summands are zero. The within-summand entries are unchanged, so the Seifert matrix of $S$ is $V_K\oplus V_J$.

The determinant of the block matrix satisfies $\det ((V_K\oplus V_J)-t(V_K\oplus V_J{)}^T)=\det (V_K-t{V}_K^T)\det (V_J-t{V}_J^T)$. Apply [F7] to $S,F,G$ to identify these three determinants with generators of their respective $E_0$ ideals, up to units. By step 2.1 and [F1], [F2], and [F8] this identity gives the formula for the specified connected sum and is unchanged by the datum, surface, basis, or representative choices.

Each plumbing has Euler characteristic $0+0-1=-1$, using a cell structure respecting the common square, since each annulus has Euler characteristic $0$ and the square has Euler characteristic $1$. Its boundary is one circle by the stipulated gluing, so $\chi =1-2g$ gives genus one. By [F3], $H_1(F;\mathbb{Z})$ is free of rank two. Choose any ordered basis $u,v$; [F2] makes its intersection matrix $J_0$ skew-symmetric and unimodular, hence $\det J_0=1$. Write the two core classes $a,b$ in this basis as the columns of an integer matrix $P$. The plumbing gives $a{\cdot }_{F}b=1$, so their intersection matrix is $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=P^T{J}_{0}P$. Taking determinants gives $1=(\det P{)}^2$, hence $\det P=\pm 1$; therefore $a,b$ is an integral basis.

Equally oriented parallel cores of an annulus transport to its two induced boundary orientations with one reversed, so their linking is minus the boundary linking. Returning a small positive normal push in the complement of the other parallel core shows that each Hopf core self-pairing is minus its boundary linking. Thus the Hopf-band signs give $V_{11}=V_{22}=-1$ for $T$ and $V_{11}=-1,V_{22}=1$ for $E$. The specified disk bounded by $a^{+}$ is disjoint from $b$, so its class vanishes in the complement of $b$ and the linking definition gives $V_{12}=0$. Now that $a,b$ is a basis, [F2] yields $V_{12}-V_{21}=a{\cdot }_{F}b=1$, hence $V_{21}=-1$. This proves the displayed matrices.

For $V_T$, the matrix $V_T-t{V}_T^T$ is $\left(\begin{array}{cc}t-1& t\\ -1& t-1\end{array}\right)$, with determinant $(1-t{)}^2+t=1-t+t^2$. For $V_E$, the matrix $V_E-t{V}_E^T$ is $\left(\begin{array}{cc}t-1& t\\ -1& 1-t\end{array}\right)$, with determinant $(t-1)(1-t)+t=-1+3t-t^2$. By [F4], these determinants represent ${\mathrm{\Delta }}_T$ and ${\mathrm{\Delta }}_E$ up to units. Substitution gives the stated values at $1$ and $-1$; every unit is $\pm t^k$, so the absolute values at both substitutions are independent of the representative. Their values $1$ at $1$ agree with [F5].

By [F7], $T\mathrm{\#}E$ is the connected sum in the fixed oriented convention. Apply [F6] to obtain ${\mathrm{\Delta }}_{T\mathrm{\#}E}\doteq {\mathrm{\Delta }}_T{\mathrm{\Delta }}_E$. Multiplying the two displayed quadratics gives $-t^4+4t^3-5t^2+4t-1$; substituting $1$ and $-1$ gives $1$ and $-15$.

By [F9], orientation-preserving ambient isotopy preserves the polynomial class, and an integral basis change is a unimodular congruence of the matrix. Its determinant changes by $(\det P{)}^2=1$, so the determinant class and the displayed unit-invariant evaluations are unchanged.