Knot Alexander normalization, symmetry, and genus bound

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.