My question is about the Alexander polynomial of a slice knot.
For a slice knot $K$,
Fox-Millnor and Terasaka proved that
$$ \Delta_{K}(t) \doteq f(t) f(t^{-1})$$
for some polynomial $f(t) \in \mathbb{Z}[t]$,
where $\Delta_{K}(t)$ is the Alexander polynomial of $K$
and $\doteq$ means up to units of $\mathbb{Z}[t, t^{-1}]$.
Let $D$ be a slice disk for $K$.
Then a folklore result states that
$$f(t) \doteq \Delta_{D}(t),$$
where $ \Delta_{D}(t)$ denotes the Alexander polynomial of $D$.
$\textbf{Question 1.} $ Is there a reference of this folklore result ?
Note that this folklore result is important
when we calculate the Alexander polynomial of a ribbon knot $R$.
Indeed, let $D$ the slice disk in the $4$-ball $B^4$
obtained from a ribbon presentation of $R$.
Then we easily obtain the presentation of $\pi_{1}(B^4 \setminus N(D))$
and can determine $ \Delta_{D}(t)$ using Fox calculus,
where $N(D)$ is a tubular neighborhood of $D$.
I heard from Akio Kawauchi (who was my adviser) that this folklore result is true, at least, for ribbon knots
and he did not know any references.
His proof is using the Blanchfield duality
(I do not follow the proof fully).
There is another question.
Here recall the Reidemeister torsion.
Let $K$ be a knot in $S^3$ and $\tau_{\alpha}(S^3 \setminus N(K))$
the Reidemeister torsion
associated to the abelian map $\alpha$ of $\pi_{1}(S^3 \setminus N(K) )$.
(Precisely, $\alpha : \pi_{1}(S^3 \setminus N(K) ) \to GL(1; \mathbb{Q}(t)).$)
Milnor's theorem states that
$$ \tau_{\alpha}(S^3 \setminus N(K)) \doteq \dfrac{\Delta_K(t)}{t-1}$$
Let $D$ be a slice disk for some knot
and $\tau_{\alpha}((B^4 \setminus N(D)))$ be the Reidemeister torsion of a slice disk $D$
associated to the abelian map $\alpha$ of $\pi_{1}(B^4 \setminus N(D) )$.
$\textbf{Question 2.} $ Is it true that
$$ \tau_{\alpha}(B^4 \setminus N(D))\doteq \dfrac{\Delta_D(t)}{t-1} \ ?$$
It seems that,
if the Whitehead group of $\pi_{1}(B^4 \setminus N(D))$ is trivial,
then Question 2 is true.
$\textbf{Question 3.} $ Is the Whitehead group of $\pi_{1}(B^4 \setminus N(D))$ trivial ?
Finally, note that I am not familiar with the Reidemeister torsion
and the Whitehead group, and therefore there might exist some wrong descriptions.
If so, please tell me !
Best Answer
It is in general not true, that for a knot $K$ with slice disk $D$ we have $\Delta_K(t)\equiv f(t)\cdot f(t^{-1})$ with $f(t)=\Delta_D(t)$. For example, even if $K$ is the trivial knot, then we can take a slice disk which is given by connect sum of the trivial disk with a knotted $S^2$ in $D^4$. Then $\Delta_D(t)$ will be the Alexander polynomial of the 2-knot, which can be non-trivial. The statement is true though if $D$ is a ribbon disk. The argument is as follows. Let $\Lambda=\Bbb{Z}[t,t^{-1}]$. We denote the zero-framed surgery on $K$ by $N$ and we denote by $X$ the exterior of $D$. We then have a long exact sequence $H_2(X;\Lambda)\to H_2(N,X;\Lambda)\to H_1(N;\Lambda)\to H_1(X;\Lambda)\to 0$. Note that the last map is indeed surjective, since $\pi_1(N)$ surjects onto $\pi_1(X)$. (Here we use that $D$ is a ribbon disk.) By Poincare duality and universal coefficient theorem we have $H_2(N,X;\Lambda)\cong \overline{H_1(X;\Lambda)}$ and $H_2(X;\Lambda)=0$. We thus have a short exact sequence $0\to\overline{H_1(X;\Lambda)}\to H_1(N;\Lambda)\to H_1(X;\Lambda)\to 0$ and the result follows from basis properties of orders of $\Lambda$-modules.