Alexander polynomial evaluated at -1 and p-colorability

Question

I have been working with a group studying generalizations of Fox's $p$-colorability / $p$-labelability and I've seen stated in several places that a knot is $p$-colorable if and only if the Alexander polynomial ∆($t$) evaluated at −1 is divisible by $p$. If someone could explain this or point me towards a resource that gives a proof, that would be very helpful. Thanks!

Answer 1

This question is basically a duplicate of Existence of Fox $n$-colorings, but since that was just me answering my own question, I'll give some more details here.

Recall that an $n$-coloring is an assignment to each arc in a diagram a color from the set $\mathbb{Z}/n\mathbb{Z}$ such that at each crossing $a+b\equiv 2c\pmod{n}$, where $c$ is the label for the overstrand. A nontrivial coloring is one for which at least two different colors are used.

This is a homogeneous system of linear equations over $\mathbb{Z}/n\mathbb{Z}$, so the set of all $n$-colorings is closed under addition and scalar multiplication by $\mathbb{Z}/n\mathbb{Z}$. (In fact, if $n$ is a prime, they form a vector space).

We only care about nontrivial colorings. Choosing one of the arcs, if a given $n$-coloring has color $c$ on the chosen arc, then subtracting the all-$c$ trivial $n$-coloring yields an $n$-coloring where that arc has color $0$. Nontrivial colorings remain nontrivial. Therefore, we can consider only those $n$-colorings such that chosen arc is colored with $0$ for the question of whether a nontrivial $n$-coloring exists.

It turns out that there is a linear dependence on the equations themselves, and any one of them can be removed. If there are $m$ crossings in the diagram (which is equal to the number of arcs), then we have $m-1$ equations in $m-1$ unknowns (since one of the arcs was set to have color $0$). If we take the $m\times m$ matrix for the original system of equations, we have effectively crossed off one column and one row, yielding an $(m-1)\times (m-1)$ matrix. Letting $D$ be the determinant of this matrix, which is known to be nonzero in general for knots, we see that if $n$ divides $D$, then $D\equiv 0\pmod{n}$, which implies that there is a nontrivial solution over $\mathbb{Z}/n\mathbb{Z}$, so there is a nontrivial $n$-coloring of the knot.

How this relates to the Alexander polynomial is that the matrix you get using Fox calculus from the Wirtinger presentation can be thought of as corresponding to a system of equations of the form $a-tb=(1-t)c$. When $t=-1$, this is $a+b=c$. The Alexander module presentation matrix is from crossing off one row and one column of the matrix, whose determinant is the Alexander polynomial. The knot determinant $\lvert \Delta_K(-1)\rvert$ is therefore $D$ from before, so it determinants $n$-colorability.

Taking this system of equations seriously leads to Alexander quandles, which is a generalization of $n$-colorability.