Does the determinant of a Knot bound the number of primes for which the mod p rank is nonzero?

Question

Definition. If $V$ is a Seifert matrix for a $\operatorname{knot} K$, then the determinant of $K$, denoted $\operatorname{det}(K)$, is the absolute value of the determinant of the symmetrization of the Seifert matrix: \begin{align} \det(K)=|\det(V+V^T)|. \end{align}

Definition. The $\bmod p$ rank of a knot is the dimension of the solution space of the system of linear equations attributed to a $\bmod p$ labeling of its knot diagram.

Out of pure curiosity, I am wondering whether these two numerical invariants from knot theory have anything to do with each other. Since both are defined using matrices, I presume there might be an inequality that can be derived? Similar to how the unknotting number $U(K)$ and the signature $\sigma(K)$ relate via $2 U(K) \geq|\sigma(K)|$, or how the mod $p$ ranks and the bridge index $\operatorname{brg}(K)$ relate via $\bmod p$ rank $\leq \operatorname{brg}(K) - 1$.

Context: I have studied the first five chapters of Livingston's Knot Theory book last year in a course, so I remember a portion of the material.

Answer 1

I'm assuming that in your definition the mod $p$ rank is the dimension of the solution space of non-trivial mod $p$ labelings. Sometimes the trivial (constant-labeling) ones are included. I'm making this assumption because of the bridge number bound you cite.

It's known that the determinant $n=\det(K)$ of a knot $K$ is such that there is a mod $p$ labeling of $K$ if and only if $p|n$. This is implied by the analysis in this answer, since mod $p$ labelings correspond in some way to solutions to $(V + V^T)x=0$, and there are non-trivial solutions modulo $p$ iff $\det(V+V^T)$ is congruent to $0$ modulo $p$. (I don't have a quick reason for you for why $V + V^T$ is an equivalent matrix to the matrix for the reduced system of equations for labelings, other than going through the whole theory for why $tV+V^T$ is a presentation matrix for the Alexander module.)

To summarize, the mod $p$ rank is non-zero iff $p$ divides the knot determinant.

I'm not sure how much better we can do here. We could analyze what happens when we put $V+V^T$ into Smith normal form, which is a diagonal matrix with the property that each diagonal entry divides the next. If $p$ is prime, the $p$ rank is the number of diagonal entries divisible by $p$. If the knot determinant has repeated prime factors, then the factors can either be distributed across multiple entries or be all concentrated at one entry. It's possible to get large mod $p$ ranks since, for example, the connect sum of $k$ copies of a trefoil knot has determinant $3^k$ and mod $3$ rank equal to $k$.

One thing is for sure: $\operatorname{ord}_p(\det(K))$ (the multiplicity of $p$ in the prime factorization of the knot determinant) is an upper bound for mod $p$ rank.

Answer 2

A knot with rank $n$ mod $p$ has $p(p^n-1)$ nontrivial colourings with $p$ colours (this is Theorem 2 in Counting m-coloring classes of knots and links by Brownell, O'Neil and Taalman) so rank mod $p$ is bound by $v_p(\det K)$, the exponent of $p$ in factorisation of $\det K$.

It is known that you can't calculate rank of a knot directly from the determinant because $\det(8_{18}) = \det (9_{24}) = 45$. However their mod 3 ranks are 1 and 2 respectively.