Distinct knots with same $A$-polynomial

Question

Are there two non-isotopic knots $K,K'$ in $S^3$ with the same $\mathrm{SL}_2(\mathbb C)$ $A$-polynomials? If it's an open problem, has anyone suggested a method for finding them, or a reason why no such pair should exist (i.e., why the $A$-polynomial should be a complete knot invariant)?

My first guess for how to find such a pair $K, K'$ would be to try mutation, but I wasn't able to find much about (or figure out on my own) how mutation affects the $A$-polynomial.

Answer 1

The torus knots $T_{7,15}$ and $T_{3,35}$ have the same A-polynomials. In general, if $p,q>1$ are coprime and odd then $T_{p,q}$ has A-polynomial $(L-1)(LM^{pq}+1)(LM^{pq}-1)$, which only depends on the product $pq$. This follows from a few observations:

1. The original paper on the A-polynomial by Cooper-Culler-Gillet-Long-Shalen asserts (Proposition 2.7) that the A-polynomial of $T_{p,q}$ is a multiple of $LM^{pq}+1$.
2. That factor comes from a family of representations $\pi_1(S^3\setminus T_{p,q})\to SL_2(\mathbb{C})$ sending $\mu^{pq}\lambda$ to $-I$, and if we multiply all of these by the nontrivial character $\pi_1(S^3\setminus T_{p,q}) \to \{\pm1\}$ then (since $pq$ is odd) we get a different family for which $\mu^{pq}\lambda \mapsto +I$, contributing another factor of $LM^{pq}-1$.
3. Any irreducible representation $\rho: \pi_1(S^3\setminus T_{p,q}) \to SL_2(\mathbb{C})$ with $\rho(\mu),\rho(\lambda)$ diagonal has to send $\mu^{pq}\lambda$ to $\pm I$, so every irreducible is accounted for by these two factors. This is because the complement $S^3\setminus T_{p,q}$ is Seifert fibered, with $\mu^{pq}\lambda$ the class of a generic fiber, and so $\mu^{pq}\lambda$ is central in the knot group.

So in fact you can get arbitrarily many odd torus knots with the same A-polynomial. Namely, if $p_1,\dots,p_n$ are distinct odd primes, then you can get $2^{n-1}-1$ different such knots of the form $T_{r,s}$ by letting $r$ and $s$ be the products of any two complementary, nonempty subsets of $\{p_1,\dots,p_n\}$. (This gives $2^n-2$ pairs $(r,s)$, but only half as many torus knots because $T_{r,s}$ is isotopic to $T_{s,r}$.) These all have the same A-polynomial, determined completely by the product $p_1\dots p_n$.

Answer 2

I quote (page 753) from:

Cooper, D., Long, D. D., Representation theory and the A-polynomial of a knot., Knot theory and its applications. Chaos Solitons Fractals 9 (1998), no. 4-5, 749-763.

"It is known that the $A$-polynomial is not a complete invariant of knots: it is shown in [10], that there are different knots with the same A-polynomial."

Here is the reference [10]:

Cooper, D., Long, D. D., Remarks on the A-polynomial of a knot., J. Knot Theory Ramifications 5 (1996), no. 5, 609-628.

Remark: According to the MathSciNet review of the above paper ([10]): "It is not known whether the $A$-polynomial ever distinguishes mutant knots." But that was in 1996 and I do not know the state of the art.

Remark: Marc Culler has made available a census of $A$-polynomials here. Presumably you can use this census to find many examples if you download the tarball and run a script to compare them.

Remark: It might be useful to use the algorithm Ashley, Burelle and I wrote in Rank 1 character varieties of finitely presented groups to do these computations (note the algorithm is implemented in SnapPy).

Remark: You can find the above papers by Cooper & Long here: https://web.math.ucsb.edu/~cooper/Publications.html