Unimodality of the normalized Jones-polynomial defect

Prove that for every irreducible fraction a>1, the polynomial I_a(q)=(J_a(q)^\vee-J_a(q))/(1-q), with I_a(q)=0 when the normalized Jones polynomial J_a(q) is palindromic, is unimodal except for the two explicitly specified types: 1+q^n for n\geq2, and polynomials whose coefficient sequence is (1,2,\ldots,k,k-1,k,k-1,k-2,\ldots,2,1) for some k\geq2; in particular, establish that I_a(q) is at most bimodal.

Background

For an irreducible fraction a>1, the normalized Jones polynomial J_a(q) of the associated rational link is compared with its reciprocal polynomial. When J_a(q) is not palindromic, the paper defines the palindromic polynomial I_a(q) by dividing J_a(q)\vee-J_a(q) by 1-q and choosing the normalization with positive constant term. The paper proves that I_a(q) has nonnegative coefficients and constant term 1.

The conjecture is derived from the circular-fence-poset trace conjecture stated as Conjecture 4.9. It predicts a precise classification of the possible failures of unimodality for I_a(q), and the authors note that the result would imply that every such polynomial is at most bimodal.

References

Conjecture 4.13. Ia(q) is unimodal except for the following two types. (1) 1 + q" for some n ≥ 2. (2) Zi-o aiqª with (ao, ... , @2k+1) = (1,2, ... , k, k-1, k, k-1, k-2, ... , 2,1) for some k ≥ 2. Especially, Ia(q) is at most bimodal.

Transposes in the $q$-deformed modular group and their applications to $q$-deformed rational numbers  (2502.02974 - Ren et al., 5 Feb 2025) in Conjecture 4.13, Section 4