关于双曲纽结群分类及亚历山大多项式相关问题的技术问询

Hey there, let's unpack your questions since you're digging into hyperbolic knot groups and Alexander polynomials—super interesting territory!

First, let's tackle your core questions one by one:

1. Can we classify all knots using a given Alexander Polynomial? #

Short answer: No, the Alexander polynomial isn't a "complete invariant" for knots, meaning multiple distinct knots can share the same Alexander polynomial.

Here's why:

• The Alexander polynomial only captures specific algebraic information about the knot complement's fundamental group, and it's symmetric (satisfies $\Delta(t) = \Delta(t^{-1})$). For example, the Conway knot and Kinoshita-Terasaka knot are distinct hyperbolic knots, but their Alexander polynomials are identical: $(t^2 - t + 1)^2(t2 + t + 1)^2$.
• Even simpler cases: Mirror images of non-amphicheiral knots (like right-hand and left-hand trefoils) have identical Alexander polynomials, but they're distinct knots unless the knot is amphicheiral (like the figure-eight knot).
• For hyperbolic knots specifically, you'll need stronger invariants to distinguish them—things like hyperbolic volume, the Jones polynomial, or knot Floer homology often fill in the gaps where the Alexander polynomial falls short.

2. All knots with Alexander Polynomial equal to 1 #

Knots with $\Delta_K(t) = 1$ are often called "Alexander one knots," and they have some key properties:

• The most obvious example is the unknot (trivial knot).
• There are non-trivial examples too—like the Whitehead double of the unknot (both positive and negative versions). This is a satellite knot (not hyperbolic) that's non-trivial but has an Alexander polynomial of 1.
• A critical result: All Alexander one knots are slice knots (meaning they bound a smooth disk in the 4-ball). Conversely, slice knots have Alexander polynomials that are squares of integer polynomials, so 1 (being $1^2$) fits this criteria.
• Importantly, there are no non-trivial hyperbolic knots with Alexander polynomial 1. Hyperbolic knot complements are hyperbolic 3-manifolds, and their Alexander polynomials are non-trivial (they have degree at least 2, like the figure-eight knot's $\Delta(t) = t - 2 + t^{-1}$). Any knot with $\Delta(t)=1$ is either trivial, a satellite knot, or a connected sum of such knots.

If you're exploring finite index subgroups of hyperbolic knot groups, keep in mind these groups are torsion-free and word-hyperbolic—their finite index subgroups inherit these properties, so you might want to dive into subgroup growth or congruence subgroup results for specific knot families!

备注：内容来源于stack exchange，提问作者T ghosh