What do variables $A$ and $t$ represent in the normalized bracket polynomial and jones polynomial respectively?

Question

I am putting together a presentation on the Jones polynomial and some of its applications. I have been using The Knot Book by Colin C. Adams as my primary reference and have been supplementing it using various other articles and videos such as this one. In the trial run of my presentation, I have received a few questions that I don't feel I have a sufficient answer for.

1. What does the variable "$A$" in the normalized bracket polynomial actually signify? In my textbook, "$A$" is simply introduced as a placeholder variable with no meaning attached to it.

2. What does "$t$" signify in the Jones polynomial? More specifically what is the significance of letting $A = t^{-1/4}$ when going from the normalized bracket polynomial to the Jones polynomial?

3. What does the Jones polynomial tell us about the link? Currently, I have three answers to this question but I was curious if I was missing any. I note that the Span of the polynomial is less than or equal to the number of crossings. I also state that plugging in $t = 1$ then taking the absolute value gives the number of components in the link. Finally, I state that by evaluating the polynomial at $t= t^{-1}$ can determine the chirality of the link.

Answer 1

Short answer. The symbols $A$ and $t$ are just that: abstract symbols. Their meaning comes from the utility of the polynomials, and you've mentioned some of the most important applications.

Longer answer. (Forgive me as I'm likely stating many things that you already know.) The entire cottage industry of constructing link invariants amounts to making use of the observation that we care about links but we compute with link projections. As you know, when we construct a map $$ J: \{ \text{link projections} \} \longrightarrow \{ \text{algebraic stuff} \}, $$ and by some small miracle it happens to be invariant on all projections of a given link, then we get a "shadow" of the link in the algebraic world (the image under the map). Crucially, it is here that we can distinguish things. We can unambiguously show that two polynomials are distinct or that two numbers are distinct; whereas, it was much more difficult to conclude that back in the world of links.

There's an implied equivalence relation on any set that you map out of. In our situation, two links are equivalent iff they give the same value for a given invariant: $$ L_1 \sim L_2 \quad \Longleftrightarrow \quad J(L_1) = J(L_2) $$ We don't mean that they're isotopic, mind you. We're just declaring them equivalent by fiat! The question is how much more crude is this equivalence than the one we care about: isotopy.

There's a sort of conservation law at work. The better the map is at distinguishing links (smaller equivalence classes but more of them), the more difficult it is to construct/compute the map so that it's invariant. Conversely, with larger equivalence classes but fewer of them, the invariant map tends to be easier to construct/compute but it's not as good at distinguishing links. Broadly speaking, you can see this at work when you compare invariants that take values in a (Laurent) polynomial ring vs. just integers.

Somehow, Alexander, Jones, Kauffman, the HOMFLYPT authors, etc. have managed to find a sweet spot, constructing invariants that can tell many links apart and are somehow computable. But where this story gets a lot more interesting is when you begin to see the entire category of links: not just the topological embeddings themselves, but the maps between them, in the form of cobordisms: surfaces who have the links as boundaries.

These polynomial invariants get "lifted" to more sophisticated invariants of the entire categories, such as Khovanov cohomology, where a cobordism from $L_1 \to L_2$ gives a map from $J(L_1) \to J(L_2)$. With not just polynomials, but maps between them, there's a lot more information to be ascertained, often in the form of a negative statement. If this link could be embedded in this-or-that way, then there would be such a polynomial map. But you can check that such map is impossible.