Why is the Jones polynomial for a knot a Laurent polynomial?

Question

We know that the Jones polynomial for a link $L$ is defined as $$V_L(t):=(-1)^{w(L)}\langle L\rangle,$$ where $w(L)$ denotes the writhing number of $L$.

Question:

Why, for a knot $K$, is $V_K(t)\in\mathbb Z[t,t^{-1}]$?

Answer 1

I would note that the title of the question is a bit misleading: The Jones polynomial of any link is in $\mathbb{Z}[t^{1/2}, t^{-1/2}]$, which is also a Laurent polynomial ring; it just happens to be Laurent polynomials in variables that come with fractional powers. It is not uncommon to substitute $t = A^{-4}$ in which case you get rid of fractional powers altogether.

That being said, let us show that if $L$ has an even number of components, then all powers that appear in the Jones polynomial of $L$ are non-integral, whereas if $L$ has an odd number of components (which in particular covers the case where $L$ is a knot), then all powers that appear in the Jones polynomial of $L$ are integral.

To show this, we first recall that if $L_+$, $L_-$ and $L_0$ are three link diagrams that are equal except for at a single crossing, where the crossing is positive, negative, or resolved in the direction of the orientation, then one has the skein relation

$$(t^{1/2} - t^{-1/2}) V_{L_0}(t) = t^{-1}V_{L_+}(t) - tV_{L_-}(t).$$

Now, suppose that $L_+$ has $n$ components. Then $L_-$ also has $n$ components (which can be seen by considering the two cases where the strands belong to the same component, or different components), and $L_0$ will have either $n+1$ or $n-1$ components.

At this point, we proceed by a double induction on the crossing number and the unknotting number of the links. The base cases are the links with zero crossing number or zero unknotting number. Those are exactly the $n$-component unlinks (i.e. any link with crossing number 0). Those have Jones polynomial $(t^{1/2} + t^{-1/2})^{n-1}$ which is of the desired form.

Assume that the claim has been proven for all links whose crossing number is less than $k$, as well as for all links whose crossing number is $k$ but whose unknotting numbers are at most $m$, and let $L$ be a link whose unknotting and crossing numbers are $m+1$ and $k$ respectively.

Then there is a crossing, positive, say, in $L = L_+$ so that changing that crossing reduces the unknotting number. That is, $L_-$ has unknotting number at most $m-1$ and at most $k$ crossings, so its Jones polynomial satisfies the claim. Similarly, $L_0$ has at most $k-1$ crossings and thus also satisfies the claim, no matter what its unknotting number might end up being. At this point, the claim for $L_+$ follows directly from the skein relation.