Question about Conway Polynomials of oriented links Hi I've got a few questions on Conway polynomials in preparation for an exam this Saturday that I don't know how to do:
Let $L$ be an oriented link.
(a) If $\mu(L)=1$ then $C(L)\in 1+z^2\mathbb{Z}[z]$
(b) If $\mu(L)=3$ then $C(L)\in z^2\mathbb{Z}[z]$
(c) Suppose $\mu(L)=2$ and let $L_{1},L_{2}$ denote the components of L. Then $C(L)\in zlk(L_{1}, L_{2})+z^3\mathbb{Z}[z]$
(d) Prove $C(L)\in z^{\mu(L)-1}\mathbb{Z}[z]$. What is the coefficient of $z^{\mu(L)-1}$ in $C(L)$. 
$C(L)$ is the Conway polynomial of $L$, $\mu(L)$ is the number of components of link $L$ and $lk$ is the linking number.
If anyone could go through them I'd be so grateful.
 A: The Conway polynomial satisfies the following skein relation (which I write out in two forms as a small convenience):

That is, for $\epsilon=\pm 1$, we have $C(L_\epsilon) = C(L_{-\epsilon}) + \epsilon z C(L_0)$, where $L_{+}$ and $L_{-}$ are right- and left-handed crossings, respectively, and $L_0$ is the no-crossing version of the diagram.
The relations in this form suggest the concept of a resolution tree: given a link diagram, choose a crossing that, when changed, gives a "more unknotted" diagram, then for the computation of $C$ there are two branches corresponding to each of the terms appearing in the relevant form of the skein relation.  Eventually, the leaves of the resolution tree are diagrams that are equivalent to split diagrams of unknots.  It is an easy exercise to use the skein relation to show that $C(\coprod_{i=1}^n\mathrm{unknot})$ is $1$ if $n=1$ and otherwise $0$ if $n>1$.
(That finite resolution trees exist is not so hard to prove.  Each component is given an origin point labeled with a distinct number, so the components themselves are in order.  You want to switch crossings so that each component is totally "on top of" the next component, and along any particular component it is always "going up" while walking along from the origin.)
(a) Let's first take things modulo $z$.  Then all the terms of the resolution tree that involve multiplying by $z$ are congruent to $0$.  Hence, $C(L_\epsilon)\equiv C(L_{-\epsilon})\pmod {z}$.  Since $C(\mathrm{unknot})=1$, it follows that $C(K)\equiv 1\pmod{z}$ for $K$ a knot.  For a link $L$ with more than one component, $C(L)\equiv C(\coprod_{i=1}^{\mu(L)}\mathrm{unknot})\pmod{z}$, so $C(L)\equiv 0\pmod{z}$.  Notice that the $K_0$ resolution of the crossing gives a link with two components, so $C(K)\equiv 1\pm zC(K_0)\equiv 1\pmod{z^2}$.
(b) It's roughly the same difficulty to show the stronger statement $C(L)\in z^{\mu(L)-1} \mathbb{Z}[z]$.  We already showed this for $\mu(L)=1$, and we will proceed by induction on "unknottedness."  In the case $L$ is completely unknotted, then $C(L)=0\in z^{\mu(L)-1}\mathbb{Z}[z]$.  Otherwise, when we resolve a crossing, the $L_0$ case either has one more or one fewer component than $L=L_{\epsilon}$ does.
Then by induction $C(L_0)\equiv 0\pmod{z^{\mu(L)-1}}$, and so $C(L)\equiv C(L_{-\epsilon}) \pm z C(L_0)\equiv C(L_{-\epsilon})\pmod{z^{\mu(L)-1}}$.  By induction again, $C(L_{-\epsilon})\equiv 0\pmod{z^{\mu(L)-1}}$ since $\mu(L_{-\epsilon})=\mu(L)$.  Therefore $C(L)\equiv 0\pmod{z^{\mu(L)-1}}$.
(c) In the above argument, whether the number of components in $L_0$ increased or decreased was a matter of whether the two strands at the crossing were from the same or different components, respectively.  Choose a resolution tree where inter-component crossings are dealt with last (so first flip crossings to make each component an unknot).  For these crossings, $C(L_{\epsilon})\equiv C(L_{-\epsilon})\pmod{z^{\mu(L)+1}}$ since $C(L_0)\equiv 0\pmod{z^{\mu(L)}}$ as $L_0$ has one more component.  So, modulo $z^{\mu(L)+1}$ we can change intra-component crossings so that each component of $L$ is an unknot.  For the inter-component crossings, $C(L_{\epsilon})\equiv C(L_{-\epsilon})+\epsilon z C(L_0)\pmod{z^{\mu(L)+1}}$ with $L_0$ having one fewer component.
Consider a two-component link, so $L_0$ is a knot and $C(L_0)\equiv 1\pmod{z^2}$.
Hence, $C(L_{\epsilon})\equiv C(L_{-\epsilon}) + \epsilon z\pmod{z^3}$.
Going through all the crossings, we reach the point where the link is split and we find that we got $C(L)\equiv \operatorname{lk}(L_1,L_2) z\pmod{z}^3$ from the definition of linking numbers where we add up $+1$ for each right-handed inter-component crossing and $-1$ for each left-handed one among a set of crossings to change to separate the components.
(d) We already went through the first part, and we have almost identified the second part.  For every link $L$, let $a(L)$ denote the coefficient of the $z^{\mu(L)-1}$ term.  That is, $C(L)\equiv a(L)z^{\mu(L)-1}\pmod {z^{\mu(L)}}$ (and this should be true modulo $z^{\mu(L)+1}$, too).
Consider $C(L_{\epsilon})\equiv C(L_{-\epsilon})+\epsilon z C(L_0)\pmod{z^{\mu(L)+1}}$ for an inter-component crossing again.  Since $L_0$ has one fewer component, we have $a(L_{\epsilon})=a(L_{-\epsilon})+\epsilon a(L_0)$.
Let's take stock in what we know about this $a$ invariant:


*

*Changing an intra-component crossing leaves $a$ invariant: $a(L_+)=a(L_-)$.

*Inter-component crossings satisfy the skein relation $a(L_+)-a(L_-)=a(L_0)$.

*$a(K)=1$ for a knot and $a(L)=\operatorname{lk}(L_1,L_2)$ for a two-component link.

*$a(L)=0$ if $C(L)=0$.  For example, if $L$ is a split link.


Given two components $L_i$ and $L_j$ that are linked, then by repeatedly applying the skein relation to split them we have $a(L)=a(L')+\operatorname{lk}(L_i,L_j)a(L'')$ where $L'$ is the version of $L$ where all the crossings between $L_i$ and $L_j$ are changed to split them, and $L''$ is the version where components $L_i$ and $L_j$ are merged into a single component.  Since $a$ is invariant under intra-component crossing change, $a(L'')=a(L_0)$ for each $L_0$ that shows up.
It's not too hard to see that the invariant only depends on the linking numbers between components.  There is a reasonably straightforward way to calculate $a$ using the above formula and a linking matrix $L_{ij}=\operatorname{lk}(L_i,L_j)$ for $L$.
I'm not sure what this invariant is, but I calculated that for a three-component link, with $x,y,z$ being the linking numbers between every pair of components, $a(L)=xy+xz+yz$.
Edit: After a more thorough literature search, I found that Hoste proved that $a(L)$ is given by a determinant of a matrix that comes from the linking numbers. Levine, "The Conway polynomial of an algebraically split link" has a description in Proposition 3.2.
