Skein Tree: Conway Polynomial I am trying to learn about the skein relation, but I don't understand what is being done here. Can anyone help me with this?
And how is $1+z^2$ as the final result obtained?
Additional: This is the relation given by Conway,

 A: The skein relation
$$
\nabla_{D_+}(z) - \nabla_{D_-}(z) = z \cdot \nabla_{D_0}(z)
$$
relates the Alexander-Conway polynomials of any three diagrams that are identical outside of a small disk, and have the three local forms inside the disk.
The trick to using the skein relation effectively is to find a, say, positive crossing in the diagram (call this $D_+$), and to look at the diagrams that you get by replacing the positive crossing by a negative crossing ($D_-$) and a smoothing ($D_0$), respectively.
The polynomials corresponding to each of the three diagrams are related by the skein formula.  This is only useful if the resulting diagrams are simpler or their Alexander-Conway polynomials are known.

In your example, with $D_+$ a three crossing diagram for the right-handed trefoil knot, switching a crossing results in the diagram $D_-$ that is isotopic to the unknot.  (We're done there, since we know that $\nabla_{\text{unknot}}(z) = 1$.)
On the other hand, replacing the positive crossing by a smoothing, results in a diagram $D_0$ which is the two component Hopf link.  What is $\nabla_{\text{Hopf}}(z)$?  We don't know just yet.  But, we can use the skein relation again!
This time, choose another positive crossing in the diagram, and consider this to be $D_+$ in the relation.  The other two diagrams resolve nicely.  For $D_-$ we get the two component unlink, which has $\nabla(z) = 0$?  (Do you know why the existence of an unlinked component forces $\nabla(z) = 0$?)  And for $D_0$, we get another unknot.
Now, calculate up the skein tree.  For the Hopf link,
$$
\begin{align}
\nabla_{\text{Hopf}}(z) &= 1 \cdot \nabla_{\text{2-comp unlink}}(z) + z \cdot \nabla_{\text{unknot}}(z) \\
&= 1 \cdot 0 + z \cdot 1 \\
&= z
\end{align}
$$
And for the original trefoil,
$$
\begin{align}
\nabla_{\text{trefoil}}(z) &= 1 \cdot \nabla_{\text{unknot}}(z) + z \cdot \nabla_{\text{Hopf}}(z) \\
&= 1 \cdot 1 + z \cdot z \\
&= 1 + z^2
\end{align}
$$

Does this make sense?  Do you understand how the tree branches from any given diagram to form two others?  Can you calculate back up the tree to find the polynomial for the original diagram?
