Alexander polynomial and orientation

Question

This should be a simple question, but for some reason I can't seem to find an answer anywhere. My question is: is the Alexander polynomial defined as an invariant of oriented knots or of unoriented knots? That is to say, is the Alexander polynomial of a knot always equal to the Alexander polynomial of its inverse?

Answer 1

It's an invariant of oriented links -- the construction of the infinite cyclic cover depends on the orientation of each component, and different choices can in principle give you different coverings.

For 1-component links, the orientation does not matter. Change in orientation corresponds to substituting $t^{-1}$ for $t$ in the polynomial, and the Alexander polynomial is symmetric (modulo the group of units for $\mathbb{Z}[t^{\pm 1}]$).

However, for multi-component links the orientation does matter. The following link is L9n4{1} according to LinkInfo, and its Alexander polynomial is $-t^5+t^4-t+1$. If you reverse the orientation of the small loop, you get L9n4{0}, which has Alexander polynomial $-t^7+t^6-t+1$.