Why the Alexander polynomial of the unknot (trivial knot ) is the constant polynomial $1?$

Question

The book said this:

Why the Alexander polynomial of the unknot (trivial knot ) is the constant polynomial $1?$ could anyone explain this for me please?

And is the subscript $k$ like the winding number?

Answer 1

A few ways to see that the Alexander polynomial of the trivial knot is the constant polynomial 1:

1. The Alexander polynomial can be calculated as the determinant of an $n\times n$ matrix where $n$ is the number of crossings in a presentation of the knot. The obvious presentation of the unknot has no crossings, so you're taking the determinant of a $0\times0$ matrix, which equals $1$.
2. The Alexander polynomial $\Delta$ is a reparameterization of the Alexander-Conway polynomial $\nabla$, which is defined by a sort of "recurrence relation" whose base case is $\nabla(\textrm{unknot})=1$.
3. Another way to calculate the Alexander polynomial begins by finding a presentation for the knot group (fundamental group of the complement of the knot); for an unknot it's obvious that this is just $\Bbb{Z}$ (generated by a loop winding once around the knot). In general one gets $n$ generators and $n-1$ relations for some $n$, and then constructs an $n\times n-1$ matrix of (approximately) formal derivatives of relations by generators, and the $n-1\times n-1$ minors of this matrix generate an ideal which turns out to be a principlalideal whose generator is the Alexander polynomial. The only bit of that that actually matters right now is that you're taking the determinants of $n-1\times n-1$ matrices, and here we can take $n=1$ so again the determinants are all (vacuously) $1$ and therefore that principal ideal is $(1)$ whose generator is $1$.

In case it isn't obvious that the determinant of an empty matrix should be taken to be 1, one low-tech way to see it is as follows: $|A|=\sum_\sigma\mathrm{sign}(\sigma)\prod_ia_{i\sigma(i)}$, and when the matrix is of size $0$ there is just one permutation $\sigma$, for which the sign is obviously $1$ and the (empty) product is also $1$. (There are other ways of thinking about what the determinant is and they all also lead to the conclusion that $1$ is the right value for an empty determinant.)