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

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?

• There is a whole sequence of Alexander polynomials, where $\Delta_k$ is the GCD of the $(n-k)\times(n-k)$ minors of the Alexander matrix. Each one is divisible by the next. Lots of knots have $\Delta_2=1$, but for example $8_{18}$ has $\Delta_2\neq 1$ Commented Aug 20, 2019 at 2:43
• @KyleMiller and Why the Alexander polynomial of the unknot (trivial knot ) is the constant polynomial 1?
– user591668
Commented Aug 20, 2019 at 18:04

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.)
• Maybe one view is: If we have matrices $T:V \to V$ and $S:W \to W$ (positive dimensional vector spaces), we have $\det(T \oplus S) = \det(T)\det(S)$. We'll also like this to be true when $W = \{0\}$ which means we need $\det(S) = 1$. Commented May 6, 2022 at 15:45