The book said this:

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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?

  • 1
    $\begingroup$ 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 $\Delta_2\neq 1$ $\endgroup$ Commented Aug 20, 2019 at 2:43
  • $\begingroup$ @KyleMiller and Why the Alexander polynomial of the unknot (trivial knot ) is the constant polynomial 1? $\endgroup$
    – user591668
    Commented Aug 20, 2019 at 18:04

1 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.)

  • $\begingroup$ 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 $\det(S) = 1$. $\endgroup$
    – inkievoyd
    Commented May 6, 2022 at 15:45

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