What is the Alexander polynomial of knot $6_{1}$?

Question

My Alexander polynomial for $6_{1}$ knot is $t^4−5t^3+8t^2−5t+1$ , so its span is 4, so the lower bound of its genus is (1/2 span of its Alexander polynomial) so the lower bound of the genus is equal to 2. Which turns out to be wrong (from wikipedia ) and from my drawing (with the help of @KyleMiller) of the seifert surface.

I learned how to calculate the Alexander polynomial of a knot from this video https://www.youtube.com/watch?v=dwz9GvJk49k&list=PLOROtRhtegr4c1H1JaWN1f6J_q1HdWZOY&index=3

Could anyone help me in detecting my mistakes?

An Edit to show my detailed calculations:

The matrix that I get is the following one (where the columns represent th six arcs in our knot and the rows represents the six crossings in our knot and the orientation I have chosen is the counterclockwise one and the shape of the knot I used is in the attached picture below where I named the crossings on the picture by Roman letters $I, II, III, IV,...$ and the arcs by $x_{i}'s$ where $i = 1,2,...,6$ ) :

\begin{pmatrix} 1-t & t & 0 & 0 & 0 & -1 \\ -1 & 1-t & 0 & t & 0 & 0 \\ 0 & -1 & 1-t & t & 0 & 0 \\ 0 & 0 & t & 1-t & -1 & 0 \\ 0 & 0 & t & 0 & 1-t & -1 \\ t & 0 & 0 & 0 & -1 & 1-t \end{pmatrix}

Then I cancelled the sixth row and six column (as said in the video mentioned above), then I began to calculate the $5 \times 5$-matrix and I got the polynomial I mentioned above.

Answer 1

The exact matrix you get from the diagram depends on the particular way you write the relations down for $\pi_1(S^3-K)$. Any differences can have the effect of multiplying rows by $\pm t^{k}$ for $k\in\mathbb{Z}$, but this does not affect the resulting Alexander polynomial. I used the following presentation: \begin{equation*} \langle x_1,\dots,x_6\mid R_1,\dots,R_6\rangle \end{equation*} where \begin{align*} R_1&=x_1x_6x_1^{-1}x_2^{-1} \\ R_2&=x_2x_4x_2^{-1}x_1^{-1}\\ R_3&=x_2x_3x_4^{-1}x_3^{-1}\\ R_4&=x_5x_4x_3^{-1}x_4^{-1}\\ R_5&=x_5x_3x_5^{-1}x_6^{-1}\\ R_6&=x_6x_1x_6^{-1}x_5^{-1}. \end{align*} The procedure to construct the matrix is to take $R_i=abc^{-1}d^{-1}$ and "linearize it" to get the expression $a+tb-tc-d$, which you put in row $i$ by placing the $\pm 1$ and $\pm t$ coefficients into the columns associated with the generators corresponding to $a$, $b$, $c$, and $d$. This is the matrix I got from doing this: \begin{equation*} \left( \begin{array}{cccccc} 1-t & -1 & 0 & 0 & 0 & t \\ -1 & 1-t & 0 & t & 0 & 0 \\ 0 & 1 & t-1 & -t & 0 & 0 \\ 0 & 0 & -t & t-1 & 1 & 0 \\ 0 & 0 & t & 0 & 1-t & -1 \\ t & 0 & 0 & 0 & -1 & 1-t \\ \end{array} \right). \end{equation*} So it looks like you swapped $x_2$ and $x_6$ for the first crossing --- without knowing which relations you used I still can't tell you what your mistake was, but maybe this is enough for you to figure it out. Other than that, it is the same matrix (up to multiplying some rows by $-1$).

If I cross off the sixth row and sixth column, the resulting matrix has determinant $-2t+5t^2-2t^3$. Alexander polynomials are defined only up to multiplication by $\pm t^k$ for $k\in\mathbb{Z}$, so this can be written in symmetric form as $-2t^{-1}+5-2t$.

---

I computed this using a small Mathematica program. First, I encoded the relations like so:

rels =
 {w[x[1], x[6], x[1]^-1, x[2]^-1],
  w[x[2], x[4], x[2]^-1, x[1]^-1],
  w[x[2], x[3], x[4]^-1, x[3]^-1],
  w[x[5], x[4], x[3]^-1, x[4]^-1],
  w[x[5], x[3], x[5]^-1, x[6]^-1],
  w[x[6], x[1], x[6]^-1, x[5]^-1]}

There is no significance to w or x to Mathematica itself, but to us x[i] represents $x_i$ and w[...] represents a group word by being a list of $x_i$ and $x_i^{-1}$ terms.

The next part is a function arow to compute a row of the Alexander matrix:

ClearAll[arow];
arow[n_, w[]] := Table[0, n];
arow[n_, w[x[i_], r___]] := UnitVector[n, i] + t arow[n, w[r]];
arow[n_, w[x[i_]^-1, r___]] := -t^-1 UnitVector[n, i] + t^-1 arow[n, w[r]];

This uses the "total derivative" version of Fox derivatives. arow[n, w[...]] computes the row associated to the word w[...], given that there are n generators (columns).

The Alexander matrix from above was computed by doing

arow[6, #] & /@ rels // Simplify // MatrixForm

which runs arow[6, ...] for every word in rels. I then copied the result as LaTeX and pasted it into this answer.

When calculating the Alexander polynomial, through this method you can cross off any row and column then take the determinant. This is a shortcut for calculating the GCD of the ideal generated by the $(n-1)\times (n-1)$ minors of this matrix. Just to check that I (probably) got the correct matrix, I ran the following to do the full calculation:

GroebnerBasis[Flatten[Minors[matrix]], t, CoefficientDomain -> Integers]

and it indicates that the Alexander ideal has the following as a generating set: $\{2 t - 5 t^2 + 2 t^3\}$.