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I have been reading the book 'Grid Homology for Knots and Links' (see https://web.math.princeton.edu/~petero/GridHomologyBook.pdf) - in Section 3.3 it provided a way to compute the Alexander polynomial using grid diagrams. Specifically, it gave an explicit formula for the Alexander polynomial in Definition 3.3.4 on Page 53. However, when I tried it on a left-hand trefoil knot, it does not seem to work. Please could I have some help:

The grid diagram I have been working on is as follows.

enter image description here

From here I have obtained the matrix

$$ M(\mathbb{G})= \begin{pmatrix} 1 & 1 & t & t & t\\ 1 & t & t^2 & t^2 & t\\ 1 & t & t^2 & t & 1\\ 1 & t & t & 1 & 1\\ 1 & 1 & 1 & 1 & 1 \end{pmatrix},$$

of which the determinant is $t^6-5t^5+11t^4-14t^3+11t^2-5t+1$ (see https://www.wolframalpha.com/input/?i=det(%7B%7B1,1,t,t,t%7D,+%7B1,t,t%5E2,t%5E2,t%7D,+%7B1,t,t%5E2,t,1%7D,+%7B1,t,t,1,1%7D,%7B1,1,1,1,1%7D%7D) for verification).

Since it is a $5 \times 5$ grid, we have $n=5$, so multiplying the above expression with $(t^{-1/2}-t^{1/2})^{-4}$, which gives $(t^4-t^3+t^2)$. So by looking at the formula in the book, we would need $a(\mathbb{G})$ to be $-3$ to give the correct Alexander polynomial $\Delta(t)=t^{-1}-1+t$ (see http://mathworld.wolfram.com/TrefoilKnot.html). But I am not sure how to obtain that based on the instruction above Definition 3.3.4:

By summing these contributions for all O's and X's and dividing the result by 8, we get a number $a(\mathbb{G})$ associated to the $n \times n$ grid.

This is part of my project but unfortunately I cannot find many resources online. I have also found a master's thesis (by Nancy Scherich) on this topic (see https://nancyscherich.files.wordpress.com/2018/01/the-alexander-polynomial.pdf, where the 'Minesweeper matrix' on Page 32 would give the same determinant as $M(G)$, however I don't think the Theorem 5.8 is correct as I have tried some examples...)

Many thanks in advance!

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    $\begingroup$ Just a point of information: the Alexander polynomial is classically only determined up to a power of $t$ since it just a generator of a particular ideal, though it was known to be symmetric (so is presented symmetrically in tables). Conway figured out a way to fix the degree of the polynomial, and later through grid homology (Heegaard Floer homology) Ozsváth and Szabó found another way to fix the degree. The polynomial $t^4-t^3+t^2$ is the Alexander polynomial of the trefoil, just not the most precise kind of Alexander polynomial known. $\endgroup$ – Kyle Miller Apr 24 at 19:29
  • $\begingroup$ @KyleMiller That is very interesting. Perhaps this is a stupid question but I'm wondering what is good about this method in calculating the Alexander polynomial, since computing the determinant of a large matrix is very inefficient... $\endgroup$ – JpW May 1 at 20:32
  • $\begingroup$ While I think it's meant to be an anticipation of grid homology (as suggested by the final paragraph of section 3.3), calculating Alexander polynomials is usually from the determinant of a somewhat smaller matrix whose entries are a bit more complicated. The computational complexity of calculating determinants is only $O(n^4)$, so things could be much worse. $\endgroup$ – Kyle Miller May 2 at 4:37
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Let's compute $a(\mathbb{G})$ for your example.

The winding numbers at the intersections of horizontal and vertical lines (including the top and rightmost lines) are given by the array $$\begin{array}{r r r r r r} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & -1 & -1 & 0\\ 0 & -1 & -2 & -2 & -1 & 0\\ 0 & -1 & -2 & -1 & 0 & 0\\ 0 & -1 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}$$

I'll color code the entries in that array as follows: a black entry is not in the corner of any $X$ or $O$, a red entry is in the corner of exactly one $X$ or $O$, and a blue entry is in the corner of exactly two $X$'s or $O$'s.

$$\begin{array}{r r r r r r} 0 & \color{red}{0} & \color{red}0 & 0 & \color{red}0 & \color{red}0\\ \color{red}0 & \color{blue}0 & \color{red}{-1} &\color{red}{-1} & \color{blue}{-1} & \color{red}{0}\\ \color{red}{0} & \color{red}{-1} & \color{red}{-2} & \color{blue}{-2} & \color{blue}{-1} &\color{red}{0}\\ 0 & \color{red}{-1} & \color{blue}{-2} & \color{blue}{-1} & \color{blue}{0} & \color{red}{0}\\ \color{red}{0} & \color{blue}{-1} & \color{blue}{-1} & \color{blue}{0} & \color{red}{0} & 0\\ \color{red}{0} & \color{red}0 & \color{red}0 & \color{red}0 & 0 & 0 \end{array}$$

Then $$a(\mathbb{G}) = \frac{1}{8}\left(\sum\text{red entries} + 2\sum\text{blue entries}\right)=-3,$$ as desired.

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  • $\begingroup$ Thank you - your explanation clears it! On a side note (just need a confirmation), is this the reason behind $\epsilon (\mathbb{G})=1$ in this example: From the diagram we have $\sigma _{\mathbb{O}}=(1,2,3,4,5)=(5,1,2,3,4)$, so we need the sign of permutation which connects $\sigma _{\mathbb{O}}$ with $(5,4,3,2,1)$. We observe that the two are equal if we swap $1$ with $4$, and $2$ with $3$. So the permutation required is $(1,4)(2,3)$, which contains two transpositions, therefore the sign of permutation $\epsilon (\mathbb{G})$ is $1$? $\endgroup$ – JpW Apr 24 at 17:04
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    $\begingroup$ Their notation for permutations doesn't seem to be cycle notation, so I don't think it's correct that $(1,2,3,4,5)=(5,1,2,3,4)$. Using their notation, $(1,2,3,4,5)$ is the identity permutation. The permutation taking $(1,2,3,4,5)$ to $(5,4,3,2,1)$ is the one that swaps $1$ with $5$ and $2$ with $4$. That's an even permutation, and so $\epsilon(\mathbb{G})=1$. Does that look right? $\endgroup$ – Adam Lowrance Apr 25 at 4:06

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