Classification of pretzel links up to link homotopy using alexander quandle

Question

I am currently reading this paper where the author classifies the pretzel links up to link homotopy using a quasi-trivial quandle $\mathbb{Z}_{k}[t^{\pm 1}]\diagup_{(t-1)^{2}}$, and I find it difficult to understand the proof of lemma 5.6. Can anyone help with this proof?

Answer 1

Beginning with coloring the $k$th box using ${\mathbb{Z}_k[t^{\pm1}]}/_{(t-1)^2}$, we conclude that if the link is homotopically trivial, then $p_k$ divides all $p_i$'s.

It seems that there is a typo here: "${\mathbb{Z}_k[t^{\pm1}]}/_{(t-1)^2}$" should be "${\mathbb{Z}_{p_k}[t^{\pm1}]}/_{(t-1)^2}$". By Corollary 5.2, if we start by colouring the top $k$th box with colour $\begin{bmatrix}a\\b\end{bmatrix}$, the bottom colour will come out as $\begin{bmatrix}a\\b\end{bmatrix}$. Compare it with the colour of the box immediate to the right (it the $k$th box is rightmost, consider the first box). Both the top colour and the bottom colour start with $b$. To make sure that this is the case, using Corollary 5.2 we conclude that $p_{k+1}$ is a multiple of $p_k$. Continuing this argument to all boxes, we conclude that every $p_i$ is a multiple of $p_k$.

Thus if the $n$-Pretzel link $(p_1,p_2,\ldots,p_n)$ is homotopically trivial, then $|p_1|=\cdots=|p_n|$.

This is because for every pair $p_i,p_j$, $p_i$ divides $p_j$ and vice versa.

Now if $p_1 \neq 0$, We color the link by elements of ${\mathbb{Z}_k[t^{\pm1}]}/_{(t-1)^2}$ with $k=\max\{2,|p_1|-2\}$.

This is to make sure that the bottom colour of the first box is different from the top colour.

Let the top color vector be $(x_1, \cdots, x_n)$. Choosing $x_1=0$ and $x_2=x_3=1$ in the coloring of the $n$-pretzel link leads to a contradiction.

This contradiction is, the second element of the bottom colour of the 1st box is different from the first element of the bottom colour of the 2nd box. But they should be the same as they are in the same arc in the diagram.

Thus the number of colorings is less than $k^{2n}$ implying that the homotopy link is non-trivial.

If the homotopy link were trivial, every ordered pair $(x_1,\ldots,x_n)$ colours the link and the number of colouring should be $k^{2n}$. But now for this link, one of the colouring does not work and thus it is not homotopically trivial.