Seifert Matrix of Amphichiral Knots

Question

In Murasugi "Knot Theory and Its Applications" 5.4.7 I found an information that given the knot $K$ with the Seifert matrix $M_K$, the Seifert matrix $M_{K*}$ of its mirror image $K*$ is $S$-equivalent to $-M_K^T$. I am wondering if I am right thinking that then the Seifert matrix of an amphichiral knot, say $M$ is $S$-equivalent to $-M^T$. Thanks in advance.

Answer 1

My reference for the following is Lickorish's An introduction to knot theory.

1. Theorem 8.4 is that any two Seifert matrices of the same oriented link are $S$-equivalent.

2. The Seifert matrix $A$ of $K$ is given by $a_{ij}=\operatorname{lk}(f_i,f_j^+)$ where $\{f_i\}_i$ form a basis for the first homology of a Seifert surface $F$, and where $f_j^+$ is $f_j$ pushed off $F$ in the positive direction relative to the orientation of $F$.

3. The mirror image $K_*$ of $K$ can be thought of as changing the types of all the crossings in a knot diagram for $K$, and in particular as a reflection through the plane of the diagram. Let $G$ be the reflected Seifert surface with reversed orientation (so the orientation of $K_*$ is still induced by that of $G$) and let $\{g_i\}_i$ be a basis for $H_1(G)$, but with each $g_i$ the reflection of $f_i$. Then $$\operatorname{lk}(g_i,g^+_j)=-\operatorname{lk}(f_i,f^-_j)=-\operatorname{lk}(f^-_j,f_i)=-\operatorname{lk}(f_j,f^+_i).$$ The first equality is from "pulling the curves through the mirror," which involves negating the linking form. Hence, in this setup $K_*$ gets the Seifert matrix $-A^T$.

4. If $K$ is amphichiral, meaning $K$ is equivalent to $K_*$, by 1 we have $A$ is $S$-equivalent to $-A^T$.