= Solution
Number the seven crossings from top to bottom. Orient the strand entering the top-left arch downward. With the usual positive braid crossing convention, the crossing signs are
$$
(+,+,+,-,-,+,+).
$$
To make the diagram calculation reproducible, its oriented <Gauss code> is
$$
1_o\,2_u\,3_o\,6_u\,7_o\,1_u\,2_o\,4_u\,5_o\,7_u\,6_o\,5_u\,4_o\,3_u.
$$
Subscripts record overpassing and underpassing. In this convention a positive <trefoil knot> has <knot signature> $-2$.
Label the seven <Wirtinger generators> $x_0,\ldots,x_6$ successively between undercrossings, beginning just before the encounter $2_u$. The <Alexander matrix>, with rows ordered by crossing number, is
$$
M(t)=
\begin{pmatrix}
1-t&0&t&-1&0&0&0\\
t&-1&0&1-t&0&0&0\\
-1&1-t&0&0&0&0&t\\
0&0&0&1&-t&0&t-1\\
0&0&0&0&t-1&1&-t\\
0&t&-1&0&0&1-t&0\\
0&0&1-t&0&t&-1&0
\end{pmatrix}.
$$
At a positive crossing the <Fox derivative> entries at the overpassing, incoming and outgoing arcs are $1-t,t,-1$; at a negative crossing a unit multiple of the row has entries $t-1,1,-t$. Deleting the last row and column gives <determinant> $t(t^4-5t^3+7t^2-5t+1)$. Hence a symmetric normalization of the \b[Alexander polynomial of a knot] is
$$
\boxed{\Delta_K(t)=t^2-5t+7-5t^{-1}+t^{-2}.}
$$
In particular, $\Delta_K(1)=-1$ in this displayed normalization; multiplying by $-1$ gives normalization $1$ and does not change any conclusion.
The <Seifert algorithm> produces four <Seifert circles> and seven bands. Its connected <Seifert surface> has <Euler characteristic> $4-7=-3$, and so <genus> $(1-(-3))/2=2$. Conversely, the <Alexander breadth bound on Seifert genus> gives $2g_s(K)\geq4$. Therefore
$$
\boxed{g_s(K)=2.}
$$
For the <knot signature>, use the <alternating diagram signature formula>: for a reduced <alternating knot diagram>,
$$
\sigma(K)=s_A(D)-c_+(D)-1,
$$
where $s_A$ counts circles in the all-$A$ bracket smoothing and $c_+$ counts positive crossings. Here the all-$A$ smoothing has four circles, the all-$B$ smoothing has five, and $c_+=5$. Thus
$$
\boxed{\sigma(K)=-2.}
$$
The opposite global <knot signature> convention gives $+2$ instead.
For the <slice genus>, the <Levine-Tristram signature bound on the slice genus> at $-1$ gives $g_4(K)\geq1$. There is also an explicit <unknotting crossing>: switch crossing $3$. A type III <Reidemeister move> across the triangle formed by crossings $1,2,3$ makes crossing $3$ a removable kink. Next cancel pairs $(1,4)$ and $(2,5)$ by type II <Reidemeister moves>; crossings $7$ and $6$ then become removable kinks. This leaves the <unknot>. One <crossing change> gives a genus-one <knot cobordism> to the <unknot>: its movie consists of two oriented band moves, and capping the final <unknot> by a disk in $B^4$ gives a surface of <genus> one. Consequently
$$
\boxed{g_4(K)=1.}
$$
Back to article page
