Trefoil and figure-eight Jones spans

For the closure of ${\sigma }_1^3$, the endpoint permutation is the transposition of the two strands, hence one cycle by [F4]. The three positive crossings form a connected projection. Starting on the left bottom strand, the successive crossing visits are under, over, under; after closure the right bottom strand visits over, under, over. Thus the diagram is alternating. Its black Tait graph has two vertices and three parallel edges, while the white dual graph is a triangle. Both graphs have no loops, so no crossing repeats a region on its boundary; the diagram is reduced by [F6].

For the four-level braid with levels ${\sigma }_2^{-1},{\sigma }_1,{\sigma }_2^{-1},{\sigma }_1$, the successive endpoint orders are $[1,3,2]$, $[3,1,2]$, $[3,2,1]$, and $[2,3,1]$. The resulting endpoint permutation is the cycle $1\mapsto 3\mapsto 2\mapsto 1$, so the closure is one knot component by [F4]. Traversing it from the first bottom strand gives crossing visits under, over; from the third bottom strand gives under, over, under; and from the second gives over, under, over. The full cyclic sequence alternates. The black Tait graph is the theta graph with two parallel edges between $u,v$ and a two-edge path $u-w-v$; it has no loops or bridges. Its dual has no loops because the primal has no bridges. Thus both color pairs at every crossing are distinct and the diagram is reduced by [F6].

If exactly $k$ of the three positive crossings use the $B$-smoothing, the states and circle counts are: $k=0$: $1$ state with $2$ circles; $k=1$: $3$ states with $1$ circle; $k=2$: $3$ states with $2$ circles; $k=3$: $1$ state with $3$ circles. Stacking through pairings and cup-cap pairings and closing the two top endpoints to the corresponding bottom endpoints gives these counts. By [F1]–[F2], the grouped contributions are $A^3\delta +3A+3A^{-1}\delta +A^{-3}{\delta }^2=A^{-7}-A^{-3}-A^5$. The writhe is $3$ by [F3], so [F7] gives $V_{3_1}(t)=(-A^3{)}^{-3}(A^{-7}-A^{-3}-A^5){\mid }_{t=A^{-4}}=t+t^3-t^4$. Its span is $3$, and [F5] gives crossing number $3$.

Encode a state by four bits from bottom to top, with $0$ meaning through pairing and $1$ meaning cup-cap pairing. Let $p$ be the number of cup-cap choices at positive crossings and $q$ the number at negative crossings; [F1] gives weight exponent $2(q-p)$. Closing the pairings gives the exhaustive state data $\begin{array}{ccc}\text{state}& \text{circles}& \text{exponent}\\ 0000& 3& 0\\ 0001& 2& -2\\ 0010& 2& 2\\ 0011& 1& 0\\ 0100& 2& -2\\ 0101& 3& -4\\ 0110& 1& 0\\ 0111& 2& -2\\ 1000& 2& 2\\ 1001& 1& 0\\ 1010& 3& 4\\ 1011& 2& 2\\ 1100& 1& 0\\ 1101& 2& -2\\ 1110& 2& 2\\ 1111& 1& 0\end{array}$. Thus the five one-circle states contribute $5$, the eight two-circle states contribute $4(A^2+A^{-2})\delta$, and the three three-circle states contribute $(1+A^4+A^{-4}){\delta }^2$. Expanding $\delta =-A^2-A^{-2}$ gives $⟨D_F⟩=5+4(A^2+A^{-2})\delta +(1+A^4+A^{-4}){\delta }^2=A^8-A^4+1-A^{-4}+A^{-8}$. There are two positive and two negative crossings, so writhe is $0$ by [F3]. By [F7], $V_{4_1}(t)=t^{-2}-t^{-1}+1-t+t^2$, whose span is $4$. Applying [F5] gives crossing number $4$.