Seifert surfaces and knot genus — Examples and Counterexamples

The word ${\sigma }_1^3$ has three crossings, and its endpoint permutation is $(12{)}^3=(12)$ by [F7] and [F8], so its closure is one knot component. Smoothing each braid crossing continues its oriented strand to the adjacent lower segment, leaving the identity two-strand closure with two cycles; hence $c=3$ and $s=2$.

The word ${\sigma }_1{\sigma }_2^{-1}{\sigma }_1{\sigma }_2^{-1}$ has four crossings. Since ${\sigma }_i$ and ${\sigma }_i^{-1}$ both induce the adjacent transposition, its endpoint permutation is $(12)(23)(12)(23)$, a $3$-cycle; [F8] therefore makes its closure one knot component. Smoothing each crossing straightens its adjacent oriented strands, leaving the identity three-strand closure with three cycles. Hence $(s,c)=(3,4)$.

The kinked unknot diagram has one crossing; its oriented smoothing separates the small kink loop from the remaining circle, so $(s,c)=(2,1)$. In the zero-crossing boundary case, the round unknot has one smoothing circle and no bands, so $(s,c)=(1,0)$.

By [F3], all three outputs are connected oriented spanning surfaces. Applying [F4] gives $(1-2+3)\mathrm{/}2=1$, $(1-3+4)\mathrm{/}2=1$, and $(1-2+1)\mathrm{/}2=0$. For the round unknot the disk surface has genus $0$, and all surface genera are nonnegative by [F6]; therefore its minimum is exactly $0$. For the trefoil and figure-eight, [F5] yields only the upper bound $g(K)\le 1$, as claimed.

Let $U$ be the standard round unknot and $D$ a smooth embedded spanning disk. Remove the interiors of two disjoint small disks from $\mathrm{Int}D$ to obtain a pair of pants $P$. Join the two new boundary circles by a simple arc in a small ball on one side of $D$ and disjoint from $U$; the lateral boundary of a sufficiently thin tubular neighborhood of that arc is an embedded annulus $A$ meeting $P$ only in its two attaching circles. Glue with matching orientations. Then $F=P\cup A$ is connected, compact, smooth, embedded, oriented, and has only boundary $U$. Choose finite cell structures with both attaching circles subcomplexes. Cell-count inclusion-exclusion gives $\chi (P)=\chi (D)-2\chi (D^2)+2\chi (S^1)=1-2+0=-1$; then $\chi (F)=\chi (P)+\chi (A)-2\chi (S^1)=-1+0-0=-1$.

The surface $F$ has $b=1$. By [F3] and [A1], $-1=\chi (F)=2-2g(F)-1$, hence $g(F)=1$. The original disk $D$ has $b=1$ and $\chi (D)=1$, so [F3] gives $g(D)=0$.

By [F1], both $D$ and $F$ are admissible spanning surfaces for $U$. Therefore [F2] and nonnegativity of surface genus give $g(U)=0$, while step 2.1 gives $g(F)=1$. Thus $F$ is not minimum genus and the refuted statement is false.