Group presentation for positive trefoil $+1$ surgery

In [F1], for $0<t<1\mathrm{/}2$ the left-starting motion is $(2ht-h,-2ht)$ and the right-starting motion its negative. At the crossing the latter has greater $y$, hence is over when viewed along $+y$. Their projected tangents in coordinates $(t,x)$ are $(1,-2h)$ and $(1,2h)$, with determinant $4h>0$. Rounding the diamond corners and flattening endpoints preserves this strict sign. The downward three-crossing diagram therefore has three positive crossings. The diagram agrees with the fixed geometric closure: the change from stacked diamonds to rounded diagonal strands is a height-preserving interpolation with nonzero pair difference, followed by the fixed closure map. In particular it is not the mirror braid.

Label the initial left and right meridians $a,b$. At a positive crossing the right input is over, and the left input follows the positive over-normal to become the right output. Successive rows of the braid have input/output pairs $\begin{array}{cccc}\text{row}& \text{input }(L,R)& \text{output }(L,R)& \text{underpass relation}\\ 1& (a,b)& (b,c)& c=b^{-1}ab\\ 2& (b,c)& (c,a)& a=c^{-1}bc\\ 3& (c,a)& (a,b)& b=a^{-1}ca\end{array}$ Here the last two output labels use the closure identifications of bottom and top positions. By [F8], this is the finite three-crossing code to which the Wirtinger correctness theorem [F3] applies. These are all crossing relations by [F3]. Substituting $c=b^{-1}ab$ into $b=a^{-1}ca$ gives $aba=bab$; the remaining relation follows because $b^{-1}ab=ab{a}^{-1}$ and then $c^{-1}bc=a$. Thus ${\pi }_1(S^3\setminus T)=⟨a,b\mid aba=bab⟩$, with the stated positive meridians. In the chosen tube, the radial normal homotopy from positive radius $r<ϵ$ to $(1-s)r+sϵ$ identifies this group with ${\pi }_1(E_T)$.

Trace a small parallel starting at the initial left arc. Along the oriented component its underpasses occur in row order $1,3,2$, with over-meridians $b,a,c$; it passes over the intervening crossings. This order follows directly from the input/output table: $a$ becomes $c$ on the right in row 1, passes over in row 2, becomes $b$ in row 3, closes to the initial right position, passes over in row 1, becomes $a$ in row 2 and closes after passing over in row 3. At each underpass the access-path calculation in [F3] is ${\alpha }_{\mathrm{i}\mathrm{n}}\beta =g_o{\alpha }_{\mathrm{o}\mathrm{u}\mathrm{t}}$, where $\beta$ is the path along the small parallel; thus the three successive factors are $b,a,c$, with positive exponents. At an overpass the access path slides through its upper product chart and contributes no factor. On the crossing-free exterior closure arcs use the same product charts, again contributing no factor. Concatenating gives the based parallel word $w=bac$. These path identities can be made on the upper side of the normal circles. The upper normal section and the blackboard section are homotopic without zero: in the normal quotient the blackboard vector and viewing vector form a positively oriented basis, so the interpolation rotating the former through a fixed quarter-turn to the latter is everywhere nonzero. This proves that the traced parallel has the blackboard framing and oriented degree $+1$, rather than identifying a peripheral word only from its exponent sum. By [F4] and step 1.1 its self-linking is $+3$.

The boundary torus has product coordinates from this actual parallel: its degree-one section is $w$ and its positive fibre is $a$. A section making three negative fibre turns is an embedded degree-one curve with based word $w{a}^{-3}$; the torus rectangle homotopy makes the fibre and section commute. In the abelianization of step 2.1 the relation $aba=bab$ gives $[a]=[b]$, and sending each to $1$ supplies an inverse to the cyclic generator map. Thus $H_1(E_T)=\mathbb{Z}$ on $[a]$ by [F6], and $[w]=[b]+[a]+[c]=3[a]$. Therefore $\lambda =w{a}^{-3}$ has zero exterior homology and degree $+1$, precisely [F5]. Moreover $c=ab{a}^{-1}$, so $w{a}^3=b{a}^{2}b{a}^2=ba(bab)a=(ba{)}^3=(ab{)}^3\mathrm{.}$ Both last cubes equal $(aba{)}^2$, which commutes with $a,b$ because conjugation by $aba$ exchanges them. This proves $\lambda =(ab{)}^3{a}^{-6}$ with the fixed positive meridian and the correct sign.

The $+1$ slope is the primitive boundary curve $\lambda \mu$, since its coefficients in the oriented meridian/longitude basis are $(1,1)$. In [F7] the solid torus has cyclic fundamental group; a complementary boundary generator identifies its core generator and eliminates it, while its meridian kills precisely $\lambda \mu$. Thus filling adds $w{a}^{-2}=1$. Substituting $w=bac$ and $c=b^{-1}ab$ gives $ba{b}^{-1}ab{a}^{-2}=1$. Equivalently step 4.1 gives $(ab{)}^3=a^5$. These are the asserted presentations of the same positive-trefoil $+1$ filling.