The preferred lift of a half twist shifts the bigrading by chi(-1,1)

The deck element exists and is unique. Since $\tau (c)=c$, the preferred lift sends the bigrading $\tilde{c}$ of $c$ to a bigrading $\tilde{\tau }(\tilde{c})$ of the same curve $c$; by [L2] there is a unique $(r_1,r_2)\in {\mathbb{Z}}^2$ with $\tilde{\tau }(\tilde{c})=\chi (r_1,r_2)\tilde{c}$, and it is independent of the chosen bigrading because the deck group is abelian and acts freely. The whole content of the lemma is the computation of $(r_1,r_2)$.

The test loop in $P$ and its class. Use the standard rotational half-twist representative along $c$, conjugated from a round support disk; its midpoint is its unique fixed point on $c$, its derivative there is $-I$, and it is the identity near $\mathrm{\partial }D$. Choose the standard embedded path $\beta :[0,1]\to D\setminus \mathrm{\Delta }$ from the boundary to that midpoint, as in source Figure 9, with nonzero endpoint tangents. Let $\pi \text{\hspace{0.17em}⁣}:[0,2]\to P$ be the closed path $\pi (t):=\mathbb{R}{\beta }^{\mathrm{\prime }}(t)(0\le t\le 1),\pi (t):=D\tau (\mathbb{R}{\beta }^{\mathrm{\prime }}(2-t))(1\le t\le 2),$ where $\mathbb{R}v$ denotes the tangent line spanned by $v$; the two halves match at $t=1$ because $D\tau =-I$ at the midpoint fixes every projective tangent line, and the endpoint lines at the boundary match because $\tau$ is the identity nearby, and $\pi$ is a loop in $P$. For this standard path, the source's Figure 9 computation gives $[\pi ]=-[\mathbb{R}{P}^1\times \mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}]-[\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\times {\lambda }_z]$ for one endpoint $z$ of $c$; this is the literature calculation in Khovanov--Seidel Lemma half-twist, printed pp. 24--25. Transport by the support-disk coordinates preserves its value under the covering class because all positive puncture loops have monodromy $(-2,1)$.

The value of the deck element. By the definition of the local index and the preferred lift, the deck element $(r_1,r_2)$ comparing $\tilde{\tau }(\tilde{c})$ with $\tilde{c}$ is obtained by evaluating the classifying class $C$ on the loop of tangent lines swept by the preferred lift of $\tau$ along $c$, which is the class $[\pi ]$ of step 1.2; hence $(r_1,r_2)=-C([\pi ])=C([\mathbb{R}{P}^1\times \text{point}])+C([\text{point}\times {\lambda }_z])=(1,0)+(-2,1)=(-1,1)$. This proves the main formula.

The consequence for the string tables. The half twist ${\tau }_k$ along $b_k$ is the preferred lift acting on bigraded curves, so by [L3] and the main formula $I^{\mathrm{b}\mathrm{i}\mathrm{g}\mathrm{r}}({\tilde{b}}_k,{\tilde{\tau }}_k(\tilde{c}))=I^{\mathrm{b}\mathrm{i}\mathrm{g}\mathrm{r}}({\tilde{\tau }}_k^{-1}{\tilde{b}}_k,\tilde{c})=I^{\mathrm{b}\mathrm{i}\mathrm{g}\mathrm{r}}(\chi (1,-1){\tilde{b}}_k,\tilde{c})=(q_1^{-1}{q}_2)I^{\mathrm{b}\mathrm{i}\mathrm{g}\mathrm{r}}({\tilde{b}}_k,\tilde{c}),$ where the first equality uses the invariance of $I^{\mathrm{b}\mathrm{i}\mathrm{g}\mathrm{r}}$ under the preferred lifts and the second uses that ${\tilde{\tau }}_k^{-1}$ acts on ${\tilde{b}}_k$ by $\chi (1,-1)$ by the main formula applied at index $k$. Applied to a $k$-string, this is the factor $(q_1^{-1}{q}_2{)}^u$ of the source's table. For other representatives of the same half-twist class, the same identities hold on bigraded isotopy classes by the homotopy-lifting and freeness suppliers. AC is inherited for the supplied bigraded intersection-number invariance in this consequence.