Oriented kink shifts for braid diagrams

The complex of the type IA curl. By [F1] and [F3], setting $x_3=x_2$ in the Koszul forms (8), (9) presents $C(D_2)\{0,2\}$ as the two-term complex whose two terms are the factorization $X\otimes (0,0)$ with $X=(a,x_1-x_4)$, and whose differential is the morphism $\mathrm{I}\mathrm{d}\otimes \psi (x_4-x_2)$ of [F3]. Indeed both rows of the two matrices become $(a,x_1-x_4)$ and $(0,0)$, and the flip morphism $\psi (x_4-x_2)$ has components the identity on the first term and multiplication by $x_4-x_2$ on the middle term, where the shift $R\{-1,3\}\to R\{-1,1\}$ makes the total bidegree $(0,0)$; the potential of the curl diagrams is $w=a(x_1-x_4)$ and $x_2$ is the label of the internal mark.

Splitting off the first row. The differential is the identity on the $X$ factor; in the two-term complex with terms $X\otimes (0,0)$ and differential $\mathrm{I}\mathrm{d}\otimes \psi (x_4-x_2)$, the summand on which $\psi$ acts by the identity splits off as the contractible complex $0\to X\stackrel{1}{\to }X\to 0$. What remains is the tensor product of $X$ with the two-term complex $0\to R\{-1,3\}\stackrel{x_4-x_2}{\to }R\{-1,1\}\to 0$, with the surviving row in odd inner parity, exactly the splitting displayed in section 4 of the source. Tensoring $X$ with this odd scalar row yields $\mathrm{\Pi }X$ with the indicated internal shifts, not the even scalar tensor unit.

The type IB computation. Relabeling its external ends to have potential $a(x_1-x_4)$, the positive curl has the same specialization $x_3=x_2$ in the two Koszul forms, but uses ${\mathrm{I}\mathrm{d}}_X\otimes {\psi }^{\mathrm{\prime }}(x_4-x_2)$ with the positive source shift $\{0,2\}$. The odd components have the identical shift $\{-1,3\}$ and the map between them is the identity, so that pair is contractible. The even components leave $0\to R\{0,2\}\stackrel{x_4-x_2}{\to }R\to 0$, in cohomological degrees $-1,0$. Polynomial division gives $R=R^{\mathrm{\prime }}\oplus (x_4-x_2)R$ as an $R^{\mathrm{\prime }}$-module; multiplication by $x_4-x_2$ is an isomorphism from the source to the second summand of the target and is homogeneous with the given shifts. Canceling that pair leaves only $R^{\mathrm{\prime }}$ in degree $0$ with zero shift. Tensoring with $X$ therefore leaves exactly the straight-strand factorization, so $C(E_1)\cong C(E_2)$ with no shift.

Excluding the internal variable. The variable $x_2$ is internal with $w=a(x_1-x_4)\in R^{\mathrm{\prime }}=\mathbb{Q}[a,x_1,x_4]$ and the surviving row is $(0,x_4-x_2)=-(0,x_2-x_4)$; by [F2], with $y=x_2$ and $\mu =x_4$, the polynomial-subspace map $Ro(x_4-x_2)R$ is invertible over $R^{\mathrm{\prime }}$ and its two-term subcomplex splits off contractibly and the remaining factorization descends to $R^{\mathrm{\prime }}$ with $x_2\mapsto x_4$. The two-term complex $0\to R\{-1,3\}\stackrel{x_4-x_2}{\to }R\{-1,1\}\to 0$ therefore reduces to the odd-parity shifted copy $R^{\mathrm{\prime }}\{-1,1\}[-1]$ of the scalar complex, and $C(D_2)\{0,2\}\cong \mathrm{\Pi }X\{-1,1\}[-1]=\mathrm{\Pi }C(\mathrm{\Gamma })\{-1,1\}[-1]$ with $\mathrm{\Gamma }$ the straight-strand diagram. Since $\mathrm{\Gamma }$ is $D_1$ with the same labels and the same potential, rearranging the shifts and using ${\mathrm{\Pi }}^2=1$ gives $C(D_1)\cong \mathrm{\Pi }C(D_2)\{1,1\}[1]$, which is the type IA conclusion.

Conclusion. Steps 1.1-3.1 establish $C(D_1)\cong \mathrm{\Pi }C(D_2)\{1,1\}[1]$ for the type IA pair, with inner parity retained separately from the three gradings; after forgetting inner parity on termwise cohomology the shift is $\{1,1\}[1]$; step 2.2 establishes $C(E_1)\cong C(E_2)$ with no shift for the type IB pair. The two shift conventions differ, so the two conclusions cannot be interchanged, and the caveat records that the labels IA and IB refer to the source's printed oriented pictures.