The positive and negative Khovanov-Rozansky crossing complexes

The positive complex is a complex of factorizations. In a two-term complex the composite of its two differentials is zero on one side because there is nothing to compose on the other, so the complex condition in ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$ is automatic; the term $C({\mathrm{\Gamma }}_0)\{0,2\}$ lies in cohomological degree $-1$ and $C({\mathrm{\Gamma }}_1)$ in degree $0$, and both terms are objects of ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$ with potential $w=a(x_1+x_2-x_3-x_4)$ because the source and target of ${\chi }_0$ have that potential. The differential ${\chi }_0$ has bidegree $(0,2)$ by [F1], and the shift $\{0,2\}$ on its source subtracts $(0,2)$ from the bidegree of the map of shifted terms, so the differential of $C_p$ has bidegree $(0,0)$ as required.

The negative complex is a complex of factorizations. Likewise the two-term complex $0\to C({\mathrm{\Gamma }}_1)\{0,-2\}\stackrel{{\chi }_1}{\to }C({\mathrm{\Gamma }}_0)\{0,-2\}\to 0$ has zero composite on the one composable side, its terms are objects of ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$, and ${\chi }_1$ has bidegree $(0,0)$ by [F1]; the overall shift $\{0,-2\}$ is applied to both terms, so it shifts the grading of both terms alike and leaves the differential bidegree-preserving. With $C({\mathrm{\Gamma }}_1)\{0,-2\}$ in degree $0$ and $C({\mathrm{\Gamma }}_0)\{0,-2\}$ in degree $1$, the two terms are exactly the cone of ${\chi }_1$.

The source conflict and the grading comparison. In the integer grading, the negative cone has Euler characteristic $q^{-2}(⟨D{e}_i⟩-⟨D⟩)$, as in arXiv section 7, whereas reversing its two terms changes the sign; moreover the printed ${\chi }_0$ with equal shifts would still have bidegree $(0,2)$ by [F1]. Thus that prose display is incompatible with both the bidegree condition and Figure 6. For the published positive cone, shifting the outer degrees $-1,0$ by $[-\frac{1}{2}]$ gives $-\frac{1}{2},\frac{1}{2}$, and adding $(-\frac{1}{2},-\frac{1}{2})$ to the internal shifts gives $(-\frac{1}{2},\frac{3}{2})$ and $(-\frac{1}{2},-\frac{1}{2})$, exactly formula (12). For the negative cone, $[\frac{1}{2}]$ moves degrees $0,1$ to $-\frac{1}{2},\frac{1}{2}$, and adding $(\frac{1}{2},\frac{1}{2})$ to both internal shifts $(0,-2)$ gives $(\frac{1}{2},-\frac{3}{2})$, exactly formula (13). The maps remain ${\chi }_0,{\chi }_1$, up to compatible shift signs; agreement with the published cones therefore requires the recorded regrading.