Pointed odd and Bar–Natan tensor products over the marked rings

A right-linear reduced decomposition. At each state let $V$ be the circle group, $Z$ its marked generator, and $U=\ker (\mathrm{aug}:V\to \mathbb{Z})$, where every circle has augmentation one. The vectors $Y_j=Z_j-Z$ for unmarked circles are a basis of $U$, and $V=U\oplus \mathbb{Z}Z$. Thus $\mathrm{\Lambda }V=\mathrm{\Lambda }U\otimes \mathbb{A}$ as a right module, by $u\otimes 1\mapsto u$, $u\otimes X\mapsto u\wedge Z$. A merge preserves augmentation and takes the marked generator to the marked generator. For a split choose the lift of its old marked generator to the new marked generator; if the marked circle is unaffected this is the usual lift. This lift also preserves augmentation, and the split difference belongs to $U$ in the target. Hence in these coordinates every edge acts as its reduced exterior quotient or split on $\mathrm{\Lambda }U$, tensored with the identity of $\mathbb{A}$. This is an equality of maps on both $u$ and $u\wedge Z$, not merely an unmarked abelian-group splitting.

The map and its exact signs. For a vertex $v_j$ set ${\sigma }_j(v_j)=(k_j(v_j)-k_j(0)+\mathrm{\mid }{v}_j\mathrm{\mid })\mathrm{/}2$. A merge leaves ${\sigma }_j$ fixed and a split increases it by one. At a joined state the reduced difference group is $U_1\oplus U_2$, by sending every $Z_i-Z_j^{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}}$ to its difference from the merged marked generator. For $u_j\in {\mathrm{\Lambda }}^{r_j}{U}_j$ and $a_j\in \mathbb{A}$, define $\mathrm{\Phi }((u_1\otimes a_1)\otimes (u_2\otimes a_2))=(-1{)}^{r_1{\sigma }_2(v_2)}(u_1\wedge u_2)\otimes a_1{a}_2$. The degree $r_1$ here is the reduced exterior degree and excludes the marked $X$-factor. The formula is balanced and right $\mathbb{A}$-linear because its sign is independent of $a_1,a_2$. Its reduced wedge basis times $1,X$ is exactly the joined basis, so it is bijective. Conjugating the tensor differential gives the following coefficients relative to the joined unsigned saddles: ${ϵ}_1$ for a first-diagram merge, $(-1{)}^{{\sigma }_2}{ϵ}_1$ for a first-diagram split, and $(-1{)}^{i(v_1)}{ϵ}_2$ for either second-diagram saddle. For the last assertion, a second split increases ${\sigma }_2$ by one, contributing $(-1{)}^{r_1}$; moving its left split difference past $u_1$ contributes the same sign, and the two cancel. The remaining sign is the tensor cochain sign. A first split increases $r_1$ by one and supplies precisely $(-1{)}^{{\sigma }_2}$. Merges preserve reduced exterior degree. These equations prove that $\mathrm{\Phi }$ is a chain isomorphism for the induced signs.

The induced signs are fully admissible. A two-face within one diagram has the same geometric coefficient as that diagram's face: insertion of the other marked arc and its passive circles changes neither the cyclic order of its saddle endpoints nor the oriented inside/outside test. On a first-diagram face the additional sign in step 2.1 has boundary exponent ${\sigma }_2$ times the sum of the split counts in its two paths. The counts agree, since each equals $(k_{\mathrm{o}\mathrm{u}\mathrm{t}}-k_{\mathrm{i}\mathrm{n}}+2)\mathrm{/}2$, so this exponent is even. On a second-diagram face the common extra sign $(-1{)}^{i(v_1)}$ occurs four times. Thus all internal face products are unchanged. For a mixed face let $b_j$ be its saddle's split indicator. The joining strip separates the two saddle disks by a two-edge cut. Their unsigned maps commute except for the interchange of two reduced split factors, giving $\kappa =(-1{)}^{b_1{b}_2}$; no alternating-endpoint zero face occurs across that cut. The induced coefficient product is $(-1{)}^{1+b_1{b}_2}$: changing a first coordinate changes $i(v_1)$ by one, while a second split changes the first split's factor $(-1{)}^{{\sigma }_2}$ by $(-1{)}^{b_1{b}_2}$. This is exactly $-\kappa$. Hence all prescribed faces, including the internal zero faces, have their required products. By [F2], any other admissible realization is related to this one by a marked vertex-sign isomorphism.

The Bar–Natan map, gradings and reduction. At a state multiply the two marked circle factors, leave every unmarked circle factor unchanged, and multiply by $(-1{)}^{n_{-}^{(1)}\mathrm{\mid }{v}_2\mathrm{\mid }}$. Balanced tensoring over $\mathbb{R}$ identifies its basis with the joined state basis, so this is an isomorphism. Associativity and the Frobenius identity in [F3] show that the unsigned multiplication commutes with every saddle. For first-diagram edges the added vertex sign is unchanged. For second-diagram edges it changes by $(-1{)}^{n_{-}^{(1)}}$, converting the tensor cochain sign $(-1{)}^{i(v_1)}$ into the joined cube sign $(-1{)}^{\mathrm{\mid }{v}_1\mathrm{\mid }}$, since $i(v_1)=\mathrm{\mid }{v}_1\mathrm{\mid }-n_{-}^{(1)}$. This proves the signed chain equation. Both maps identify two marked circles with one, so their quantum degree before the displayed shift is $-1$. Crossing counts and $n_{+},n_{-}$ add, giving the stated normalized homological and quantum degrees. In characteristic two the odd sign in step 2.1 is one, $Y_j$ corresponds to the even label operator $X_j+X_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}}$, and the merged marked generator corresponds to $X_{\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}}$. Thus the reduced wedge product times $a_1{a}_2$ is exactly the even state-label identification obtained by marked multiplication. Both theories therefore induce the same comparison square after the required quotients. This proves all assertions.