The Khovanov-Seidel grid resolutions of the vertex modules

Basis and products in the vertex projectives. By [L3] the $4m+1$ displayed classes form a $\mathbb{Z}$-basis of $A_m$, and each basis path has a source and a target, so by [F2] the module $P_j=A_m{e}_j$ has the $\mathbb{Z}$-basis consisting of $e_j$ together with the basis paths ending at $j$, namely $e_j$; $u_{j-1}=(j-1\mathrm{\mid }j)$; $d_j=(j+1\mathrm{\mid }j)$; $r_j$, a symbol being omitted when the arrow has no source or target in $\{0,\dots ,m\}$. Thus $P_j$ has basis $\{e_j,u_{j-1},d_j,r_j\}$ for $0<j<m$, basis $\{e_0,d_0\}$ for $j=0$ (since $(0\mathrm{\mid }1\mathrm{\mid }0)$ has class $0$ and $u_{-1}$ does not exist) and basis $\{e_m,u_{m-1},r_m\}$ for $j=m$ (since $d_m$ does not exist). For the multiplication we use [F1], [F2] and [L3]: $x{e}_j=x$ for $x\in P_j$, while $e_j{e}_j=e_j$; $u_{j-1}{u}_j$ is a monotone length-two path, hence $0$ by [F1], while $r_j{u}_j$ and $r_j{d}_{j-1}$ are classes of paths of length three, hence $0$; and $d_j{u}_j=(j+1\mathrm{\mid }j\mathrm{\mid }j+1)$ is the return at $j+1$, which is exactly the basis return $r_{j+1}=(j+1\mathrm{\mid }j\mathrm{\mid }j+1)$.

The staircase is a first-quadrant double complex in $A_m\text{-mod}$. The entries $C_{p,q}=P_{j(p,q)}\{p\}$ are graded left $A_m$-modules, vanish for $p<0$ or $q<0$, and are nonzero exactly on $0\le p\le m-i$, $0\le q\le i+p$; the staircase condition $q\le i+p$ means $j(p,q)\ge 0$, and $p\le m-i$ means $j(p,q)=i+p-q\le i+p\le m$, so the projective indices are in range. When both source and target entries are nonzero, right multiplication by $\alpha =(j\mathrm{\mid }j-1)$ carries $P_j$ into $P_j\alpha \subseteq P_{j-1}$, and by [L4] it is a degree-zero module map $P_j\{p\}\to P_{j-1}\{p-1\}$ because $\mathrm{deg}\alpha =1$ exactly compensates the drop of the shift; right multiplication by $\beta =(j\mathrm{\mid }j+1)$ carries $P_j$ into $P_{j+1}$ and is degree-zero as a map $P_j\{p\}\to P_{j+1}\{p\}$ because $\mathrm{deg}\beta =0$. Hence $h_{p,q}:C_{p,q}\to C_{p-1,q}$ and $v_{p,q}:C_{p,q}\to C_{p,q-1}$ are degree-zero module maps, the sign $(-1{)}^p$ being a unit of the ground ring. On $C_{p,q}$, $h^2$ is right multiplication by the monotone length-two path $(j\mathrm{\mid }j-1\mathrm{\mid }j-2)$, hence is $0$ by [F1] (and $h^2=0$ at the boundary, where one of the two symbols does not exist); similarly $v^2$ is right multiplication by the monotone length-two path $(j\mathrm{\mid }j+1\mathrm{\mid }j+2)=0$ by [F1], and $hv+vh$ is right multiplication by $(-1{)}^{p-1}(j\mathrm{\mid }j-1\mathrm{\mid }j)+(-1{)}^p(j\mathrm{\mid }j+1\mathrm{\mid }j)=(-1{)}^{p-1}[(j\mathrm{\mid }j-1\mathrm{\mid }j)-(j\mathrm{\mid }j+1\mathrm{\mid }j)]=0$ by the relation of [F1]. For the mixed relation at the staircase boundary, if $p=0$ or $q=0$ both composites have zero target. If $p,q>0$ and $j=0$, the horizontal-first composite vanishes and the vertical-first composite multiplies by $(0\mathrm{\mid }1\mathrm{\mid }0)=0$. In all other nonzero mixed squares $0<j<m$, so the displayed return relation applies. Maps with zero source or target are zero by convention; the same convention handles the boundary cases of $h^2$ and $v^2$. Thus $(C,h,v)$ satisfies [L7].

The columns are exact above their top entry. Fix $0\le j\le m-1$ and $x=\alpha e_j+\beta u_{j-1}+\gamma d_j+\delta r_j\in P_j$. By the product rules of step 1.1, $x{u}_j=\alpha u_j+\beta u_{j-1}{u}_j+\gamma d_j{u}_j+\delta r_j{u}_j=\alpha u_j+\gamma r_{j+1}$, so by basis independence $x{u}_j=0$ if and only if $\alpha =\gamma =0$, that is, if and only if $x\in \mathbb{Z}{u}_{j-1}\oplus \mathbb{Z}{r}_j$. On the other hand right multiplication by $u_{j-1}$ on $P_{j-1}$ has image $\{x^{\mathrm{\prime }}{u}_{j-1}\}$ spanned by $e_{j-1}{u}_{j-1}=u_{j-1}$ and $d_{j-1}{u}_{j-1}=r_j$, the other two basis products being $0$ by step 1.1; so the image is $\mathbb{Z}{u}_{j-1}\oplus \mathbb{Z}{r}_j$ and $\ker (x\mapsto x{u}_j)=\mathrm{im}(x^{\mathrm{\prime }}\mapsto x^{\mathrm{\prime }}{u}_{j-1})$. For $j=0$ the same computation gives $x{u}_0=\alpha u_0+\gamma r_1$ (using $r_0=0$ by [L3]); the two images are independent basis classes, so the kernel is $0=\mathrm{im}(x^{\mathrm{\prime }}\mapsto x^{\mathrm{\prime }}{u}_{-1})$, with $P_{-1}=0$. In the column $C_{p,\ast }$ the maps $v$ multiply by $u_{i+p-q}$ and the statement just proved identifies, for every $q\ge 1$, the kernel of $v:C_{p,q}\to C_{p,q-1}$ with the image of $v:C_{p,q+1}\to C_{p,q}$, since $C_{p,q+1}=P_{j-1}\{p\}$ and $C_{p,q-1}=P_{j+1}\{p\}$; hence $H_q^v(C_{p,\ast })=0$ for every $p$ and every $q>0$.

The bottom row of the assembled complex. By step 2.2 and [L7], $B_p:=H_0^v(C_{p,\ast })=C_{p,0}\mathrm{/}\mathrm{im}(v:C_{p,1}\to C_{p,0})$ for $0\le p\le m-i$, where $C_{p,0}=P_{i+p}\{p\}$ and the image is $P_{i+p-1}{u}_{i+p-1}=\mathbb{Z}{u}_{i+p-1}\oplus \mathbb{Z}{r}_{i+p}$ for $i+p\ge 1$ and $0$ for $i=p=0$; the differential $h$ descends to $B$ by the anticommutation proved in step 2.1. So $B_p$ is the quotient of $P_{i+p}$ by the $\mathbb{Z}$-span of $u_{i+p-1}$ and $r_{i+p}$, with $\mathbb{Z}$-basis the classes $[e_{i+p}]$ and $[d_{i+p}]$ when $i+p<m$, and basis $[e_m]$ when $i+p=m$. On these classes $h:B_p\to B_{p-1}$ is induced by right multiplication by $d_{i+p-1}$, and by step 1.1 it sends $[e_{i+p}]\mapsto [d_{i+p-1}]$ and $[d_{i+p}]\mapsto [d_{i+p}{d}_{i+p-1}]=0$.

The homology of the bottom row is $S_i$. Since $h_1:B_1\to B_0$ has image $\mathbb{Z}[d_i]$ (and is $0$ when $m=i$), $H_0(B)=B_0\mathrm{/}\mathbb{Z}[d_i]=\mathbb{Z}[e_i]$; the quotient $C_{0,0}=P_i$ modulo the span of $u_{i-1},d_i,r_i$ is the quotient of $P_i$ by $A_m{u}_{i-1}+A_m{d}_i$, which by step 1.1 equals the submodule generated by all positive-length paths ending at $i$, so $H_0(B)$ is $\mathbb{Z}$ in degree $0$ with $e_i$ acting as the identity and every other vertex idempotent and every positive-length path acting as zero: by [L5] this is exactly $S_i$. For $1\le p<m-i$, step 3.1 gives $\ker h_p=\mathbb{Z}[d_{i+p}]=\mathrm{im}{h}_{p+1}$, while at $p=m-i\ge 1$ the group $B_p=\mathbb{Z}[e_m]$ maps injectively to $B_{p-1}$, so its kernel is $0=\mathrm{im}{h}_{p+1}$; when $i=m$ there are no positive-degree $B_p$ at all. Hence $H_p(B)=\ker h_p\mathrm{/}\mathrm{im}{h}_{p+1}=0$ for every $p\ge 1$.

Totalisation is a finite graded projective resolution of $S_i$. Step 2.2 verifies the hypothesis of [L8], so the projection $\mathrm{Tot}(C)\to B$ is a quasi-isomorphism; by step 4.1 and [L10] therefore $H_0(\mathrm{Tot})\cong S_i$ and $H_n(\mathrm{Tot})=0$ for $n\mathrm{\ne }0$. Each ${\mathrm{Tot}}_n$ is the finite direct sum of the $C_{p,q}=P_{j(p,q)}\{p\}$ with $p+q=n$, a finite direct sum of internal shifts of vertex projectives; each $P_j$ is finite graded projective by [L4], its shift is finite graded projective because $\{p\}$ is an automorphism of $A_m\text{-mod}$ by [L4], and a finite direct sum of projective objects is projective by the lifting property applied componentwise with the universal property of the coproduct, so every term of $\mathrm{Tot}$ is finite graded projective. Only the finitely many degrees $0\le n\le m+(m-i)$ can be nonzero, and the differential satisfies $d^2=0$ by [L7]. The augmentation $\epsilon :{\mathrm{Tot}}_0=C_{0,0}=P_i\to S_i$ is the quotient map of step 4.1; since the projection of [L8], followed by the augmentation $B_0\to S_i$, restricts to this quotient on $C_{0,0}$, the augmented complex is exact at $P_i$ and at every other term, so by [L6] $\mathrm{Tot}$ is a finite graded projective resolution of $S_i$.

The cone of multiplication by $p$ resolves $S_i\mathrm{/}p{S}_i$. Multiplication by the integer $p$ is an $A_m$-linear endomorphism of each $P_j$ preserving degrees, so it defines a degree-zero chain map ${\mu }_p:{\mathrm{Tot}}_{\bullet }\to {\mathrm{Tot}}_{\bullet }$. By [L9] the cone has $\mathrm{Cone}({\mu }_p{)}_n={\mathrm{Tot}}_n\oplus {\mathrm{Tot}}_{n-1}$, a finite direct sum of finite graded projectives, and there is a degreewise split short exact sequence of complexes $0\to \mathrm{Tot}\to \mathrm{Cone}({\mu }_p)\to \mathrm{Tot}[1]\to 0$ whose long exact homology sequence reads $H_n(\mathrm{Tot})\stackrel{{\mu }_p}{\to }{H}_n(\mathrm{Tot})\to H_n(\mathrm{Cone}({\mu }_p))\to H_{n-1}(\mathrm{Tot})\stackrel{{\mu }_p}{\to }{H}_{n-1}(\mathrm{Tot})$. By step 5.1 the only nonzero homology of $\mathrm{Tot}$ is $H_0=S_i$; hence $H_0(\mathrm{Cone}({\mu }_p))=\mathrm{coker}(p:S_i\to S_i)=S_i\mathrm{/}p{S}_i$, $H_1(\mathrm{Cone}({\mu }_p))=\ker (p:S_i\to S_i)=0$ because $S_i=\mathbb{Z}$ is torsion free and $p\mathrm{\ne }0$, and $H_n(\mathrm{Cone}({\mu }_p))=0$ for $n\ge 2$ and for $n\le -1$. So the cone is a finite complex of finite graded projectives whose only homology is $S_i\mathrm{/}p{S}_i$ in degree $0$, that is, a finite graded projective resolution of $S_i\mathrm{/}p{S}_i$ in the sense of [L6].

Conclusion. The double complex $(C,h,v)$ of step 2.1 is first quadrant, its columns are exact above the top entry by step 2.2, and its total complex is the finite graded projective resolution of $S_i$ described in claim 1 (steps 4.1 and 5.1); the mapping cone of multiplication by $p$ on it is the finite graded projective resolution of $S_i\mathrm{/}p{S}_i$ described in claim 2 (step 6.1). The word "simple" in the identifier of this item asserts nothing beyond the explicit vertex modules $S_i$ of [L5].