ScalingStacks

Due to the direct sum decomposition $R={\mathop{\oplus}\limits_{k\geq 0}}c^{k}\mathbb{Z}$ of abelian groups, we have an abelian group decomposition

$$C^{i}_{j}(D)={\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i}_{j-2k}(D)$$

(172)

where, recall, $C^{i}_{j}(D)$ and ${\cal C}^{i}_{j}(D)$ were defined in Sections 4.2 and 7.1 respectively. Let us fix a $j\in\mathbb{Z}.$ Denote by $d$ the differential in the weight $j$ subcomplex of the complex $C(D)$:

$$\cdots\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i-1}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}C^{i+1}_{j}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\cdots$$

(173)

and by $\partial$ the differential in the complex

$$\cdots\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i-1}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i+1}_{j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}\cdots$$

(174)

where we suppress the dependence of $\partial$ on $k$. Under the identification (172) differential $d$ becomes a differential of the complex

$$\cdots\stackrel{{\scriptstyle d}}{{\longrightarrow}}{\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i}_{j-2k}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}{\mathop{\oplus}\limits_{k\geq 0}}{\cal C}^{i+1}_{j-2k}(D)\stackrel{{\scriptstyle d}}{{\longrightarrow}}\cdots$$

(175)

Consider a bigraded abelian group

$$C={\mathop{\oplus}\limits_{k\geq 0,i\in\mathbb{Z}}}{\cal C}^{i}_{j-2k}(D)$$

(176)

where we set the grading of ${\cal C}^{i}_{j-2k}(D)$ to $(i,-k).$ We thus have a bigraded abelian group $C$ and two maps, $d$ and $\partial$, from $C$ to $C.$ Map $\partial$ is bigraded of degree $(1,0)$ while $d$ is only graded relative to the first grading. However, we can decompose

$$d=\partial+\widetilde{\partial}$$

(177)

where $\widetilde{\partial}$ has grading $(1,-1)$ and satisfies

	$\displaystyle\widetilde{\partial}\widetilde{\partial}$	$\displaystyle=$	$\displaystyle 0,$		(178)
	$\displaystyle\partial\widetilde{\partial}+\widetilde{\partial}\partial$	$\displaystyle=$	$\displaystyle 0$		(179)

Besides, since $\partial$ is a differential, $\partial\partial=0.$ Therefore, $H^{i,j}(D)$ is equal to the $i$-th cohomology group of the total complex of the bicomplex $(C,\partial,\widetilde{\partial}).$ Group ${\cal H}^{s,j-2k}(D)$ is equal to the $s$-th cohomology group of the subcomplex (relative to the differential $\partial$)

$$\cdots\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i-1,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}{\cal C}^{i,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}C^{i+1,j-2k}(D)\stackrel{{\scriptstyle\partial}}{{\longrightarrow}}\cdots$$

(180)

of $C.$ Therefore, for each $j\in\mathbb{Z}$ we get a spectral sequence whose $E_{1}$-term is given by cohomology groups ${\cal H}^{s,j-2k}(D)$, $s\in\mathbb{Z},k\geq 0$ and which converges to cohomology groups $H^{i,j}(D).$ In the few cases where we managed to compute cohomology groups, we have $H^{i,j}(D)={\mathop{\oplus}\limits_{k\geq 0}}{\cal H}^{i,j-2k}(D)$ and,consequently, the spectral sequence degenerates at $E_{1}.$ We have no idea whether this is true for any diagram $D.$