What bigrading is used in this spectral sequence?

Question

I am reading this paper of Positselski and Vishik, in particular the main theorem: the cohomology algebra of a conilpotent algebra (i.e. the cohomology of its cobar construction) is Koszul if it satisfies certain conditions.

Consider a coaugmented coalgebra $C$ and filter $\overline C$ by its coradical filtration, denote it $\{F_p\overline C\}$. This produces a graded coalgebra $\operatorname{gr}(C)$ whose degree $i$ part is $F_iC/F_{i-1}C$. Moreover, the the cobar construction $\Omega C$ inherits a filtration so that

$$F_n\Omega^i C = \bigoplus_{\sum j_i = n} F_{j_1}\overline C\otimes \cdots \otimes F_{j_i}\overline C$$

This filtration is bounded below and exhaustive, so the spectral sequence associated to this filtration converges to $H^*(\Omega C)$ which, like in the paper, I will write just $H^*(C)$. If $C$ is graded, then write $H^{*,j}(C)$ for the space of cocycles of total weight $j$.

In the paper, the authors say that the $E_0$ page of this spectral sequence is the cobar construction of $\operatorname{gr}(C)$ and that $E_1^{i,j} = H^{i,j}(\operatorname{gr}(C))$. This doesn't fit the convention I'm used to, where the $E_0$-page has in bidegree $(p,q)$ the space $\Omega^{p+q}(\operatorname{gr}(C))$ in weight $p$ (by the very definition of the filtration given in the paper), so $E_1^{i,j}$ should instead be $H^{i+j,i}$. Thus, to obtain the claim of the authors, one would have to use the grading $(q,p-q)$.

There are still other issues: for example, I cannot see why $E_1^{1,0} = H^{1,1}(\operatorname{gr}(C))$ should coincide with $H^1(C) = E_\infty^{1,0}$. This should happen since $F_p H^1(C) = F_{p-1}H^1(C)$ for $p>1$. Indeed, $E_1^{p,1-p} = 0$ for such indices. This gives $E_\infty^{p,1-p} = 0$ for $p>1$ and $E_\infty^{1,0} = H^1(C)$, but it is not clear to me why $d_1 : E_1^{1,0}\to E_1^{2,0} = H^{2,2}(\operatorname{gr}(C))$ is zero.

Moreover, the paper claims that the differential $d_r$ has bidegree $(1,-r)$. But the differential of the spectral sequence of this filtration has differential $d_r$ of bidegree $(r,1-r)$. It is true, though, that with the regrading above at least $d_1$ now has bidegree $(1,1)$ (which is still not $(1,-1)$?), but still I cannot see why $d_1^{1,0}=0$.

Can someone clarify this? That is, what is the grading convention used for this spectral sequence? The filtration defined produces, according to what I know, a spectral sequence with a different grading convention. There are still a series of steps in the proof I cannot understand, but perhaps fixing the above grading issues helps with this.

Answer 1

This is simply a summary of the comments. The problem was I did not raise the index $p$ throughout.

Raise the index in $F_p\Omega^n$ to get a filtration $F^p\Omega^n$. The spectral sequence is now of second quadrant, with ${}^\prime\! E_0^{-p,q} = \Omega^{q-p}(\operatorname{gr}(C))_{(p)}$ and ${}^\prime\!E_1^{-p,q}= H^{q-p,p}(\operatorname{gr}(C))$. If we set $E_r^{p,q} ={}^\prime\!E_r^{-q,p+q}$, then $E_r^{p,q} = H^{p,q}(\operatorname{gr}(C))$ and the differential is

$$d_r^{p,q} :E_r^{p,q}={}^\prime\! E_r^{-q,p+q}\to {}^\prime\!E_r^{r-q,p+1+q-r}=E_r^{p+1,q-r}$$

of bidegree $(1,-r)$, as desired.

Corrections. Now $d_1^{1,1}$ lands in $E_1^{2,0}$ which vanishes. Note too that $E_r^{p,q}=0$ if $q<p$. Since $\operatorname{gr}(C)$ is $1$-cogenerated, the $p=1$ column is concentrated in degree $q$ where it is $H^{1,1}(\operatorname{gr}(C))$, so $H^1(C) = H^{1,1}(\operatorname{gr}(C))$. Moreover, this empty column and the observation on weight vs degree means $E_r^{2,2}$ receives zero differentials everywhere so it lives to $E_\infty$ and we also get that $H^{2,2}(\operatorname{gr}(C)) = F_2H^2(C)$.