3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is going on is that the notation systems in mathematics research papers vary from one paper to another, because they are chosen for convenience of presentation of the subject matter, and the subject matter in different papers is not the same. In particular, the notation system in a mathematics reseach paper you are reading may differ from the one in your favorite textbook (because the notation system in the textbook was chosen to optimize presentation of Subject Matter A, while the paper is devoted to Subject Matter B. $\endgroup$ – Leonid Positselski Jul 24 '18 at 18:37
  • $\begingroup$ @LeonidPositselski Dear Leonid, I apologise if the phrasing of the question "What is going on?" is unfortunate. I can try to fix that to the following: *What is the grading convention (say B) used for this spectral sequence, given the filtration is defined to produce a grading convention (say A) I am used to and is not convention B?". $\endgroup$ – Pedro Tamaroff Jul 24 '18 at 18:38
  • 5
    $\begingroup$ Thus, when reading a research paper, it is advisable to make a reasonable effort to understand the notation system of the author(s). That is why it is not a good idea to say that "something seems to be incorrect" when the difference between what the paper says and what you think is correct is the one of a simple notation change. Surprising as it may be (though I do not at all understand why it should be surprising), we are not using the standard spectral sequence notation $E_r^{p,q}$ in our paper. $\endgroup$ – Leonid Positselski Jul 24 '18 at 18:44
  • 1
    $\begingroup$ So you should have ${}'\!E_r^{p,q}=E_r^{i,j}$ whenever $i=p+q$ and $j=-p$. Try applying this linear change of coordinates/notation comparison consistently, and maybe your present confusion will be replaced with a more meaningful one. $\endgroup$ – Leonid Positselski Jul 24 '18 at 19:19
  • 1
    $\begingroup$ Concerning the previous question -- it is because the subalgebra $\bigoplus_i N_iH^i(C)$ in $H^*(C)$ is isomorphic to the quotient algebra of the diagonal cohomology algebra by the images of the differentials, and the diagonal cohomology algebra is isomorphic to $\mathrm{q}H^*(C)$. $\endgroup$ – Leonid Positselski Jul 24 '18 at 22:59
2
$\begingroup$

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 $H^{2,2}(\operatorname{gr}(C)) = F_2H^2(C)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.