3
$\begingroup$

I am trying to understand strong convergence for whole-plane spectral sequences in the paper by J.Boardman: https://www.uio.no/studier/emner/matnat/math/MAT9580/v12/undervisningsmateriale/boardman-conditionally-1999.pdf

I am currently confused about the convergence of the zig-zag double complex $D$ mentioned here as one of the answers: https://math.stackexchange.com/questions/1368777/double-complex-with-exact-rows

Firstly Let $C$ be the total complex of $D$. By Theorem 10.1 in Boardman's paper the spectral sequences $E^{\alpha}_2=H(H(D,\delta),d)$ and $E^{\beta}_2=H(H(D,d),\delta)$ converges conditionally to the cohomology $H(\widehat{C})$ where $\widehat{C}$ is the completion of $C$. Furthermore it is clear that $E^{\alpha}$ collapses at page 1 and $E^{\beta}$ collapses at page 2 - so by the remark after Theorem 7.1 we see that $RE_{\infty}=0$ for both spectral sequences. The fact that both spectral sequences collapse imply that the condition for Theorem 8.1 is also satisfied - which means that both of them have $W=0$. Theorem 8.2 then implies that both spectral sequences converge strongly to $H(\widehat{C})$.

Is this a contradiction? Because $E^{\alpha}$ implies that all the filter quotients of $H^n(\widehat{C})$ is zero for all $n$, whereas $E^{\beta}$ tells that that the filter quotients of $H^n(\widehat{C})$ are non-zero.

Thanks!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The spectral sequences $E^\alpha$ and $E^\beta$ arise from two different filtrations of $C$, which have different completions, say $\widehat C^\alpha$ and $\widehat C^\beta$. The first converges strongly to $H^*(\widehat C^\alpha) = 0$, and the second converges strongly to $H^*(\widehat C^\beta) \ne 0$. The apparent contradiction arises because you assume that the two completions are the same.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks very much for your answer. This clears up alot of confusion for me. $\endgroup$
    – Steve
    Commented Jan 29, 2018 at 23:24
  • $\begingroup$ May I ask if it's possible to 'complete' the double complex $D$ in the first place, say $D_1$, so that the completions of the total complex $\widehat{C_1}^\alpha$ and $\widehat{C_1}^\beta$ are the same? and that $H^*(\widehat{C_1}^\alpha)=H^*(\widehat{C_1}^\beta)=H^*(\widehat{C}^\beta)\not=0$? $\endgroup$
    – Steve
    Commented Feb 5, 2018 at 1:04
  • $\begingroup$ Sorry I meant "...so that the cohomology of the completions of the total complexes $\widehat{C_1}^\alpha$ and $\widehat{C_1}^\beta$ are the same?" $\endgroup$
    – Steve
    Commented Feb 5, 2018 at 1:12
  • $\begingroup$ I don't see how to complete the double complex $D$. You could consider a completion of the total complex $C$, at "both ends", but it would not arise from the filtrations giving rise to the two spectral sequences. $\endgroup$
    – John Rognes
    Commented Feb 5, 2018 at 19:59
  • 1
    $\begingroup$ Elements in $C^0 = \bigoplus_p D^{p,-p}$ are finite sums $\sum_p x_p$ with $x_p \in D^{p,-p}$. In the completions we discussed above you allow formal sums where $p$ is either bounded below, or bounded above (depending on the case). By completing at "both ends" I meant that you would allow formal sums, without bounds, amounting to the product $\prod_p D^{p,-p}$. Similarly for $C^1 = \bigoplus_p D^{p,1-p}$, completing to $\prod_p D^{p,1-p}$. $\endgroup$
    – John Rognes
    Commented Feb 5, 2018 at 23:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .