Timeline for Conditionally convergent spectral sequences with exiting and entering differentials
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2019 at 17:44 | vote | accept | Pavel | ||
| Jul 24, 2019 at 21:57 | history | edited | John Rognes | CC BY-SA 4.0 |
added response to question about filtered bicomplex
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| Jul 24, 2019 at 15:36 | comment | added | Pavel | awgh...I'll try once more: Let $Z^{i,j}\subset B^{i,j}$ be $d_h$-closed subspaces such that $d_h(B^{i-1,j})\subset Z^{i,j}$. Define $F_k := \bigoplus_{j-i>k} B^{ij} \oplus \bigoplus_i Z^{i,i+k}$. Now it should be correct, I apologize for confusion. | |
| Jul 24, 2019 at 14:29 | comment | added | John Rognes | Are you sure? For $k=1$ I get $F_1 = \bigoplus_{j-i>1} B^{ij} \oplus \bigoplus_i Z^{ii}$. The vertical differential $d_v$ maps $Z^{ii} \subset F_1$ into $B^{i,i+1}$, which is not in $F_1$, so the differential on $\bigoplus_{i,j} B^{ij}$ does not restrict to a differential on $F_1$. | |
| Jul 23, 2019 at 22:19 | comment | added | Pavel | Oh, I am sorry, it must be $j - i > k$ then (the graph above the function $j(i)= i +k$). | |
| Jul 23, 2019 at 22:17 | comment | added | John Rognes | Your answers to (2) and (3) suggest that $i+j>k$ is not the intended condition in the definition of $F_k$. Do you mean $j > i > k$ or something like that? | |
| Jul 23, 2019 at 22:08 | comment | added | Pavel | (1) I mean $d_h Z^{ii} = 0$. (2) It is the internal sum (I imagine replacing the diagonal with Z^{ii} and discarding all $B^{ij}$'s below). (3) The vertical differential points upstairs, and hence it preserves $F_k$, as far as I can see. Therefore, it should be invariant under the total differential. | |
| Jul 23, 2019 at 22:04 | comment | added | John Rognes | (1) By "let $Z^{ii} \subset B^{ii}$ be subspaces closed under the horizontal differential", do you mean that the horizontal differential maps $Z^{ii}$ to zero? (2) In the definition of $F_k$, is the direct sum with $Z^{ii}$ for all $i$ intended to mean the internal sum in $\bigoplus_{i,j} B^{i,j}$? (3) It looks as if the vertical differential might not map $F_k$ to itself. Don't you want $F_k$ to be a subcomplex with respect to the total differential? | |
| Jul 23, 2019 at 18:47 | comment | added | Pavel | Also, please, could you comment on the proof of your statement? Is it, in fact, important that the groups in a half-plane vanish or is the vanishing of the differentials fundamental? P.S. Thanks for the better reference! | |
| Jul 23, 2019 at 18:43 | comment | added | Pavel | Thank you very much for your answer! Switching $s$ and $d$ helps for showing strong convergence of the horizontal filtration of a cohomological right half-plane bicomplex. But I have another example: Consider the direct-sum total complex of a cohomological half-plane bicomplex $B^{ij}$ (I draw it in the right half-plane, the vertical differential points upstairs and the horizontal to the right), and let $Z^{ii} \subset B^{ii}$ be subspaces closed under the horizontal differential. Let $F_k = \bigoplus_{i + j > k} B^{ij} \oplus \bigoplus_i Z^{ii}$ be the diagonal filtration. What now? | |
| Jul 23, 2019 at 15:33 | history | edited | John Rognes | CC BY-SA 4.0 |
deleted oversimplification
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| Jul 23, 2019 at 15:26 | history | answered | John Rognes | CC BY-SA 4.0 |