Mixing
Given filtered complex $K=oplus_n K^n$ with gradation $n$ and filtration ${K_p}$ of $K$, $K_pcap K^n$ is...
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Let $K$ be a filtered graded complex s.t. $K=oplus_{nin Z}K^n$ and ${K_p}$ filtration of $K$. Define $K_p^n=K_pcap K^n$.
$textbf{Q:}$ Why ${K_p^n}_{pin Z}$ forms filtration of $K^n$? The definition of filtation is a family of subcomplexes $K=K_0supsetdotssupset K_{i-1}supset K_idots$ s.t. differential operator $D$ of $K$ has $D|_{K_i}:K_ito K_i$. In other words, why do I have $D|_{K_p^n}:K_p^nto K_p^n$. It is possible that on $K$, $D:K^nto K^{n+1}$. Then $D|_{K_pcap K^n}$ certainly will have image in $K_pcap K^{n+1}$.
Ref. Bott-Tu Differential Forms in Algebraic Topology Sec 14, The Spectral Sequence of a Filtered Complex
abstract-algebra algebraic-topology homological-algebra
asked Jan 28 at 1:42
user45765user45765
2,6792724
$endgroup$
add a comment |
$begingroup$
Let $K$ be a filtered graded complex s.t. $K=oplus_{nin Z}K^n$ and ${K_p}$ filtration of $K$. Define $K_p^n=K_pcap K^n$.
$textbf{Q:}$ Why ${K_p^n}_{pin Z}$ forms filtration of $K^n$? The definition of filtation is a family of subcomplexes $K=K_0supsetdotssupset K_{i-1}supset K_idots$ s.t. differential operator $D$ of $K$ has $D|_{K_i}:K_ito K_i$. In other words, why do I have $D|_{K_p^n}:K_p^nto K_p^n$. It is possible that on $K$, $D:K^nto K^{n+1}$. Then $D|_{K_pcap K^n}$ certainly will have image in $K_pcap K^{n+1}$.
Ref. Bott-Tu Differential Forms in Algebraic Topology Sec 14, The Spectral Sequence of a Filtered Complex
abstract-algebra algebraic-topology homological-algebra
asked Jan 28 at 1:42
user45765user45765
2,6792724
$endgroup$
$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45
add a comment |
$begingroup$
Let $K$ be a filtered graded complex s.t. $K=oplus_{nin Z}K^n$ and ${K_p}$ filtration of $K$. Define $K_p^n=K_pcap K^n$.
$textbf{Q:}$ Why ${K_p^n}_{pin Z}$ forms filtration of $K^n$? The definition of filtation is a family of subcomplexes $K=K_0supsetdotssupset K_{i-1}supset K_idots$ s.t. differential operator $D$ of $K$ has $D|_{K_i}:K_ito K_i$. In other words, why do I have $D|_{K_p^n}:K_p^nto K_p^n$. It is possible that on $K$, $D:K^nto K^{n+1}$. Then $D|_{K_pcap K^n}$ certainly will have image in $K_pcap K^{n+1}$.
Ref. Bott-Tu Differential Forms in Algebraic Topology Sec 14, The Spectral Sequence of a Filtered Complex
abstract-algebra algebraic-topology homological-algebra
asked Jan 28 at 1:42
user45765user45765
2,6792724
$endgroup$
Let $K$ be a filtered graded complex s.t. $K=oplus_{nin Z}K^n$ and ${K_p}$ filtration of $K$. Define $K_p^n=K_pcap K^n$.
$textbf{Q:}$ Why ${K_p^n}_{pin Z}$ forms filtration of $K^n$? The definition of filtation is a family of subcomplexes $K=K_0supsetdotssupset K_{i-1}supset K_idots$ s.t. differential operator $D$ of $K$ has $D|_{K_i}:K_ito K_i$. In other words, why do I have $D|_{K_p^n}:K_p^nto K_p^n$. It is possible that on $K$, $D:K^nto K^{n+1}$. Then $D|_{K_pcap K^n}$ certainly will have image in $K_pcap K^{n+1}$.
Ref. Bott-Tu Differential Forms in Algebraic Topology Sec 14, The Spectral Sequence of a Filtered Complex
abstract-algebra algebraic-topology homological-algebra
abstract-algebra algebraic-topology homological-algebra
asked Jan 28 at 1:42
user45765user45765
2,6792724
asked Jan 28 at 1:42
user45765user45765
2,6792724
asked Jan 28 at 1:42
user45765user45765
2,6792724
asked Jan 28 at 1:42
user45765user45765
2,6792724
asked Jan 28 at 1:42
user45765user45765
2,6792724
2,6792724
$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45
add a comment |
$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45
$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45
$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45
add a comment |
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$begingroup$
If you restrict to $K^n$, there is no more differentials. You don't look at a complex anymore, just a module (or whatever kind of object $K^n$ is). So $K^n$ is a filtered module with filtration $K_pcap K^nsubset K_{p-1}cap K^nsubset...subset K_0cap K^n=K^n$.
$endgroup$
– Roland
Jan 28 at 9:45