Mixing
Cellular action on CW complex
$begingroup$
Let $G$ act cellularly on a CW complex $X$. For each $nge 0$, the action induces an action on the indexing set $I_n$ for the $n$-cells.
Now look at the cellular chain complex $C_bullet(X)$. Each component $C_n(X)=mathbb{Z}langle I_nrangle$ gets the structure of a $mathbb{Z}[G]$-module by extending the action on the basis linearly. I want to see that also the differentials are $mathbb{Z}[G]$-linear.
For $ein I_n$ and $e'in I_{n-1}$ denote the incidence number by $[e:e']$ and we know that $partial e = sum [e:e']cdot e'$. My first attempt was to show $[ge:ge'] = [e:e']$ and then do a calculation like
$$partial ge = sum [ge:e']cdot e' = sum [ge:ge']cdot ge' = sum [e:e']cdot ge' = gpartial e.$$
However, $[ge:ge'] = [e:e']$ seems wrong by the following counterexample:
Consider two $0$-cells $p$ and $q$ and two $1$-cells $a$ and $b$ where $a:pto q$ and $b:qto p$. Changing $a$ and $b$ is a $mathbb{Z}/2$-action, but $[ga:gp] = [b:p] = 1 ne -1 = [a:p]$.
Maybe it is just a question of the “correct” local orientation?
homology-cohomology group-actions cw-complexes
asked Jan 18 at 9:44
FKranholdFKranhold
1987
$endgroup$
add a comment |
$begingroup$
Let $G$ act cellularly on a CW complex $X$. For each $nge 0$, the action induces an action on the indexing set $I_n$ for the $n$-cells.
Now look at the cellular chain complex $C_bullet(X)$. Each component $C_n(X)=mathbb{Z}langle I_nrangle$ gets the structure of a $mathbb{Z}[G]$-module by extending the action on the basis linearly. I want to see that also the differentials are $mathbb{Z}[G]$-linear.
For $ein I_n$ and $e'in I_{n-1}$ denote the incidence number by $[e:e']$ and we know that $partial e = sum [e:e']cdot e'$. My first attempt was to show $[ge:ge'] = [e:e']$ and then do a calculation like
$$partial ge = sum [ge:e']cdot e' = sum [ge:ge']cdot ge' = sum [e:e']cdot ge' = gpartial e.$$
However, $[ge:ge'] = [e:e']$ seems wrong by the following counterexample:
Consider two $0$-cells $p$ and $q$ and two $1$-cells $a$ and $b$ where $a:pto q$ and $b:qto p$. Changing $a$ and $b$ is a $mathbb{Z}/2$-action, but $[ga:gp] = [b:p] = 1 ne -1 = [a:p]$.
Maybe it is just a question of the “correct” local orientation?
homology-cohomology group-actions cw-complexes
asked Jan 18 at 9:44
FKranholdFKranhold
1987
$endgroup$
add a comment |
$begingroup$
Let $G$ act cellularly on a CW complex $X$. For each $nge 0$, the action induces an action on the indexing set $I_n$ for the $n$-cells.
Now look at the cellular chain complex $C_bullet(X)$. Each component $C_n(X)=mathbb{Z}langle I_nrangle$ gets the structure of a $mathbb{Z}[G]$-module by extending the action on the basis linearly. I want to see that also the differentials are $mathbb{Z}[G]$-linear.
For $ein I_n$ and $e'in I_{n-1}$ denote the incidence number by $[e:e']$ and we know that $partial e = sum [e:e']cdot e'$. My first attempt was to show $[ge:ge'] = [e:e']$ and then do a calculation like
$$partial ge = sum [ge:e']cdot e' = sum [ge:ge']cdot ge' = sum [e:e']cdot ge' = gpartial e.$$
However, $[ge:ge'] = [e:e']$ seems wrong by the following counterexample:
Consider two $0$-cells $p$ and $q$ and two $1$-cells $a$ and $b$ where $a:pto q$ and $b:qto p$. Changing $a$ and $b$ is a $mathbb{Z}/2$-action, but $[ga:gp] = [b:p] = 1 ne -1 = [a:p]$.
Maybe it is just a question of the “correct” local orientation?
homology-cohomology group-actions cw-complexes
asked Jan 18 at 9:44
FKranholdFKranhold
1987
$endgroup$
Let $G$ act cellularly on a CW complex $X$. For each $nge 0$, the action induces an action on the indexing set $I_n$ for the $n$-cells.
Now look at the cellular chain complex $C_bullet(X)$. Each component $C_n(X)=mathbb{Z}langle I_nrangle$ gets the structure of a $mathbb{Z}[G]$-module by extending the action on the basis linearly. I want to see that also the differentials are $mathbb{Z}[G]$-linear.
For $ein I_n$ and $e'in I_{n-1}$ denote the incidence number by $[e:e']$ and we know that $partial e = sum [e:e']cdot e'$. My first attempt was to show $[ge:ge'] = [e:e']$ and then do a calculation like
$$partial ge = sum [ge:e']cdot e' = sum [ge:ge']cdot ge' = sum [e:e']cdot ge' = gpartial e.$$
However, $[ge:ge'] = [e:e']$ seems wrong by the following counterexample:
Consider two $0$-cells $p$ and $q$ and two $1$-cells $a$ and $b$ where $a:pto q$ and $b:qto p$. Changing $a$ and $b$ is a $mathbb{Z}/2$-action, but $[ga:gp] = [b:p] = 1 ne -1 = [a:p]$.
Maybe it is just a question of the “correct” local orientation?
homology-cohomology group-actions cw-complexes
homology-cohomology group-actions cw-complexes
asked Jan 18 at 9:44
FKranholdFKranhold
1987
asked Jan 18 at 9:44
FKranholdFKranhold
1987
edited Jan 18 at 10:04
FKranhold
asked Jan 18 at 9:44
FKranholdFKranhold
1987
asked Jan 18 at 9:44
FKranholdFKranhold
1987
asked Jan 18 at 9:44
FKranholdFKranhold
1987
1987
add a comment |
add a comment |
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