Computing cohomology of dihedral group in detail
up vote
4
down vote
favorite
So I tried to compute the cohomology of $D_{2n}$, for n odd , $H^{k}(D_{2n}, Bbb Z)$. using Lyndon SS. I have obtained a few obstacles:
My computation, using the fact that there is a $C_2$ action $H^q(C_m, Bbb Z)$, shows that my $E_2^{pq}$ page looks like this:
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$ $$ cdots 0 cdots $$ $$ cdots 0 cdots $$
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ Bbb Z quad 0 quad Bbb Z/2 quad 0 quad Bbb Z/2 quad 0 quad cdots $$
Is this correct?
I suspect the correct $E_2$ page is more or less similar. But I am still unclear how this gives $H^k(D_{2n},Bbb Z)$.
a) How is this $E_2$ page related to $H^k(D_{2n},Bbb Z)$? I only know the case when page 2 collapses to an axis.
b) I suppose one has to show all differential are $0$ to compute $E_infty^{p,q}$. In my case this is simple, since all are $0$.
c) Even if we know the $E_infty$ page does this give $H_n$ - surely we cannot just take direct sum? **
** Supposing my calucations were correct - then we can go to c) directly: What we have here is the relation when $n equiv 0 pmod 4$
$$ 0 rightarrow C_2 rightarrow H^n rightarrow C_m rightarrow 0 $$
homology-cohomology homological-algebra group-cohomology
add a comment |
up vote
4
down vote
favorite
So I tried to compute the cohomology of $D_{2n}$, for n odd , $H^{k}(D_{2n}, Bbb Z)$. using Lyndon SS. I have obtained a few obstacles:
My computation, using the fact that there is a $C_2$ action $H^q(C_m, Bbb Z)$, shows that my $E_2^{pq}$ page looks like this:
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$ $$ cdots 0 cdots $$ $$ cdots 0 cdots $$
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ Bbb Z quad 0 quad Bbb Z/2 quad 0 quad Bbb Z/2 quad 0 quad cdots $$
Is this correct?
I suspect the correct $E_2$ page is more or less similar. But I am still unclear how this gives $H^k(D_{2n},Bbb Z)$.
a) How is this $E_2$ page related to $H^k(D_{2n},Bbb Z)$? I only know the case when page 2 collapses to an axis.
b) I suppose one has to show all differential are $0$ to compute $E_infty^{p,q}$. In my case this is simple, since all are $0$.
c) Even if we know the $E_infty$ page does this give $H_n$ - surely we cannot just take direct sum? **
** Supposing my calucations were correct - then we can go to c) directly: What we have here is the relation when $n equiv 0 pmod 4$
$$ 0 rightarrow C_2 rightarrow H^n rightarrow C_m rightarrow 0 $$
homology-cohomology homological-algebra group-cohomology
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
So I tried to compute the cohomology of $D_{2n}$, for n odd , $H^{k}(D_{2n}, Bbb Z)$. using Lyndon SS. I have obtained a few obstacles:
My computation, using the fact that there is a $C_2$ action $H^q(C_m, Bbb Z)$, shows that my $E_2^{pq}$ page looks like this:
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$ $$ cdots 0 cdots $$ $$ cdots 0 cdots $$
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ Bbb Z quad 0 quad Bbb Z/2 quad 0 quad Bbb Z/2 quad 0 quad cdots $$
Is this correct?
I suspect the correct $E_2$ page is more or less similar. But I am still unclear how this gives $H^k(D_{2n},Bbb Z)$.
a) How is this $E_2$ page related to $H^k(D_{2n},Bbb Z)$? I only know the case when page 2 collapses to an axis.
b) I suppose one has to show all differential are $0$ to compute $E_infty^{p,q}$. In my case this is simple, since all are $0$.
c) Even if we know the $E_infty$ page does this give $H_n$ - surely we cannot just take direct sum? **
** Supposing my calucations were correct - then we can go to c) directly: What we have here is the relation when $n equiv 0 pmod 4$
$$ 0 rightarrow C_2 rightarrow H^n rightarrow C_m rightarrow 0 $$
homology-cohomology homological-algebra group-cohomology
So I tried to compute the cohomology of $D_{2n}$, for n odd , $H^{k}(D_{2n}, Bbb Z)$. using Lyndon SS. I have obtained a few obstacles:
My computation, using the fact that there is a $C_2$ action $H^q(C_m, Bbb Z)$, shows that my $E_2^{pq}$ page looks like this:
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$ $$ cdots 0 cdots $$ $$ cdots 0 cdots $$
$$ Bbb Z/m quad 0 quad 0 quad 0 quad 0 quad 0 quad cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ cdots 0 cdots $$
$$ Bbb Z quad 0 quad Bbb Z/2 quad 0 quad Bbb Z/2 quad 0 quad cdots $$
Is this correct?
I suspect the correct $E_2$ page is more or less similar. But I am still unclear how this gives $H^k(D_{2n},Bbb Z)$.
a) How is this $E_2$ page related to $H^k(D_{2n},Bbb Z)$? I only know the case when page 2 collapses to an axis.
b) I suppose one has to show all differential are $0$ to compute $E_infty^{p,q}$. In my case this is simple, since all are $0$.
c) Even if we know the $E_infty$ page does this give $H_n$ - surely we cannot just take direct sum? **
** Supposing my calucations were correct - then we can go to c) directly: What we have here is the relation when $n equiv 0 pmod 4$
$$ 0 rightarrow C_2 rightarrow H^n rightarrow C_m rightarrow 0 $$
homology-cohomology homological-algebra group-cohomology
homology-cohomology homological-algebra group-cohomology
edited Nov 15 at 8:45
asked Nov 15 at 0:02
CL.
2,0352822
2,0352822
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Your $E_2$ page is correct.
To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.
Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, Bbb Z)$ have a filtration $0 = F^0 supset cdots supset F^i supset F^{i+1} supset cdots$ and the associated graded $$text{gr}^p H^{p+q}(D_{2n}, Bbb Z) = E^{p,q}_infty.$$
For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_infty^{4n+2,0}$.
On the lines with $p + q = 4n >0$, we have two terms. There is $E_infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_infty^{0,4n}.$
In particular, we have a short exact sequence $0 to E_infty^{0,4n} to H^{4n} to E_infty^{4n,0} to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_infty^{0,4n}$ and walking down the line.
In this particular case, $E_infty^{4n,0} = Bbb Z/2$, and $E_infty^{0,4n} = Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = Bbb Z/2 oplus Bbb Z/m cong Bbb Z/2m$ in this case.
If you are also interested in the multiplicative structure, one can write this ring as $Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your $E_2$ page is correct.
To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.
Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, Bbb Z)$ have a filtration $0 = F^0 supset cdots supset F^i supset F^{i+1} supset cdots$ and the associated graded $$text{gr}^p H^{p+q}(D_{2n}, Bbb Z) = E^{p,q}_infty.$$
For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_infty^{4n+2,0}$.
On the lines with $p + q = 4n >0$, we have two terms. There is $E_infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_infty^{0,4n}.$
In particular, we have a short exact sequence $0 to E_infty^{0,4n} to H^{4n} to E_infty^{4n,0} to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_infty^{0,4n}$ and walking down the line.
In this particular case, $E_infty^{4n,0} = Bbb Z/2$, and $E_infty^{0,4n} = Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = Bbb Z/2 oplus Bbb Z/m cong Bbb Z/2m$ in this case.
If you are also interested in the multiplicative structure, one can write this ring as $Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
add a comment |
up vote
3
down vote
accepted
Your $E_2$ page is correct.
To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.
Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, Bbb Z)$ have a filtration $0 = F^0 supset cdots supset F^i supset F^{i+1} supset cdots$ and the associated graded $$text{gr}^p H^{p+q}(D_{2n}, Bbb Z) = E^{p,q}_infty.$$
For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_infty^{4n+2,0}$.
On the lines with $p + q = 4n >0$, we have two terms. There is $E_infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_infty^{0,4n}.$
In particular, we have a short exact sequence $0 to E_infty^{0,4n} to H^{4n} to E_infty^{4n,0} to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_infty^{0,4n}$ and walking down the line.
In this particular case, $E_infty^{4n,0} = Bbb Z/2$, and $E_infty^{0,4n} = Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = Bbb Z/2 oplus Bbb Z/m cong Bbb Z/2m$ in this case.
If you are also interested in the multiplicative structure, one can write this ring as $Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your $E_2$ page is correct.
To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.
Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, Bbb Z)$ have a filtration $0 = F^0 supset cdots supset F^i supset F^{i+1} supset cdots$ and the associated graded $$text{gr}^p H^{p+q}(D_{2n}, Bbb Z) = E^{p,q}_infty.$$
For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_infty^{4n+2,0}$.
On the lines with $p + q = 4n >0$, we have two terms. There is $E_infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_infty^{0,4n}.$
In particular, we have a short exact sequence $0 to E_infty^{0,4n} to H^{4n} to E_infty^{4n,0} to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_infty^{0,4n}$ and walking down the line.
In this particular case, $E_infty^{4n,0} = Bbb Z/2$, and $E_infty^{0,4n} = Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = Bbb Z/2 oplus Bbb Z/m cong Bbb Z/2m$ in this case.
If you are also interested in the multiplicative structure, one can write this ring as $Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.
Your $E_2$ page is correct.
To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.
Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, Bbb Z)$ have a filtration $0 = F^0 supset cdots supset F^i supset F^{i+1} supset cdots$ and the associated graded $$text{gr}^p H^{p+q}(D_{2n}, Bbb Z) = E^{p,q}_infty.$$
For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_infty^{4n+2,0}$.
On the lines with $p + q = 4n >0$, we have two terms. There is $E_infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_infty^{0,4n}.$
In particular, we have a short exact sequence $0 to E_infty^{0,4n} to H^{4n} to E_infty^{4n,0} to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_infty^{0,4n}$ and walking down the line.
In this particular case, $E_infty^{4n,0} = Bbb Z/2$, and $E_infty^{0,4n} = Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = Bbb Z/2 oplus Bbb Z/m cong Bbb Z/2m$ in this case.
If you are also interested in the multiplicative structure, one can write this ring as $Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.
edited Nov 16 at 16:27
answered Nov 15 at 9:47
Mike Miller
35.2k467132
35.2k467132
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
add a comment |
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
Thanks a lot. Just curious, is there a reason why your notation went from $E_infty $ to $E^infty$?
– CL.
Nov 16 at 7:38
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
@CL. I normally think about homology spectral sequences and so I am more used to writing $E^infty$. So I reverted by mistake halfway through. I will edit.
– Mike Miller
Nov 16 at 16:25
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998997%2fcomputing-cohomology-of-dihedral-group-in-detail%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown