Mixing
“2â€-group cohomology
Clash Royale CLAN TAG#URR8PPP
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In order to define the cohomology
of a topological group G, we first have to introduce the concept of a classifying space. A classifying space BG is the base space of a principal G bundle EG. The EG is the universal bundle: Any principal G bundle E over a manifold M allows a bundle map into the universal bundle, and any two such morphisms are smoothly homotopic. Given
$$
gamma: M to BG
$$ the induced map of the base manifolds, it is the so-called classifying map.
The topology of the bundle E is completely determined by the homotopy class of the classifying map $gamma$. That is, the different components of the space Map(M,BG) correspond to the different bundles E over M. It can be shown that up to homotopy BG is uniquely determined by requiring EG to be contractible. That is, any contractible space with a free action of G is a realization of EG. In general, the classifying space BG of a compact group is an infinite-dimensional space.
Attempt 1:
Naively, we can have a generalized statement by modifying the map to a product of Eilenberg–MacLane space:
$$
gamma: M to K(G_1,1) times K(G_2,2)
$$
where $G_1$ and $G_2$ are two different groups. The $G_1$ can be non-abelian.
How can we be more precise to state the similar structure, defining a "2"-group cohomology, by generalizing the relation between the "group cohomology" and the "topological cohomology of classifying space"?
Is it necessary to have $G_2$ be abelian?
Attempt 2
- A more general classifying map may be:
$$
gamma: M to BG'
$$
where $BG'$ is the possible fibration
$$1 to K(G_2,2)to BG' to K(G_1,1) to 1,$$
where we may classify the fibration by Postinikov class $$[a] in H^d(K(G_1,1), G_2).$$ Is this formulation precise?
- How can we be sure that $d=3$ is the only solution and $[a] in H^d(K(G_1,1), G_2)$ classifies all sensible "2" group cohomology?
general-topology algebraic-topology homology-cohomology classifying-spaces eilenberg-maclane-spaces
asked Jul 25 at 21:11
wonderich
1,62421226
add a comment |Â
up vote
1
down vote
favorite
In order to define the cohomology
of a topological group G, we first have to introduce the concept of a classifying space. A classifying space BG is the base space of a principal G bundle EG. The EG is the universal bundle: Any principal G bundle E over a manifold M allows a bundle map into the universal bundle, and any two such morphisms are smoothly homotopic. Given
$$
gamma: M to BG
$$ the induced map of the base manifolds, it is the so-called classifying map.
The topology of the bundle E is completely determined by the homotopy class of the classifying map $gamma$. That is, the different components of the space Map(M,BG) correspond to the different bundles E over M. It can be shown that up to homotopy BG is uniquely determined by requiring EG to be contractible. That is, any contractible space with a free action of G is a realization of EG. In general, the classifying space BG of a compact group is an infinite-dimensional space.
Attempt 1:
Naively, we can have a generalized statement by modifying the map to a product of Eilenberg–MacLane space:
$$
gamma: M to K(G_1,1) times K(G_2,2)
$$
where $G_1$ and $G_2$ are two different groups. The $G_1$ can be non-abelian.
How can we be more precise to state the similar structure, defining a "2"-group cohomology, by generalizing the relation between the "group cohomology" and the "topological cohomology of classifying space"?
Is it necessary to have $G_2$ be abelian?
Attempt 2
- A more general classifying map may be:
$$
gamma: M to BG'
$$
where $BG'$ is the possible fibration
$$1 to K(G_2,2)to BG' to K(G_1,1) to 1,$$
where we may classify the fibration by Postinikov class $$[a] in H^d(K(G_1,1), G_2).$$ Is this formulation precise?
- How can we be sure that $d=3$ is the only solution and $[a] in H^d(K(G_1,1), G_2)$ classifies all sensible "2" group cohomology?
general-topology algebraic-topology homology-cohomology classifying-spaces eilenberg-maclane-spaces
asked Jul 25 at 21:11
wonderich
1,62421226
What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
1
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
1
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In order to define the cohomology
of a topological group G, we first have to introduce the concept of a classifying space. A classifying space BG is the base space of a principal G bundle EG. The EG is the universal bundle: Any principal G bundle E over a manifold M allows a bundle map into the universal bundle, and any two such morphisms are smoothly homotopic. Given
$$
gamma: M to BG
$$ the induced map of the base manifolds, it is the so-called classifying map.
The topology of the bundle E is completely determined by the homotopy class of the classifying map $gamma$. That is, the different components of the space Map(M,BG) correspond to the different bundles E over M. It can be shown that up to homotopy BG is uniquely determined by requiring EG to be contractible. That is, any contractible space with a free action of G is a realization of EG. In general, the classifying space BG of a compact group is an infinite-dimensional space.
Attempt 1:
Naively, we can have a generalized statement by modifying the map to a product of Eilenberg–MacLane space:
$$
gamma: M to K(G_1,1) times K(G_2,2)
$$
where $G_1$ and $G_2$ are two different groups. The $G_1$ can be non-abelian.
How can we be more precise to state the similar structure, defining a "2"-group cohomology, by generalizing the relation between the "group cohomology" and the "topological cohomology of classifying space"?
Is it necessary to have $G_2$ be abelian?
Attempt 2
- A more general classifying map may be:
$$
gamma: M to BG'
$$
where $BG'$ is the possible fibration
$$1 to K(G_2,2)to BG' to K(G_1,1) to 1,$$
where we may classify the fibration by Postinikov class $$[a] in H^d(K(G_1,1), G_2).$$ Is this formulation precise?
- How can we be sure that $d=3$ is the only solution and $[a] in H^d(K(G_1,1), G_2)$ classifies all sensible "2" group cohomology?
general-topology algebraic-topology homology-cohomology classifying-spaces eilenberg-maclane-spaces
asked Jul 25 at 21:11
wonderich
1,62421226
In order to define the cohomology
of a topological group G, we first have to introduce the concept of a classifying space. A classifying space BG is the base space of a principal G bundle EG. The EG is the universal bundle: Any principal G bundle E over a manifold M allows a bundle map into the universal bundle, and any two such morphisms are smoothly homotopic. Given
$$
gamma: M to BG
$$ the induced map of the base manifolds, it is the so-called classifying map.
The topology of the bundle E is completely determined by the homotopy class of the classifying map $gamma$. That is, the different components of the space Map(M,BG) correspond to the different bundles E over M. It can be shown that up to homotopy BG is uniquely determined by requiring EG to be contractible. That is, any contractible space with a free action of G is a realization of EG. In general, the classifying space BG of a compact group is an infinite-dimensional space.
Attempt 1:
Naively, we can have a generalized statement by modifying the map to a product of Eilenberg–MacLane space:
$$
gamma: M to K(G_1,1) times K(G_2,2)
$$
where $G_1$ and $G_2$ are two different groups. The $G_1$ can be non-abelian.
How can we be more precise to state the similar structure, defining a "2"-group cohomology, by generalizing the relation between the "group cohomology" and the "topological cohomology of classifying space"?
Is it necessary to have $G_2$ be abelian?
Attempt 2
- A more general classifying map may be:
$$
gamma: M to BG'
$$
where $BG'$ is the possible fibration
$$1 to K(G_2,2)to BG' to K(G_1,1) to 1,$$
where we may classify the fibration by Postinikov class $$[a] in H^d(K(G_1,1), G_2).$$ Is this formulation precise?
- How can we be sure that $d=3$ is the only solution and $[a] in H^d(K(G_1,1), G_2)$ classifies all sensible "2" group cohomology?
general-topology algebraic-topology homology-cohomology classifying-spaces eilenberg-maclane-spaces
asked Jul 25 at 21:11
wonderich
1,62421226
asked Jul 25 at 21:11
wonderich
1,62421226
asked Jul 25 at 21:11
wonderich
1,62421226
asked Jul 25 at 21:11
wonderich
1,62421226
1,62421226
What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
1
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
1
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58
add a comment |Â
What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
1
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
1
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58
What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
1
1
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
1
1
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58
add a comment |Â
1 Answer
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up vote
1
down vote
I am not sure if the following gives the answer you want, but there is a notion of crossed module $mathcal M =(mu: M to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $mathcal M$ has a classifying space $B mathcal M$. with first and second homotopy groups Coker $mu$, Ker $mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.
This book does not contain much on the topological or Lie case.
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
thanks for the comment +1
– wonderich
Jul 26 at 14:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I am not sure if the following gives the answer you want, but there is a notion of crossed module $mathcal M =(mu: M to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $mathcal M$ has a classifying space $B mathcal M$. with first and second homotopy groups Coker $mu$, Ker $mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.
This book does not contain much on the topological or Lie case.
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
thanks for the comment +1
– wonderich
Jul 26 at 14:38
add a comment |Â
up vote
1
down vote
I am not sure if the following gives the answer you want, but there is a notion of crossed module $mathcal M =(mu: M to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $mathcal M$ has a classifying space $B mathcal M$. with first and second homotopy groups Coker $mu$, Ker $mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.
This book does not contain much on the topological or Lie case.
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
thanks for the comment +1
– wonderich
Jul 26 at 14:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I am not sure if the following gives the answer you want, but there is a notion of crossed module $mathcal M =(mu: M to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $mathcal M$ has a classifying space $B mathcal M$. with first and second homotopy groups Coker $mu$, Ker $mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.
This book does not contain much on the topological or Lie case.
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
I am not sure if the following gives the answer you want, but there is a notion of crossed module $mathcal M =(mu: M to P)$ (due to JHC Whitehead) where $M,P$ are groups and $P$ acts on $M$ satisfyng two rules, which I won't detail here. These crossed modules classify pointed homotopy $2$-types. A crossed module $mathcal M$ has a classifying space $B mathcal M$. with first and second homotopy groups Coker $mu$, Ker $mu $ respectively. Full details are in the book Nonabelian Algebraic Topology.
This book does not contain much on the topological or Lie case.
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
answered Jul 26 at 11:54
Ronnie Brown
11.6k12938
11.6k12938
thanks for the comment +1
– wonderich
Jul 26 at 14:38
add a comment |Â
thanks for the comment +1
– wonderich
Jul 26 at 14:38
thanks for the comment +1
– wonderich
Jul 26 at 14:38
thanks for the comment +1
– wonderich
Jul 26 at 14:38
add a comment |Â
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What do you mean by "2-group cohomology"? Note that $K(G, n)$ only makes sense for $n > 1$ if $G$ is abelian, because higher homotopy groups are abelian.
– Mike Miller
Jul 26 at 7:41
1
I'm afraid its not at all clear what you are looking for. If you are looking for a cohomology theory of 2-groups, then these already exist. For example a google search for "crossed module cohomology" will give you results. Even searching for 2-group cohomology will give you a paper on Arxiv by Ginot.
– Tyrone
Jul 26 at 9:40
1
As for your first comment note that $K(G_1,1)times K(G_2,2)simeq BG_1^deltatimes BK(G_2,1)simeq B(G_2^deltatimes K(G_2,1))$, where $G_1^delta$ is the (possibly non-abelian) group $G_1$ given the discrete topology, and $G_2$ is assumed abelian. Eilenberg-Mac lane spaces only give information on the underlying discrete group, and ignore any topological information it may hold.
– Tyrone
Jul 26 at 9:41
Tyrone - thanks +1
– wonderich
Jul 29 at 17:58