semi-direct product between manifolds
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1
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question 1: Are there mathematical definition of the semi-direct product between manifolds
$$
M^{d_1} rtimes V^{d_2}?
$$
For example, is it defined as a fibration such that $M^{d_1}$ is the fiber and the $V^{d_2}$ is the base, so the total space is a bundle with the following relation
$$
M^{d_1} hookrightarrow M^{d_1} rtimes V^{d_2} to V^{d_2}
$$
question 2: Can $M^{d_1} rtimes V^{d_2}$ be a mapping torus?
question 3: What is the mapping class group of $M^{d_1} rtimes V^{d_2}$?
Partial answers are very welcome! Thanks!
general-topology manifolds geometric-topology fiber-bundles surgery-theory
add a comment |
up vote
1
down vote
favorite
question 1: Are there mathematical definition of the semi-direct product between manifolds
$$
M^{d_1} rtimes V^{d_2}?
$$
For example, is it defined as a fibration such that $M^{d_1}$ is the fiber and the $V^{d_2}$ is the base, so the total space is a bundle with the following relation
$$
M^{d_1} hookrightarrow M^{d_1} rtimes V^{d_2} to V^{d_2}
$$
question 2: Can $M^{d_1} rtimes V^{d_2}$ be a mapping torus?
question 3: What is the mapping class group of $M^{d_1} rtimes V^{d_2}$?
Partial answers are very welcome! Thanks!
general-topology manifolds geometric-topology fiber-bundles surgery-theory
No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
2
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
question 1: Are there mathematical definition of the semi-direct product between manifolds
$$
M^{d_1} rtimes V^{d_2}?
$$
For example, is it defined as a fibration such that $M^{d_1}$ is the fiber and the $V^{d_2}$ is the base, so the total space is a bundle with the following relation
$$
M^{d_1} hookrightarrow M^{d_1} rtimes V^{d_2} to V^{d_2}
$$
question 2: Can $M^{d_1} rtimes V^{d_2}$ be a mapping torus?
question 3: What is the mapping class group of $M^{d_1} rtimes V^{d_2}$?
Partial answers are very welcome! Thanks!
general-topology manifolds geometric-topology fiber-bundles surgery-theory
question 1: Are there mathematical definition of the semi-direct product between manifolds
$$
M^{d_1} rtimes V^{d_2}?
$$
For example, is it defined as a fibration such that $M^{d_1}$ is the fiber and the $V^{d_2}$ is the base, so the total space is a bundle with the following relation
$$
M^{d_1} hookrightarrow M^{d_1} rtimes V^{d_2} to V^{d_2}
$$
question 2: Can $M^{d_1} rtimes V^{d_2}$ be a mapping torus?
question 3: What is the mapping class group of $M^{d_1} rtimes V^{d_2}$?
Partial answers are very welcome! Thanks!
general-topology manifolds geometric-topology fiber-bundles surgery-theory
general-topology manifolds geometric-topology fiber-bundles surgery-theory
asked Nov 3 at 1:48
annie heart
682718
682718
No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
2
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32
add a comment |
No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
2
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32
No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
2
2
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
I don't know any standard way to define this semidirect product of manifolds, and I don't see how we can invent one.
Your intuition is correct that a fiber bundle of manifolds is like a product manifold but with more complex structure, roughly in the same way that a semidirect product of groups is like a product group but with more complex structure.
However, I think the analogy breaks down. In the case of groups, when we have $G = N rtimes_phi H$, we are also given an action $phi$ of $H$ on $N$. In the case of manifolds $M$ and $V$, in order for $V$ to act on $M$, $V$ needs some algebraic structure. We could require $V$ to be a Lie group (basically, a group which is also a manifold), but I still don't see how to define the total space.
A standard mapping torus on $M$ is a fiber bundle with fiber $M$, but its base space is a circle whereas you want it to be a more general $V$.
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I don't know any standard way to define this semidirect product of manifolds, and I don't see how we can invent one.
Your intuition is correct that a fiber bundle of manifolds is like a product manifold but with more complex structure, roughly in the same way that a semidirect product of groups is like a product group but with more complex structure.
However, I think the analogy breaks down. In the case of groups, when we have $G = N rtimes_phi H$, we are also given an action $phi$ of $H$ on $N$. In the case of manifolds $M$ and $V$, in order for $V$ to act on $M$, $V$ needs some algebraic structure. We could require $V$ to be a Lie group (basically, a group which is also a manifold), but I still don't see how to define the total space.
A standard mapping torus on $M$ is a fiber bundle with fiber $M$, but its base space is a circle whereas you want it to be a more general $V$.
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
add a comment |
up vote
1
down vote
I don't know any standard way to define this semidirect product of manifolds, and I don't see how we can invent one.
Your intuition is correct that a fiber bundle of manifolds is like a product manifold but with more complex structure, roughly in the same way that a semidirect product of groups is like a product group but with more complex structure.
However, I think the analogy breaks down. In the case of groups, when we have $G = N rtimes_phi H$, we are also given an action $phi$ of $H$ on $N$. In the case of manifolds $M$ and $V$, in order for $V$ to act on $M$, $V$ needs some algebraic structure. We could require $V$ to be a Lie group (basically, a group which is also a manifold), but I still don't see how to define the total space.
A standard mapping torus on $M$ is a fiber bundle with fiber $M$, but its base space is a circle whereas you want it to be a more general $V$.
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't know any standard way to define this semidirect product of manifolds, and I don't see how we can invent one.
Your intuition is correct that a fiber bundle of manifolds is like a product manifold but with more complex structure, roughly in the same way that a semidirect product of groups is like a product group but with more complex structure.
However, I think the analogy breaks down. In the case of groups, when we have $G = N rtimes_phi H$, we are also given an action $phi$ of $H$ on $N$. In the case of manifolds $M$ and $V$, in order for $V$ to act on $M$, $V$ needs some algebraic structure. We could require $V$ to be a Lie group (basically, a group which is also a manifold), but I still don't see how to define the total space.
A standard mapping torus on $M$ is a fiber bundle with fiber $M$, but its base space is a circle whereas you want it to be a more general $V$.
I don't know any standard way to define this semidirect product of manifolds, and I don't see how we can invent one.
Your intuition is correct that a fiber bundle of manifolds is like a product manifold but with more complex structure, roughly in the same way that a semidirect product of groups is like a product group but with more complex structure.
However, I think the analogy breaks down. In the case of groups, when we have $G = N rtimes_phi H$, we are also given an action $phi$ of $H$ on $N$. In the case of manifolds $M$ and $V$, in order for $V$ to act on $M$, $V$ needs some algebraic structure. We could require $V$ to be a Lie group (basically, a group which is also a manifold), but I still don't see how to define the total space.
A standard mapping torus on $M$ is a fiber bundle with fiber $M$, but its base space is a circle whereas you want it to be a more general $V$.
answered Nov 15 at 1:04
Hew Wolff
1,995716
1,995716
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
add a comment |
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
1
1
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
The analogy is actually very solid: extensions $1 to N to G to G/N to 1$ of groups correspond exactly to fibrations $BN to BG to B(G/N)$ of Eilenberg-MacLane spaces. If $G/N cong mathbb{Z}$ we get mapping tori. In the other direction, a fiber bundle $F to E to B$ of connected spaces is like an extension $Omega F to Omega E to Omega B$ of loop spaces, and all such extensions can be thought of as arising from a generalized semidirect product coming from an action of $Omega B$ on $Omega F$, in a suitable sense.
– Qiaochu Yuan
Nov 15 at 1:45
thanks +1, for the help
– annie heart
Nov 15 at 18:38
thanks +1, for the help
– annie heart
Nov 15 at 18:38
add a comment |
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No, this is not a thing. And you will have extreme difficulty in calculating anything about mapping class groups in dimension bigger than 3.
– Mike Miller
Nov 3 at 2:26
"What" is not a thing ?.
– annie heart
Nov 3 at 2:30
2
The first question you ask is "Is there a definition of semidirect product of manifolds". The answer is no.
– Mike Miller
Nov 3 at 2:32