semi-direct product between manifolds











up vote
1
down vote

favorite
2












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!










share|cite|improve this question






















  • 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















up vote
1
down vote

favorite
2












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!










share|cite|improve this question






















  • 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













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





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!










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










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 $V$.






share|cite|improve this answer

















  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2982445%2fsemi-direct-product-between-manifolds%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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 $V$.






share|cite|improve this answer

















  • 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















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 $V$.






share|cite|improve this answer

















  • 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













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 $V$.






share|cite|improve this answer












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 $V$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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














  • 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


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2982445%2fsemi-direct-product-between-manifolds%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Plaza Victoria

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...