semidirect product of isometry group
I am doing exercise about semidirect product. Here is the question:
Prove that the isometry group of Euclidean space $R^n$ is $O(n)rtimes R^n$.
I was stucking. Any ideas?
semidirect-product
asked Sep 16 '16 at 19:19
nellsynellsy
93
add a comment |
I am doing exercise about semidirect product. Here is the question:
Prove that the isometry group of Euclidean space $R^n$ is $O(n)rtimes R^n$.
I was stucking. Any ideas?
semidirect-product
asked Sep 16 '16 at 19:19
nellsynellsy
93
1
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25
add a comment |
I am doing exercise about semidirect product. Here is the question:
Prove that the isometry group of Euclidean space $R^n$ is $O(n)rtimes R^n$.
I was stucking. Any ideas?
semidirect-product
asked Sep 16 '16 at 19:19
nellsynellsy
93
I am doing exercise about semidirect product. Here is the question:
Prove that the isometry group of Euclidean space $R^n$ is $O(n)rtimes R^n$.
I was stucking. Any ideas?
semidirect-product
semidirect-product
asked Sep 16 '16 at 19:19
nellsynellsy
93
asked Sep 16 '16 at 19:19
nellsynellsy
93
asked Sep 16 '16 at 19:19
nellsynellsy
93
asked Sep 16 '16 at 19:19
nellsynellsy
93
asked Sep 16 '16 at 19:19
nellsynellsy
93
93
1
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25
add a comment |
1
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25
1
1
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25
add a comment |
1 Answer
1
active
oldest
votes
The notation is a problem : the orthogonal group $O(n)$ acts on the translations group. Thus $E(n) = {mathbb R}^n rtimes O(n)$, not the other way round. The translation group $T(n) cong {mathbb R}^n$ is a abelian normal subgroup of $E(n)$ with $E(n)/T(n) cong O(n)$.
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
add a comment |
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',
autoActivateHeartbeat: false,
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
});
}
});
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%2f1929585%2fsemidirect-product-of-isometry-group%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
The notation is a problem : the orthogonal group $O(n)$ acts on the translations group. Thus $E(n) = {mathbb R}^n rtimes O(n)$, not the other way round. The translation group $T(n) cong {mathbb R}^n$ is a abelian normal subgroup of $E(n)$ with $E(n)/T(n) cong O(n)$.
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
add a comment |
The notation is a problem : the orthogonal group $O(n)$ acts on the translations group. Thus $E(n) = {mathbb R}^n rtimes O(n)$, not the other way round. The translation group $T(n) cong {mathbb R}^n$ is a abelian normal subgroup of $E(n)$ with $E(n)/T(n) cong O(n)$.
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
add a comment |
The notation is a problem : the orthogonal group $O(n)$ acts on the translations group. Thus $E(n) = {mathbb R}^n rtimes O(n)$, not the other way round. The translation group $T(n) cong {mathbb R}^n$ is a abelian normal subgroup of $E(n)$ with $E(n)/T(n) cong O(n)$.
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
The notation is a problem : the orthogonal group $O(n)$ acts on the translations group. Thus $E(n) = {mathbb R}^n rtimes O(n)$, not the other way round. The translation group $T(n) cong {mathbb R}^n$ is a abelian normal subgroup of $E(n)$ with $E(n)/T(n) cong O(n)$.
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
answered Jan 6 at 12:14
SiddharthaSiddhartha
215
215
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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%2f1929585%2fsemidirect-product-of-isometry-group%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
1
It is obvious that the $O(n)rtimes R^n$ is in $text{Iso}(mathbb R^n)$. You just need to check that all isometry are of this form.
– user99914
Sep 16 '16 at 19:25