semidirect product of isometry group
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
1
$begingroup$
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.
$endgroup$
– user99914
Sep 16 '16 at 19:25
add a comment |
$begingroup$
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
$endgroup$
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
$begingroup$
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.
$endgroup$
– user99914
Sep 16 '16 at 19:25
add a comment |
1
$begingroup$
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.
$endgroup$
– user99914
Sep 16 '16 at 19:25
1
1
$begingroup$
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.
$endgroup$
– user99914
Sep 16 '16 at 19:25
$begingroup$
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.
$endgroup$
– user99914
Sep 16 '16 at 19:25
add a comment |
1 Answer
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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
416
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$begingroup$
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
416
$endgroup$
add a comment |
$begingroup$
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
416
$endgroup$
add a comment |
$begingroup$
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
416
$endgroup$
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
416
answered Jan 6 at 12:14
SiddharthaSiddhartha
416
answered Jan 6 at 12:14
SiddharthaSiddhartha
416
answered Jan 6 at 12:14
SiddharthaSiddhartha
416
416
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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.
$endgroup$
– user99914
Sep 16 '16 at 19:25