Why the isometry group is not the orthogonal group?
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I found the following result:
If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:Vto V$ is an isometry iff its matrix respect to $B$ is orthogonal.
The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then
begin{equation*}
X^top mathbb{I} Y =langle x, y rangle =langle f(x),f(y)rangle =(AX)^top mathbb{I} (AY)=X^top A^top A Y,
end{equation*}
where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^top A=mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.
Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.
linear-algebra group-theory isometry
edited Nov 19 at 22:46
darij grinberg
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
add a comment |
up vote
2
down vote
favorite
I found the following result:
If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:Vto V$ is an isometry iff its matrix respect to $B$ is orthogonal.
The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then
begin{equation*}
X^top mathbb{I} Y =langle x, y rangle =langle f(x),f(y)rangle =(AX)^top mathbb{I} (AY)=X^top A^top A Y,
end{equation*}
where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^top A=mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.
Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.
linear-algebra group-theory isometry
edited Nov 19 at 22:46
darij grinberg
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
4
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I found the following result:
If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:Vto V$ is an isometry iff its matrix respect to $B$ is orthogonal.
The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then
begin{equation*}
X^top mathbb{I} Y =langle x, y rangle =langle f(x),f(y)rangle =(AX)^top mathbb{I} (AY)=X^top A^top A Y,
end{equation*}
where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^top A=mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.
Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.
linear-algebra group-theory isometry
edited Nov 19 at 22:46
darij grinberg
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
I found the following result:
If $V$ is a euclidean vector space of finite dimension $n$ and $B$ is an orthonormal basis of $V$, then an endomorphism $f:Vto V$ is an isometry iff its matrix respect to $B$ is orthogonal.
The proof is not hard. If $f$ is an isometry and $A$ is its matrix respect to $B$, then
begin{equation*}
X^top mathbb{I} Y =langle x, y rangle =langle f(x),f(y)rangle =(AX)^top mathbb{I} (AY)=X^top A^top A Y,
end{equation*}
where $X$ and $Y$ are the coordinates of the vectors $x$ and $y$ with respect to $B$, respectively. As the vectors were arbitrarily chosen, it follows that $A^top A=mathbb{I}$, so $A$ is orthonormal. The reciprocal statement is obvious from this.
Then, I thought that the orthogonal group can be equal to the isometry group, but I read that this is not true: the isometry group has order $n(n+1)/2$ (because include the traslations) and the orthogonal group has order $n(n-1)/2$. However, I cannot see why is this way.
linear-algebra group-theory isometry
linear-algebra group-theory isometry
edited Nov 19 at 22:46
darij grinberg
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
edited Nov 19 at 22:46
darij grinberg
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
edited Nov 19 at 22:46
darij grinberg
10.2k33061
edited Nov 19 at 22:46
darij grinberg
10.2k33061
edited Nov 19 at 22:46
darij grinberg
10.2k33061
10.2k33061
asked Nov 19 at 22:33
user326159
1,1391722
asked Nov 19 at 22:33
user326159
1,1391722
asked Nov 19 at 22:33
user326159
1,1391722
1,1391722
4
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37
add a comment |
4
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37
4
4
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
3
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
1
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37
add a comment |
1 Answer
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Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.
answered Nov 19 at 23:31
Chris Custer
10.1k3724
add a comment |
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Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.
answered Nov 19 at 23:31
Chris Custer
10.1k3724
add a comment |
up vote
1
down vote
Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.
answered Nov 19 at 23:31
Chris Custer
10.1k3724
add a comment |
up vote
1
down vote
up vote
1
down vote
Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.
answered Nov 19 at 23:31
Chris Custer
10.1k3724
Translations aren't linear (they don't take $0$ to $0$); but are isometries. There are $n$ dimensions of translations in dimension $n$, accounting for the difference.
answered Nov 19 at 23:31
Chris Custer
10.1k3724
answered Nov 19 at 23:31
Chris Custer
10.1k3724
answered Nov 19 at 23:31
Chris Custer
10.1k3724
answered Nov 19 at 23:31
Chris Custer
10.1k3724
10.1k3724
add a comment |
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4
Do you know what "include the translations" means?
– Eric Wofsey
Nov 19 at 22:41
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
Nov 19 at 22:41
What does it mean, @EricWofsey?
– user326159
Nov 19 at 22:49
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
Nov 19 at 23:19
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
Nov 20 at 2:37