Mixing
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 2 days ago
darij grinberg
9,90532961
asked 2 days ago
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 2 days ago
darij grinberg
9,90532961
asked 2 days ago
user326159
1,1391722
4
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago
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 2 days ago
darij grinberg
9,90532961
asked 2 days ago
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 2 days ago
darij grinberg
9,90532961
asked 2 days ago
user326159
1,1391722
edited 2 days ago
darij grinberg
9,90532961
asked 2 days ago
user326159
1,1391722
edited 2 days ago
darij grinberg
9,90532961
edited 2 days ago
darij grinberg
9,90532961
edited 2 days ago
darij grinberg
9,90532961
9,90532961
asked 2 days ago
user326159
1,1391722
asked 2 days ago
user326159
1,1391722
asked 2 days ago
user326159
1,1391722
1,1391722
4
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago
add a comment |
4
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago
4
4
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
3
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
1
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago
add a comment |
1 Answer
1
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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 2 days ago
Chris Custer
8,6692623
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
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 2 days ago
Chris Custer
8,6692623
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 2 days ago
Chris Custer
8,6692623
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 2 days ago
Chris Custer
8,6692623
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 2 days ago
Chris Custer
8,6692623
answered 2 days ago
Chris Custer
8,6692623
answered 2 days ago
Chris Custer
8,6692623
answered 2 days ago
Chris Custer
8,6692623
8,6692623
add a comment |
add a comment |
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4
Do you know what "include the translations" means?
– Eric Wofsey
2 days ago
3
Not all isometries are endomorphisms; only those that fix the origin are.
– Travis
2 days ago
What does it mean, @EricWofsey?
– user326159
2 days ago
I was thinking about it. The orthogonal group is the linear isometry group?
– user326159
2 days ago
1
Presumably "order" in the question was supposed to be "dimension".
– Andreas Blass
2 days ago