How does one prove that $text{Isom}(mathbb{S}^2 times mathbb{R}) = text{Isom}(mathbb{S}^2) times...
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How does one prove that $text{Isom}(mathbb{S}^2 times mathbb{R}) = text{Isom}(mathbb{S}^2) times text{Isom}(mathbb{R})$?
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Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $mathbb{S}^2 times mathbb{R}$ and $mathbb{H}^2 times mathbb{R}$ are $4$ dimensional and equal to $ text{Isom}(mathbb{S}^2) times text{Isom}(mathbb{R})$ and $ text{Isom}(mathbb{H}^2) times text{Isom}(mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?
differential-geometry
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
$endgroup$
add a comment |
$begingroup$
Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $mathbb{S}^2 times mathbb{R}$ and $mathbb{H}^2 times mathbb{R}$ are $4$ dimensional and equal to $ text{Isom}(mathbb{S}^2) times text{Isom}(mathbb{R})$ and $ text{Isom}(mathbb{H}^2) times text{Isom}(mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?
differential-geometry
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
$endgroup$
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12
add a comment |
$begingroup$
Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $mathbb{S}^2 times mathbb{R}$ and $mathbb{H}^2 times mathbb{R}$ are $4$ dimensional and equal to $ text{Isom}(mathbb{S}^2) times text{Isom}(mathbb{R})$ and $ text{Isom}(mathbb{H}^2) times text{Isom}(mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?
differential-geometry
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
$endgroup$
Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $mathbb{S}^2 times mathbb{R}$ and $mathbb{H}^2 times mathbb{R}$ are $4$ dimensional and equal to $ text{Isom}(mathbb{S}^2) times text{Isom}(mathbb{R})$ and $ text{Isom}(mathbb{H}^2) times text{Isom}(mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?
differential-geometry
differential-geometry
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,385418
1,385418
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12
add a comment |
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $f$ be an isometry of $Bbb S^2times Bbb R$.
Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$.
Now $Bbb S^2times {0}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $pi$. We conclude that $f$ maps $Bbb S^2times{0}$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
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1 Answer
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$begingroup$
Let $f$ be an isometry of $Bbb S^2times Bbb R$.
Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$.
Now $Bbb S^2times {0}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $pi$. We conclude that $f$ maps $Bbb S^2times{0}$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
$begingroup$
Let $f$ be an isometry of $Bbb S^2times Bbb R$.
Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$.
Now $Bbb S^2times {0}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $pi$. We conclude that $f$ maps $Bbb S^2times{0}$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
$begingroup$
Let $f$ be an isometry of $Bbb S^2times Bbb R$.
Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$.
Now $Bbb S^2times {0}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $pi$. We conclude that $f$ maps $Bbb S^2times{0}$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
Let $f$ be an isometry of $Bbb S^2times Bbb R$.
Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$.
Now $Bbb S^2times {0}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $pi$. We conclude that $f$ maps $Bbb S^2times{0}$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
$begingroup$
Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $mathbb{H}^2 times mathbb{R}$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
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$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ is not true in general, see here.
$endgroup$
– Dietrich Burde
Mar 11 at 19:52
$begingroup$
@DietrichBurde I'm aware of that, but I know that in this case it's true.
$endgroup$
– Matheus Andrade
Mar 11 at 19:54
$begingroup$
So you have a link? Then you could add it.
$endgroup$
– Dietrich Burde
Mar 11 at 19:58
$begingroup$
@DietrichBurde books.google.com.br/…
$endgroup$
– Matheus Andrade
Mar 11 at 20:12