How does one prove that $textIsom(mathbbS^2 times mathbbR) = textIsom(mathbbS^2) times textIsom(mathbbR)$?Isometry group of a product of isomorphic manifoldsWhat are some examples of $textIsom(M)$ and $textConf(M)$?Is there a geometric explanation for why principal curvature directions are orthogonal?Proof that the $mathbb R P^n$ is HausdorffMinimal-dimension example of (open) subset of $mathbbR^n$ with trivial first cohomology but nontrivial fundamental groupComplex symplectic group and geometry?How to prove that $Gamma$ is finitely generated?The relationship between the Weyl group and Isometries of a Maximal FlatCan we characterise concircular vector fields by their flow?Is $mathbbS^2 times mathbbR$ homeomorphic to any subset of $mathbbR^3$?(Total) volume preserving transformations?
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How does one prove that $textIsom(mathbbS^2 times mathbbR) = textIsom(mathbbS^2) times textIsom(mathbbR)$?
Isometry group of a product of isomorphic manifoldsWhat are some examples of $textIsom(M)$ and $textConf(M)$?Is there a geometric explanation for why principal curvature directions are orthogonal?Proof that the $mathbb R P^n$ is HausdorffMinimal-dimension example of (open) subset of $mathbbR^n$ with trivial first cohomology but nontrivial fundamental groupComplex symplectic group and geometry?How to prove that $Gamma$ is finitely generated?The relationship between the Weyl group and Isometries of a Maximal FlatCan we characterise concircular vector fields by their flow?Is $mathbbS^2 times mathbbR$ homeomorphic to any subset of $mathbbR^3$?(Total) volume preserving transformations?
$begingroup$
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 $mathbbS^2 times mathbbR$ and $mathbbH^2 times mathbbR$ are $4$ dimensional and equal to $ textIsom(mathbbS^2) times textIsom(mathbbR)$ and $ textIsom(mathbbH^2) times textIsom(mathbbR)$, 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
1,380418
$endgroup$
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 $mathbbS^2 times mathbbR$ and $mathbbH^2 times mathbbR$ are $4$ dimensional and equal to $ textIsom(mathbbS^2) times textIsom(mathbbR)$ and $ textIsom(mathbbH^2) times textIsom(mathbbR)$, 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
1,380418
$endgroup$
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ 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 $mathbbS^2 times mathbbR$ and $mathbbH^2 times mathbbR$ are $4$ dimensional and equal to $ textIsom(mathbbS^2) times textIsom(mathbbR)$ and $ textIsom(mathbbH^2) times textIsom(mathbbR)$, 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
1,380418
$endgroup$
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 $mathbbS^2 times mathbbR$ and $mathbbH^2 times mathbbR$ are $4$ dimensional and equal to $ textIsom(mathbbS^2) times textIsom(mathbbR)$ and $ textIsom(mathbbH^2) times textIsom(mathbbR)$, 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
1,380418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,380418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,380418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,380418
asked Mar 11 at 19:44
Matheus AndradeMatheus Andrade
1,380418
1,380418
$begingroup$
$Isom(X times Y)cong Isom(X)times Isom(Y)$ 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
active
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 $Bbb S^2times0$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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 $mathbbH^2 times mathbbR$ case?
$endgroup$
– Matheus Andrade
Mar 11 at 21:31
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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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 $Bbb S^2times0$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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 $mathbbH^2 times mathbbR$ 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 $Bbb S^2times0$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
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 $mathbbH^2 times mathbbR$ 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 $Bbb S^2times0$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
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 $Bbb S^2times0$ to itself.
The rest is then easy.
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
answered Mar 11 at 19:59
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
$begingroup$
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 $mathbbH^2 times mathbbR$ 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 $mathbbH^2 times mathbbR$ 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 $mathbbH^2 times mathbbR$ 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 $mathbbH^2 times mathbbR$ 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