Does ${rm Iso}(Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ hold for product metrics?












2












$begingroup$


I know that ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ is not generally correct where $M$ and $N$ are smooth Riemannian manifolds; but




Does ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ hold for product metrics?




what about warp-product metrics?






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:42






  • 1




    $begingroup$
    This seems relevant.
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:48
















2












$begingroup$


I know that ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ is not generally correct where $M$ and $N$ are smooth Riemannian manifolds; but




Does ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ hold for product metrics?




what about warp-product metrics?






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:42






  • 1




    $begingroup$
    This seems relevant.
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:48














2












2








2





$begingroup$


I know that ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ is not generally correct where $M$ and $N$ are smooth Riemannian manifolds; but




Does ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ hold for product metrics?




what about warp-product metrics?






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$




I know that ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ is not generally correct where $M$ and $N$ are smooth Riemannian manifolds; but




Does ${rm Iso} (Mtimes N)={rm Iso}(M)times {rm Iso}(N)$ hold for product metrics?




what about warp-product metrics?






differential-geometry riemannian-geometry




differential-geometry riemannian-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 7:47









C.F.GC.F.G

1,4301821




1,4301821












  • $begingroup$
    What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:42






  • 1




    $begingroup$
    This seems relevant.
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:48


















  • $begingroup$
    What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:42






  • 1




    $begingroup$
    This seems relevant.
    $endgroup$
    – Anthony Carapetis
    Dec 6 '18 at 9:48
















$begingroup$
What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
$endgroup$
– Anthony Carapetis
Dec 6 '18 at 9:42




$begingroup$
What do you mean by "for product metrics"? $M$ and $N$ should both be non-trivial Riemannian products of non-flat factors?
$endgroup$
– Anthony Carapetis
Dec 6 '18 at 9:42




1




1




$begingroup$
This seems relevant.
$endgroup$
– Anthony Carapetis
Dec 6 '18 at 9:48




$begingroup$
This seems relevant.
$endgroup$
– Anthony Carapetis
Dec 6 '18 at 9:48










1 Answer
1






active

oldest

votes


















6












$begingroup$

Example: $M=N=mathbb{R}$ with the standard metric. Then $operatorname{Isom}mathbb{E}^2$ is a 3-dimensional Lie group, so obviously cannot be the Cartesian square of $operatorname{Isom}mathbb{E}^1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. What can be say in compact case?
    $endgroup$
    – C.F.G
    Dec 7 '18 at 5:00






  • 1




    $begingroup$
    For example, see the link given by @AnthonyCarapetis
    $endgroup$
    – user10354138
    Dec 7 '18 at 15:25











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Example: $M=N=mathbb{R}$ with the standard metric. Then $operatorname{Isom}mathbb{E}^2$ is a 3-dimensional Lie group, so obviously cannot be the Cartesian square of $operatorname{Isom}mathbb{E}^1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. What can be say in compact case?
    $endgroup$
    – C.F.G
    Dec 7 '18 at 5:00






  • 1




    $begingroup$
    For example, see the link given by @AnthonyCarapetis
    $endgroup$
    – user10354138
    Dec 7 '18 at 15:25
















6












$begingroup$

Example: $M=N=mathbb{R}$ with the standard metric. Then $operatorname{Isom}mathbb{E}^2$ is a 3-dimensional Lie group, so obviously cannot be the Cartesian square of $operatorname{Isom}mathbb{E}^1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. What can be say in compact case?
    $endgroup$
    – C.F.G
    Dec 7 '18 at 5:00






  • 1




    $begingroup$
    For example, see the link given by @AnthonyCarapetis
    $endgroup$
    – user10354138
    Dec 7 '18 at 15:25














6












6








6





$begingroup$

Example: $M=N=mathbb{R}$ with the standard metric. Then $operatorname{Isom}mathbb{E}^2$ is a 3-dimensional Lie group, so obviously cannot be the Cartesian square of $operatorname{Isom}mathbb{E}^1$.






share|cite|improve this answer









$endgroup$



Example: $M=N=mathbb{R}$ with the standard metric. Then $operatorname{Isom}mathbb{E}^2$ is a 3-dimensional Lie group, so obviously cannot be the Cartesian square of $operatorname{Isom}mathbb{E}^1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 8:44









user10354138user10354138

7,4322925




7,4322925












  • $begingroup$
    Thank you for your answer. What can be say in compact case?
    $endgroup$
    – C.F.G
    Dec 7 '18 at 5:00






  • 1




    $begingroup$
    For example, see the link given by @AnthonyCarapetis
    $endgroup$
    – user10354138
    Dec 7 '18 at 15:25


















  • $begingroup$
    Thank you for your answer. What can be say in compact case?
    $endgroup$
    – C.F.G
    Dec 7 '18 at 5:00






  • 1




    $begingroup$
    For example, see the link given by @AnthonyCarapetis
    $endgroup$
    – user10354138
    Dec 7 '18 at 15:25
















$begingroup$
Thank you for your answer. What can be say in compact case?
$endgroup$
– C.F.G
Dec 7 '18 at 5:00




$begingroup$
Thank you for your answer. What can be say in compact case?
$endgroup$
– C.F.G
Dec 7 '18 at 5:00




1




1




$begingroup$
For example, see the link given by @AnthonyCarapetis
$endgroup$
– user10354138
Dec 7 '18 at 15:25




$begingroup$
For example, see the link given by @AnthonyCarapetis
$endgroup$
– user10354138
Dec 7 '18 at 15:25


















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