Existence of a relatively flat surface of $mathbbS^3$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Consider $mathbbS^3$ with the standard round metric.
Is there an embedded surface $S subseteq mathbbS^3$ with the following property:
$R^mathbbS^3(X,Y)=0$ for every two tangent vector $X,Y in TS$.
That is, I require $R^mathbbS^3(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$.
Here $R^mathbbS^3$ is the Riemann curvature tensor of $mathbbS^3$, not the curvature tensor of the induced Riemannian manifold $S$.
Is this condition related to other geometric properties like $S$ being a totally geodesic submanifold or a minimal submanifold?
riemannian-geometry surfaces curvature spheres
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
add a comment |Â
up vote
1
down vote
favorite
Consider $mathbbS^3$ with the standard round metric.
Is there an embedded surface $S subseteq mathbbS^3$ with the following property:
$R^mathbbS^3(X,Y)=0$ for every two tangent vector $X,Y in TS$.
That is, I require $R^mathbbS^3(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$.
Here $R^mathbbS^3$ is the Riemann curvature tensor of $mathbbS^3$, not the curvature tensor of the induced Riemannian manifold $S$.
Is this condition related to other geometric properties like $S$ being a totally geodesic submanifold or a minimal submanifold?
riemannian-geometry surfaces curvature spheres
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
1
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
1
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider $mathbbS^3$ with the standard round metric.
Is there an embedded surface $S subseteq mathbbS^3$ with the following property:
$R^mathbbS^3(X,Y)=0$ for every two tangent vector $X,Y in TS$.
That is, I require $R^mathbbS^3(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$.
Here $R^mathbbS^3$ is the Riemann curvature tensor of $mathbbS^3$, not the curvature tensor of the induced Riemannian manifold $S$.
Is this condition related to other geometric properties like $S$ being a totally geodesic submanifold or a minimal submanifold?
riemannian-geometry surfaces curvature spheres
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
Consider $mathbbS^3$ with the standard round metric.
Is there an embedded surface $S subseteq mathbbS^3$ with the following property:
$R^mathbbS^3(X,Y)=0$ for every two tangent vector $X,Y in TS$.
That is, I require $R^mathbbS^3(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$.
Here $R^mathbbS^3$ is the Riemann curvature tensor of $mathbbS^3$, not the curvature tensor of the induced Riemannian manifold $S$.
Is this condition related to other geometric properties like $S$ being a totally geodesic submanifold or a minimal submanifold?
riemannian-geometry surfaces curvature spheres
riemannian-geometry surfaces curvature spheres
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
edited Sep 4 at 15:54
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
asked Sep 3 at 8:44
Asaf Shachar
4,7363836
4,7363836
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
1
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
1
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50
add a comment |Â
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
1
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
1
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
1
1
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
1
1
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903661%2fexistence-of-a-relatively-flat-surface-of-mathbbs3%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you asking whether one can find an embedded surface in $S^3$ such that, when considered with the induced metric from the standard round metric it becomes flat? (has constant zero curvature)
– levap
Sep 3 at 15:01
No, that is not the question. I am using the curvature tensor of the ambient manifold $mathbbS^3$, not the curvature tensor of the induced metric.
– Asaf Shachar
Sep 4 at 8:27
1
Do you want to require that $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TS$ or $R(X,Y)Z = 0$ for all $X,Y in TS$ and $Z in TmathbbS^3$?
– levap
Sep 4 at 11:09
1
Ah, sorry. Now I understood you. I want to require $R(X,Y)Z=0$ for all $X,Y in TS$ and $Z in TmathbbS^3$. By the way I guess both versions might be interesting. Do you see nice sufficient/necessary conditions for any of them? (or have any intuition?).
– Asaf Shachar
Sep 4 at 15:50