Ricci Tensor in an Einstein Manifold
$begingroup$
I must prove that an hypersurface $M$ on $mathbb{R}^{n+1}$ that is Einstein and compact can be only the $n-$dimensional sphere when $n>2$
The Einstein condition we permits to say that scalar curvature of $M$ is costant because $n>2$.
The fact that it is an hypersfurface of $mathbb{R}^{n+1}$ can be use to consider the gauss equation and the Codazzi-Mainardi equation. The fist equation we say that
$H^2-|h|^2=cost=R^M$
Where $H$ is the mean curvature and $h$ is the second fundamental form of $M$.
In the case in which $n=3$ we have that
$cost=H^2-|h|^2=2lambdamu=2det(h)$
So the Gaussian curvature of $M$ is costant but $M$ is compact so it is the 2-sphere on $mathbb{R}^3$.
How can conclude in the case in which $n>3$?
geometry riemannian-geometry riemann-surfaces
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
$endgroup$
add a comment |
$begingroup$
I must prove that an hypersurface $M$ on $mathbb{R}^{n+1}$ that is Einstein and compact can be only the $n-$dimensional sphere when $n>2$
The Einstein condition we permits to say that scalar curvature of $M$ is costant because $n>2$.
The fact that it is an hypersfurface of $mathbb{R}^{n+1}$ can be use to consider the gauss equation and the Codazzi-Mainardi equation. The fist equation we say that
$H^2-|h|^2=cost=R^M$
Where $H$ is the mean curvature and $h$ is the second fundamental form of $M$.
In the case in which $n=3$ we have that
$cost=H^2-|h|^2=2lambdamu=2det(h)$
So the Gaussian curvature of $M$ is costant but $M$ is compact so it is the 2-sphere on $mathbb{R}^3$.
How can conclude in the case in which $n>3$?
geometry riemannian-geometry riemann-surfaces
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
$endgroup$
add a comment |
$begingroup$
I must prove that an hypersurface $M$ on $mathbb{R}^{n+1}$ that is Einstein and compact can be only the $n-$dimensional sphere when $n>2$
The Einstein condition we permits to say that scalar curvature of $M$ is costant because $n>2$.
The fact that it is an hypersfurface of $mathbb{R}^{n+1}$ can be use to consider the gauss equation and the Codazzi-Mainardi equation. The fist equation we say that
$H^2-|h|^2=cost=R^M$
Where $H$ is the mean curvature and $h$ is the second fundamental form of $M$.
In the case in which $n=3$ we have that
$cost=H^2-|h|^2=2lambdamu=2det(h)$
So the Gaussian curvature of $M$ is costant but $M$ is compact so it is the 2-sphere on $mathbb{R}^3$.
How can conclude in the case in which $n>3$?
geometry riemannian-geometry riemann-surfaces
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
$endgroup$
I must prove that an hypersurface $M$ on $mathbb{R}^{n+1}$ that is Einstein and compact can be only the $n-$dimensional sphere when $n>2$
The Einstein condition we permits to say that scalar curvature of $M$ is costant because $n>2$.
The fact that it is an hypersfurface of $mathbb{R}^{n+1}$ can be use to consider the gauss equation and the Codazzi-Mainardi equation. The fist equation we say that
$H^2-|h|^2=cost=R^M$
Where $H$ is the mean curvature and $h$ is the second fundamental form of $M$.
In the case in which $n=3$ we have that
$cost=H^2-|h|^2=2lambdamu=2det(h)$
So the Gaussian curvature of $M$ is costant but $M$ is compact so it is the 2-sphere on $mathbb{R}^3$.
How can conclude in the case in which $n>3$?
geometry riemannian-geometry riemann-surfaces
geometry riemannian-geometry riemann-surfaces
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
asked Jan 14 at 20:20
Federico FalluccaFederico Fallucca
1,94219
1,94219
add a comment |
add a comment |
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