Ricci Tensor and Einstein Manifolds












0












$begingroup$


What can we say about an hypersurface Einstein manifolds on $mathbb{R}^{n+1}$ when $ngeq 3$ ?



The manifolds is Einstein so Ricci tensor is a multiply of the metric $g$ on manifold:



$Ric=lambda g$ where $lambdain C^infty(M)$



If $S$ is the scalar curvature of $M$ then the relationship between $S$ and $lambda$ is



$lambda=frac{S}{n}$



Because the scalar curvature of $M$ is the trace of Ricci tensor.



So using the condition that $ngeq 3$ and using a contraction of the second Bianchi Identity we have that the scalar curvature of $M$ is costant so $lambda$ is costant.



Another condition that we have is that $M$ is an hypersurface of $mathbb{R}^{n+1}$ so we can use also the Codazzi-Mainardi Equation ti get some equation between the mean curvature $H$ of $M$, the metric $g$ and the second fondamental formula of $M$.
What can other we say about $M$, it is possible classify it?










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$endgroup$












  • $begingroup$
    Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
    $endgroup$
    – Moishe Cohen
    Jan 13 at 18:06










  • $begingroup$
    What I must search in this book? There are a lot of sections
    $endgroup$
    – Federico Fallucca
    Jan 13 at 18:34
















0












$begingroup$


What can we say about an hypersurface Einstein manifolds on $mathbb{R}^{n+1}$ when $ngeq 3$ ?



The manifolds is Einstein so Ricci tensor is a multiply of the metric $g$ on manifold:



$Ric=lambda g$ where $lambdain C^infty(M)$



If $S$ is the scalar curvature of $M$ then the relationship between $S$ and $lambda$ is



$lambda=frac{S}{n}$



Because the scalar curvature of $M$ is the trace of Ricci tensor.



So using the condition that $ngeq 3$ and using a contraction of the second Bianchi Identity we have that the scalar curvature of $M$ is costant so $lambda$ is costant.



Another condition that we have is that $M$ is an hypersurface of $mathbb{R}^{n+1}$ so we can use also the Codazzi-Mainardi Equation ti get some equation between the mean curvature $H$ of $M$, the metric $g$ and the second fondamental formula of $M$.
What can other we say about $M$, it is possible classify it?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
    $endgroup$
    – Moishe Cohen
    Jan 13 at 18:06










  • $begingroup$
    What I must search in this book? There are a lot of sections
    $endgroup$
    – Federico Fallucca
    Jan 13 at 18:34














0












0








0





$begingroup$


What can we say about an hypersurface Einstein manifolds on $mathbb{R}^{n+1}$ when $ngeq 3$ ?



The manifolds is Einstein so Ricci tensor is a multiply of the metric $g$ on manifold:



$Ric=lambda g$ where $lambdain C^infty(M)$



If $S$ is the scalar curvature of $M$ then the relationship between $S$ and $lambda$ is



$lambda=frac{S}{n}$



Because the scalar curvature of $M$ is the trace of Ricci tensor.



So using the condition that $ngeq 3$ and using a contraction of the second Bianchi Identity we have that the scalar curvature of $M$ is costant so $lambda$ is costant.



Another condition that we have is that $M$ is an hypersurface of $mathbb{R}^{n+1}$ so we can use also the Codazzi-Mainardi Equation ti get some equation between the mean curvature $H$ of $M$, the metric $g$ and the second fondamental formula of $M$.
What can other we say about $M$, it is possible classify it?










share|cite|improve this question









$endgroup$




What can we say about an hypersurface Einstein manifolds on $mathbb{R}^{n+1}$ when $ngeq 3$ ?



The manifolds is Einstein so Ricci tensor is a multiply of the metric $g$ on manifold:



$Ric=lambda g$ where $lambdain C^infty(M)$



If $S$ is the scalar curvature of $M$ then the relationship between $S$ and $lambda$ is



$lambda=frac{S}{n}$



Because the scalar curvature of $M$ is the trace of Ricci tensor.



So using the condition that $ngeq 3$ and using a contraction of the second Bianchi Identity we have that the scalar curvature of $M$ is costant so $lambda$ is costant.



Another condition that we have is that $M$ is an hypersurface of $mathbb{R}^{n+1}$ so we can use also the Codazzi-Mainardi Equation ti get some equation between the mean curvature $H$ of $M$, the metric $g$ and the second fondamental formula of $M$.
What can other we say about $M$, it is possible classify it?







general-topology geometry analysis riemannian-geometry riemann-surfaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 at 15:33









Federico FalluccaFederico Fallucca

1,959110




1,959110












  • $begingroup$
    Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
    $endgroup$
    – Moishe Cohen
    Jan 13 at 18:06










  • $begingroup$
    What I must search in this book? There are a lot of sections
    $endgroup$
    – Federico Fallucca
    Jan 13 at 18:34


















  • $begingroup$
    Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
    $endgroup$
    – Moishe Cohen
    Jan 13 at 18:06










  • $begingroup$
    What I must search in this book? There are a lot of sections
    $endgroup$
    – Federico Fallucca
    Jan 13 at 18:34
















$begingroup$
Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
$endgroup$
– Moishe Cohen
Jan 13 at 18:06




$begingroup$
Take a look here: researchgate.net/profile/Bang_Yen_Chen/publication/…
$endgroup$
– Moishe Cohen
Jan 13 at 18:06












$begingroup$
What I must search in this book? There are a lot of sections
$endgroup$
– Federico Fallucca
Jan 13 at 18:34




$begingroup$
What I must search in this book? There are a lot of sections
$endgroup$
– Federico Fallucca
Jan 13 at 18:34










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