Explicitly confirming invariance of curvatures under isometry

I would love to recognize just how to clearly confirm that Riemann Curvature, Ricci Curvature, Sectional Curvature and also Scalar Curvature remain stable under an isometry.

I can not see this clarified in the majority of publications I have actually considered. They atmost clarify conservation of the link.

I presume doing a specific evidence for the sectional curvature should suffice (and also most convenient? ) given that all the remainder can be created in regards to it.

Offered Akhil's reply I assume I need to attempt to recognize the link invariance evidence far better and also below goes my partial effort.

Allow $\nabla$ be the link on the manifold $(M,g)$ and also $\nabla '$ be the Riemann link on the manifold $(M',g')$ and also in between these 2 allowed $\phi$ be the isometry. After that one intends to show 2 points,

  • $D\phi [\nabla _ X Y] = \nabla ' _{D\phi[X]} D\phi [Y]$
  • $R(X,Y)Z = R'(D\phi [X],D\phi [Y]) D\phi [Z]$

Where $R$ and also $R'$ are the Riemann link on $(M,g)$ and also $(M',g')$ specifically.

One specifies the map $\nabla ''$ on M which maps 2 vector areas on M to an additional vector area making use of $\nabla '' _X Y = D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]$. By the individuality of the Riemann link the evidence is full if one can show that this $\nabla ''$ pleases all the problems of being a Riemann link on M.

I am obtaining stuck after a couple of actions while attempting to show the Lebnitz building of $\nabla ''$. Allow $f$ be some smooth function on M and afterwards one would love to show that,. $\nabla '' _X fY = X(f)Y + f\nabla '' _X Y$ which amounts revealing that,. $D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [fY]) = X(f) + f D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y])$ recognizing that $\nabla '$ pleases the Leibniz building on $M'$.

Some just how I am not having the ability to unpack the above to confirm this. I can get the 2nd regard to the formula yet not the first one.

Confirming proportion of $\nabla ''$ is very easy yet once more confirming statistics compatibility is stuck for me. If $X,Y,Z$ are 3 vector areas on M after that one would certainly intend to show that,

$Xg(Y,Z) = g(\nabla ''_X Y,Z) + g(Y, \nabla '' _X Z)$

which amounts revealing that,

$Xg(Y,Z) = g(D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]),Z) + g(Y,D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Z]) )$

recognizing that $\nabla'$ satisfies statistics compatibility formula on $M'$

It would certainly be handy if a person can aid me complete the actions.

After that one is entrusted confirming the curvature endomorphism formula.

2019-05-07 14:52:55
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Answers: 1

Take an appearance at the last large presented formula under "official definition" below. It reveals you Gauss's specific kind for a Levi - Civita link in regards to the statistics. Given that you recognize just how the statistics changes under an isometry, and also just how a Lie brace changes under a diffeomorphism, exercising just how the link changes under an isometry totals up to placing those active ingredients with each other.


2019-05-10 08:49:41