# 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.

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.

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