Explicitly proving invariance of curvatures under isometry

I would like to know how to explicitly prove that Riemann Curvature,Ricci Curvature, Sectional Curvature and Scalar Curvature are left invariant under an isometry.

I can't see this explained in most books I have looked at. They atmost explain preservation of the connection.

I guess doing an explicit proof for the sectional curvature should be enough (and easiest?) since all the rest can be written in terms of it.

Given Akhil's reply I think I should try to understand the connection invariance proof better and here goes my partial attempt.

Let $\nabla$ be the connection on the manifold $(M,g)$ and $\nabla '$ be the Riemann connection on the manifold $(M',g')$ and between these two let $\phi$ be the isometry. Then one wants to show two things,

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

Where $R$ and $R'$ are the Riemann connection on $(M,g)$ and $(M',g')$ respectively.

One defines the map $\nabla ''$ on M which maps two vector fields on M to another vector field using $\nabla '' _X Y = D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y]$. By the uniqueness of the Riemann connection the proof is complete if one can show that this $\nabla ''$ satisfies all the conditions of being a Riemann connection on M.

I am getting stuck after a few steps while trying to show the Lebnitz property of $\nabla ''$. Let $f$ be some smooth function on M and then one would like to show that, $\nabla '' _X fY = X(f)Y + f\nabla '' _X Y$ which is equivalent to showing that, $D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [fY]) = X(f) + f D\phi ^{-1} (\nabla' _{D\phi[X]} D\phi [Y])$ knowing that $\nabla '$ satisfies the Leibniz property on $M'$.

Some how I am not being able to unwrap the above to prove this. I can get the second term of the equation but not the first one.

Proving symmetry of $\nabla ''$ is easy but again proving metric compatibility is stuck for me. If $X,Y,Z$ are 3 vector fields on M then one would want to show that,

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

which is equivalent to showing that,

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

knowing that $\nabla'$ satisfies metric compatibility equation on $M'$

It would be helpful if someone can help me fill in the steps.

Then one is left with proving the curvature endomorphism equation.

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All of this follows from the preservation of the curvature tensor under an isometry; indeed, if you pull back the Levi-Civita connection by an isometry, you get an invariant connection on the other manifold, so by uniqueness, the isometry preserves the connection. Since the curvatures are all derived from the connection, the claim follows. – Akhil Mathew Aug 9 '10 at 18:59
@Akhil Thanks for your reply. Then I think I should try to understand the proof of invariance of the connection better. I have typed in above some of the steps of the proof as I can imagine it. It would be very helpful if you can fill in the missing steps. – Anirbit Aug 12 '10 at 16:05
Use the fact that $D\phi(fY) = f\circ \phi^{-1} D\phi(Y)$. Now apply the Leibniz rule for $\nabla'$. I suggest that you spend more time playing around with $\phi$ and understanding how it and its various incarnations (such as $D\phi$) interact with various constructions. For example, in trying to prove metric compatibility, you will need to use the fact that $\phi$ is an isometry. Try to phrase this condition in terms of a formula (involving $g, g', D\phi$ and tangent vectors). – Matt E Aug 12 '10 at 16:44
@Matt Thanks for the help. I couldn't really understand your comment about the rewriting you suggested for the metric compatibility proof. Can you suggest some references along these lines? – Anirbit Aug 13 '10 at 17:37
@Matt $\phi ^{-1}$ is a map from the target manifold to the domain manifold. How does that act on $D\phi [Y]$? Do you mean a pull-back of $D\phi [Y]$ along $\phi$? Then how does a $f$ compose with the pulled back vector as you seem to indicate with the $\circ$ symbol? I am very confused by your notation. – Anirbit Aug 14 '10 at 8:46