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.

Solution 1:

Take a look at the last big displayed equation under "formal definition" here. It shows you Gauss's explicit form for a Levi-Civita connection in terms of the metric. Since you know how the metric transforms under an isometry, and how a Lie bracket transforms under a diffeomorphism, working out how the connection transforms under an isometry amounts to putting those ingredients together.

http://en.wikipedia.org/wiki/Levi-Civita_connection