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.