Let $M$, $M'$ be riemann manifolds with levi-civita connection $\nabla$,$\nabla'$. If $\phi$ is an isometry (global so diffeomorphism too) I want to show:
$
\nabla'_{X'} Y'=D\phi (\nabla_X Y)
$
where $X'=D\phi X$ and Y' similarly.
So what I am doing is the following, following the lines of a similar post I found here, I am trying to show that $D\phi (\nabla_X Y)=\nabla''_{X'}Y'$ is a connection and its metric compatible and torsion free. Then by uniqueness I am done.
However I get stuck. This is my attempt:
for $f\in C^\infty(M')$
$
\nabla''_{fX'}Y'=D\phi (\nabla_{D\phi^{-1}(fX')}Y)=D\phi ((f\circ\phi)\nabla_XY)=fD\phi(\nabla_XY)=f\nabla''_XY
$
Leibniz rule
$
\nabla''_{X'}fY'=D\phi (\nabla_X(f\circ\phi)Y) = D\phi(X(f\circ\phi)Y+(f\circ\phi)\nabla_XY)=D\phi(X(f\circ\phi)Y)+f\nabla''_{X'}Y'
$
In my head the first part of this is correct, about Leibniz rule I dont understand how to continue… Can someone please help? Please dont give me a hint since I tried a lot already and I am really tired and this is a last resort, if I have written something that doesnt make sense please explain to me why it doesnt. In particular if what I have written is correct, then shouldnt like $X(f\circ \phi)$ be a function? so then shouldnt I get $D\phi X(f\circ \phi)=X(f)$?? but of course this doesnt even make sense cause f is in M' and X is in M… Please help me…
Best Answer
$d\phi (X(f \circ \phi)Y)=X'(f) d\phi(Y)=X'(f)Y'$
as desired. where $X(f\circ \phi) =d(f\circ \phi)(X) =(df \circ d\phi)(X) =df(d\phi(X)) =df(X')=X'(f)$
You have to notice that $(df \circ d\phi)(X)$ is a function on M , while $df(X')$ is (often) seen as a function on M' The different views come from $df(d\phi(X))$