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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...

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$d\phi (X(f \circ \phi)Y)=X'(f) d\phi(Y)=X'(f)Y'$

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 $df(d\phi(X))$

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  • $\begingroup$ can you just explain to me this please. $X(f\circ\phi)$ is a function so when you pull it out of $D\phi$ shouldnt you compose it with $\phi^{-1}$? $\endgroup$
    – sifsa
    May 22 '16 at 15:37
  • $\begingroup$ @sifsa Your point is another viewpoint. (Inside the $D\phi$) $X(f\circ\phi)$ "equals" (Outside the $D\phi$) $X'(f\circ\phi\circ{\phi}^{-1})$ "equals" (Outside the $D\phi$) $X'(f)$. Where I change X to X' is because $f\circ\phi$ is a function on M but $f\circ\phi\circ{\phi}^{-1}$ is a function on M' Actually X and X' have almost the same effect on functions (If you identify the functions on M and M' by $\phi). The difference is that the function followed by X should be a function on M, and similar for the X' case. $\endgroup$
    – user341158
    May 23 '16 at 6:56
  • $\begingroup$ Notice that $X_p(f \circ \phi) \in \mathbb{R}$ for $p \in M$. So because the Differential $D\phi$ is $\mathbb{R}$-linear, you can just pull it out. Consider $q = \phi(p)$. $D\phi_p(X_p(f \circ \phi)Y_p) = X_p(f \circ \phi)D\phi_p(Y_p) \in T_{F(p)}M'$. In your specific setting, $\phi$ is an isometry, thus a bijection, which is why you can write the former tangent vector as a vector field in $M'$: $(X(f \circ \phi) \circ \phi^{-1})D\phi(Y)$ because $((X(f \circ \phi) \circ \phi^{-1})D\phi(Y))_q = (X(f \circ \phi) \circ \phi^{-1}(q))(D\phi(Y))_q = X_p(f \circ \phi)D\phi_p(Y_p)$. $\endgroup$
    – oac
    May 8 '21 at 7:30

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