# Proving that Levi-Civita connection is preserved by isometries

I am trying to prove that given two Riemannian submanifolds $S,S'$ with Levi-Civita connections $\nabla , \nabla'$ and an isometry $f$, then $$Df(\nabla_XY)=\nabla'_{X'}Y'$$ where, $X',Y'=Df(X),Df(Y)$. The argument is that if $\nabla''_XY=D f^{-1}(\nabla '_{X'}Y')$ torsion free and metric then it is the Levi-Civita connection on $S$. I was trying to prove this is metric and this is what I get: $$g(\nabla''_XY,Z)+g(Y,\nabla''_XZ)=g(D f^{-1}(\nabla '_{X'}Y'),Z)+g(Y,D f^{-1}(\nabla '_{X'}Z'))$$ $$=g'(\nabla '_{X'}Y',Z)+g'(Y',\nabla '_{X'}Z')$$ as $f$ is isometry, and as $\nabla'$ is Levi-Civita, $$=X'g'(Y',Z')=X'g(Y,Z)$$ which should have been $Xg(Y,Z)$... Can someone explain what I am doing wrong? also is there a more efficient way to show what I want to prove instead of checking that $\nabla''$ is metric and torsion free (and also that it is a connection)?

This is not a duplicate I showed my working and I want to know why my working does not work.

If $f:(S,g)\to (S',g')$ is an isometry, then define $\nabla_{X'}Y':=df\ \nabla_XY$

Show that this is LC-connection :

(1) Compatibility condition : First show that $$X'(Y',Z')=X(Y,Z)$$

Proof : If $\frac{d}{dt}p(t)=X,\ p(0)=p$ then $$df_p X(df_p Y, df_p Z) =\frac{d}{dt} (df Y, df Z)_{f(p(t))} = \frac{d}{dt} (Y,Z)_{p(t)}$$ since $f$ is an isometry And $\frac{d}{dt} (Y,Z)_{p(t)}= X(Y,Z)$

So $$(\nabla_{X'}Y',Z')+(Y',\nabla_{X'}Z')=f^\ast g'( \nabla_XY,Z) + f^\ast g' (Y,\nabla_XZ) = X(Y,Z) =X'(Y',Z')$$

(2) Symmetry condition : $$\nabla_{X'}Y' -\nabla_{Y'}X'=df (\nabla_XY-\nabla_YX)=df[X,Y]=[X',Y']$$

• okay sorry but can you have a look at what I wrote in the question and tell me why it is wrong? Mar 30, 2016 at 16:14
• oh but I wrote let me copy paste for you Mar 30, 2016 at 16:17
• X′,Y′=Df(X),Df(Y) Mar 30, 2016 at 16:17
• ∇′′XY=Df−1(∇′X′Y′) Mar 30, 2016 at 16:17
• $\nabla''$ is a second connection which I define to be what I wrote Mar 30, 2016 at 16:18