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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.

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

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

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