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I'm completely stuck proving the naturality of the pullback connection.

The strategy suggested is a follows:

We let $\phi: (M,g) \to (\tilde{M}, \tilde{g})$ be an isometry, with connections $\nabla$ and $\tilde{\nabla}$ respectively.

Define the "pullback connection": $(\phi^* \tilde{\nabla})_X Y=\phi_*^{-1} ( \tilde{\nabla}_{\phi_* X}(\phi_* Y)). $ Now show that it is symmetric and compatible with the metric, and hence agrees with the unique Riemannian connection on $M.$

I am completely stuck with showing metric compatibility.

According to the definition of a symmetric connection, I need to show $(\phi^* \tilde{\nabla})_X \langle Y, Z \rangle= \phi_*^{-1} ( \tilde{\nabla}_{\phi_* X}(\phi_* \langle Y,Z \rangle)) = \langle\phi_*^{-1} (\tilde{\nabla}_{\phi_* X}(\phi_* Y)),Z \rangle + \langle Y,\phi_*^{-1} (\tilde{\nabla}_{\phi_* X}(\phi_* Z)) \rangle. $

I am willing to believe that this is just a matter of unwinding definitions. However, I don't even understand how $\phi_* \langle Y, Z \rangle$ should even be defined.

Could anyone help get me started by perhaps doing the first few terms of the computation to show me how it should go? (or else guide me to somewhere in the literature where this is done?)

JonHerman
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user142700
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    The left-hand side of the metric compatibility equation is just the vector field $X$ applied to the function $\left<Y,Z\right>$; it doesn't depend on the connection at all. (That's how the "covariant derivative" of a $0$-tensor is defined: $\nabla _X f = Xf$ by definition.) – Jack Lee Oct 02 '14 at 22:23
  • The pullback metric $\phi_\langle X, Y \rangle $ is defined by pushing forward vectors $ X$ and $ Y $. Consider curves $\gamma_X $ and $\gamma_Y $ such that $\gamma_X (0) =\gamma_Y(0) =p $, $\gamma_X'(0)=X $, and $\gamma_Y'(0) = Y $. The curves push forward to give new curves $\phi \circ \gamma_X $ and $\phi \circ \gamma_Y $. The pull back metric gives the angle between these two new curves. This is equal to $\langle\phi_ X, \phi_* Y\rangle_{\phi (p)} $. – Matthew McGonagle Oct 03 '14 at 06:34

1 Answers1

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You need to show that $Xg(Y,Z)=g((\phi^\ast\tilde\nabla)_XY,Z)+g(Y,(\phi^\ast\tilde\nabla)_XZ)$.

Try showing that for $p\in M$, $$g_p(((\phi^\ast\tilde\nabla)_XY)_p,Z_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\tilde\nabla_{\phi_\ast X}\phi_\ast Y)_{\phi(p)},(\phi_\ast Z))_{\phi(p)}\right)$$ and similarily $$g_p(Y_p,((\phi^\ast\tilde\nabla)_XZ)_p)=((\phi^{-1})^\ast g)_{\phi(p)}\left((\phi_\ast Y)_{\phi(p)},(\tilde\nabla_{\phi_\ast X}\phi_\ast Z)_{\phi(p)}\right)$$

Add these terms and use the fact that $(\phi^{-1})^\ast g=\tilde g$ and the metric compatibility of $\tilde\nabla$.

You should get that there sum at $p$ is $$ \left((\phi_\ast X)(\tilde g(\phi_\ast Y,\phi_\ast Z)\right)(\phi(p)).$$ Then try showing that this is equal to $\left(Xg(Y,Z)\right)(p)$

JonHerman
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    Thanks! I did it (and I'd never have figured it out without you). – user142700 Oct 03 '14 at 08:04
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    @JonHerman It seems to me one could prove your second equation by using the definition of $\phi^* \tilde\nabla$ with pure linear algebra; in particular, that $\phi_=d\phi_p : T_p M \to T_{\phi(p)} \tilde M$ is an isometry in the linear algebraic sense, so that $g_p(\phi_^{-1}(\tilde\nabla_{\phi_X}(\phi_Y))p,Z_p) = \tilde g{\phi(p)}((\tilde\nabla_{\phi_X}(\phi_Y))p,\phi*Z_p)$. Am I correct? – D Ford Jun 15 '17 at 23:08
  • looks good to me. – JonHerman Jun 16 '17 at 14:28