Naturality of the pullback connection

Question

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?)

Answer 1

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 $\left(Xg(Y,Z)\right)(p)$

Answer 2

I thought this was a nice exercise so I thought I would give more details, if that is okay. I will try to work entirely with tensor fields. Another way to state the naturality is to say that under the isometry $\phi:M\to\tilde M$, we have that $\nabla=\phi^*\tilde\nabla$. There is no other Levi-Civita connection on $M$. The approach is thus to prove that the right hand side satisfies the properties of a connection on $M$, which is torsion-free and compatible with the metric $g$. Uniqueness of the connection on $M$ satisfying these properties then gives $\nabla=\phi^*\tilde\nabla$.

• First note that $\phi^*\tilde\nabla$ is a map $\Gamma(TM)\times\Gamma(TM)\to \Gamma(TM)$. Then we want to check linearity in the first entry and the characteristic Leibniz rule in the second. Recall that $\phi_*(fX)=(f\circ\phi^{-1})\ \phi_*X$. Let for notation's sake $\psi:=\phi^{-1}:N\to M$, $D:=\phi^*\tilde\nabla$ and the application of a derivation $X$ on a function $h$ be $X.h=Xh$. Then for $f\in C^{\infty}(M)$

\begin{align} D_{fX}Y &= \psi_*\tilde\nabla_{\phi_*(fX)}\phi_*Y = \psi_*\tilde\nabla_{f\circ\psi\ \phi_*X}\phi_*Y \\ &= \psi_*(f\circ\psi\ \tilde\nabla_{\phi_*X}\phi_*Y) = (f\circ\psi\circ\phi)\ \psi_*\tilde\nabla_{\phi_*X}\phi_*Y \\ &= fD_X Y \end{align}

and recalling for $g\in C^{\infty}(N)$, $(\phi_*X).g = (X.(f\circ\phi))\circ\psi$ as functions on $N$,

\begin{align} D_{X}(fY) &= \psi_*\tilde\nabla_{\phi_*X}\phi_*(fY) = \psi_*\tilde\nabla_{\phi_*X}(f\circ\psi)\phi_*Y \\ &= \psi_*\left( (\phi_*X).(f\circ\psi)\phi_*Y + f\circ\psi\ \tilde\nabla_{\phi_*X}\phi_*Y \right) \\ &= \psi_*((X.(f\circ\psi\circ\phi))\circ\psi)\phi_*Y + f\psi_*\tilde\nabla_{\phi_*X}\phi_*Y \\ &= \psi_*(\underbrace{(X.f)\circ\psi)}_{\in C^{\infty}(N)}\phi_*Y + fD_{X}Y \\ &=(X.f) Y+ fD_{X}Y \end{align}

• Next we show that it is torsion free, using the fact that $[\phi_* X,\phi_* Y]=\phi_*[X,Y]$. We have from torsion-freeness of $\tilde\nabla$ that $\tilde\nabla_{\phi_* X}\phi_* Y - \tilde\nabla_{\phi_* Y}\phi_* X = [\phi_* X,\phi_* Y] = \phi_*[X,Y]$, whereby $D_XY-D_YX = \psi_*\phi_*[X,Y]=[X,Y]$ since pushforwards are isomorphisms.
• Lastly we use compatibility of $\tilde \nabla$ with $\tilde g$ to get it for $D$ with $g$. Recall $g(X,Y)=g_{Id}(X,Y)=\tilde g_{\phi}(\phi_*X,\phi_*Y)$ since $\phi:M\to N$ is an isometry. Let $\Gamma(TM)\ni Z=\psi_*\xi$ for $\xi\in\Gamma(TN)$. Then since the expressions need to be equal as functions on $M$: \begin{align} g(D_XY,Z) &= g(\psi_*\tilde\nabla_{\phi_*X}\phi_*Y,\psi_*\xi) = \tilde g_\phi(\tilde\nabla_{\phi_*X}\phi_*Y,\xi) = \tilde g_\phi(\tilde\nabla_{\phi_*X}\phi_*Y,\phi_* Z) \\ &= ((\phi_*X).\tilde g_{Id}(\phi_*Y,\phi_*Z))\circ\psi - \tilde g_\phi (\phi_*Y,\tilde\nabla_{\phi_*X}\phi_*Z) \\ &= X.\tilde g_\phi(\phi_* Y,\phi_* Z) - g(Y,\psi_*\tilde\nabla_{\phi_*X}\phi_*Z) \\ &= X.g(Y,Z) - g(Y,D_XZ). \end{align}

Thus $\nabla=\phi^*\tilde\nabla$. As a bonus we can nicely see from this that geodesics get mapped by an isometry $\phi:M\to N$. If $c:I\to M$ is a geodesic then the corresponding curve $\gamma:=\phi\circ c:I\to N$ is as well. (This curve is the flow $\tilde\theta_t(n)=\gamma(t)$ with $\gamma(0)=n$ from the vector field $\phi_*c'(t)$ with $c(0)=\psi(n)$.) $$ \psi_*\tilde\nabla_{\gamma'}\gamma' = \psi_*\tilde\nabla_{\phi_*c'}\phi_*c' = D_{c'}c'=\nabla_{c'}c'=0 $$ so $\tilde\nabla_{\gamma'}\gamma'=0$ since $\psi_*$ is an isomorphism. (Note $\gamma'(t) = \frac{d}{dt}(\phi\circ c(t))=T_{c(t)}\phi c'(t)=(\phi_*c')(t)$ where $T_{c(t)}\phi:T_{c(t)}M\to T_{\gamma(t)}N$ is the tangent map or differential.)