Property of pullback connections

Question

I'm trying to prove Proposition 4.38 in Lee's Intro to Riemannian Manifolds which is left for the reader:

Suppose $M$ and $\tilde M$ are smooth manifolds without boundary, and $\phi: M\to \tilde M$ is a diffeomorphism. Let $\tilde \nabla$ be a connection in $T\tilde M$ and let $\nabla= \phi^*\tilde \nabla$ be the pullback connection in $TM$. Suppose $\gamma: I\to M$ is a smooth curve.

(a) $\phi$ takes covariant derivatives along curves to covariant derivatives along curves: if $V$ is a smooth vector field along $\gamma$, then $$d\phi \circ D_t V = \tilde D_t(d\phi \circ V),$$ Where $D_t$ is covariant differentiation along $\gamma$ with respect to $\nabla$, and $\tilde D_t$ is covariant differentiation along $\phi\circ \gamma$ with respect to $\tilde \nabla$.

I've tried expanding both side using the component expression: $$D_t V = \dot V^i(t)\partial_i + V^j\nabla_{\gamma'(t)} \partial_j$$ But the expression grew really fast and seem to be taking me nowhere. How can I prove this result? Any help is appreciated.

Answer 1

I would try to show $\tilde D_t(\tilde V)=d\phi\circ(D_t(d\phi^{-1}\circ \tilde V))$ for all vectorfields $\tilde V$ along $\phi\circ\gamma$ by showing that the operator on the right hand side fullfills the three properties by which $\tilde D_t$ is uniquely determined $(\mathbb R$-linearity, product rule, agreement with $\tilde\nabla_{{(\phi\circ\gamma )}\dot{}}$ on extendible vectorfields).

Answer 2

If we treat a connection as an exterior covariant derivative, we have a map $$\nabla: \Gamma(TM)\to \Gamma(TM)\otimes\Gamma(T^*M),$$ that is to say $\nabla Y$ is a $TM$ valued 1-form.

Examining the definition of the Liebniz rule, we see that $\nabla = \mathrm{d}+A$, where $\mathrm{d}$ is the natural extension of the exterior derivative to sections of the tangent bundle and $A$ is the connection matrix (matrix with entries that are differential forms).

So, given a connection $\tilde \nabla$ on $\tilde M$, written as $\tilde \nabla= \mathrm{d}_{\tilde M} + A_{\tilde M}$, we can calculate the covariant derivative of a vectorfield pushed forward by the diffeomorphism $\phi: M \to \tilde M$.

Let $Y \in \Gamma(TM)$, its pushfoward onto $\tilde M$ is defined as $$ \tilde Y(x) =\phi_* Y=\mathrm{d} _M\phi|_{\phi^{-1}(x)}(Y(\phi^{-1}(x))).$$ Calculating the exterior derivative, we have

$$ \mathrm{d}_{\tilde M} \tilde Y = \mathrm{d}_M \phi \circ \mathrm{d}_M Y \circ \mathrm{d}_{\tilde M}\phi^{-1}.$$ (This is effectively the definition of the pullback of a vector field valued 1-form: reading from right to left we pushforward a vector along $\phi^{-1}$, act on it by the vector valued 1-form $\mathrm{d}_M Y$ and then pushforward the resulting vector along $\phi$).

Finally, we calculate \begin{align} \tilde \nabla_{\phi_* X}\phi_* Y &= \mathrm{d}_{\tilde M} \phi_* Y (\phi_* X)+ A_{\tilde M} \phi_* Y (\phi_* X)\\ &=\mathrm{d}_M \phi \circ \mathrm{d}_M Y \circ \mathrm{d}_{\tilde M}\phi^{-1} (\mathrm{d}_M \phi (X))+A_{\tilde M} \phi_* Y (\phi_* X)\\ &= \mathrm{d}_M\phi (\mathrm{d}_M Y (X) +\tilde A_MY(\phi_* X) ) \\ &=\mathrm{d}_M \phi(\mathrm{d}_M Y (X) +A_MY( X) ) \\ &= \mathrm{d}_M \phi(\nabla_XY).\end{align}

Where by $\tilde A_M$ I mean the 1-form on $T^* \tilde M$ that takes values in $TM$ (i.e. transformation on the 'matrix aspect' but not the 1-form aspect).

The manipulation may be easier to see treating the connection matrix instead as a section of $Hom(TM,TM)\otimes T^*M$, but what I've written should be coherent!

Answer 3

I will make use of property (iii) of the covariant derivative along a curve $\gamma$:

If $V\in\mathfrak X(\gamma)$ is extendible, then for every extension $\tilde V$ of $V$ we have $D_t V(t)=\nabla_{\gamma'(t)}\tilde V$.

Let $t\in I$. Choose a local frame $E_i$ around $p=\gamma(t)$. Locally we can write $$ V=f^i \gamma^*E_i, $$ with $f^i$ smooth functions. Using property (iii) (and also property (ii), i.e., the product rule) we obtain $$ d\phi D_t(V)=d\phi(\dot f^i \gamma^*E_i+f^i D_t(\gamma^* E_i)=\dot f^i d\phi(\gamma^*E_i))+f^i d\phi(D_t(\gamma^* E_i)). $$ We also have $$ \tilde D_t(d\phi\circ V)=\tilde D_t(f^i d\phi(\gamma^* E_i))=\dot f^i d\phi(\gamma^*E_i)+f^i\tilde D_i(d\phi(\gamma^* E_i)). $$ So we are done if we show that $$ d\phi(D_t(\gamma^* E_i))=\tilde D_t(d\phi(\gamma^*E_i)). $$

Since $\gamma^* E_i$ is an extendible vector field along $\gamma$, the above equality will follow from the following

Lemma. If $V\in\mathfrak X(\gamma)$ is extendible, then $d\phi\circ D_t V=\tilde D_t(d\phi\circ V)$.

Proof. Let $t\in I$. By property (iii) of the covariant derivative we have $$ (D_t V)(t)=\nabla_{\gamma'(t)}\tilde V. $$ Since $\nabla=\phi^*\tilde\nabla$, we find that $$ \nabla_{\gamma'(t)}\tilde V=d\phi^{-1}\tilde\triangledown_{d\phi_p\gamma'(t)}\phi_*\tilde V=d\phi^{-1}\tilde\triangledown_{(\phi\circ\gamma)'(t)}\phi_*\tilde V, $$ and hence $d\phi\circ D_tV=\tilde\nabla_{(\phi\circ\gamma)'(t)}\phi_*\tilde V$. Note also that $$ (\phi\gamma)^*\phi_*\tilde V=d\phi\tilde V\phi^{-1}\phi\gamma=\gamma^*d\phi\tilde V, $$ so $\phi_*\tilde V$ extends $T\phi V$. Using property (iii) again we find that $\tilde D_t(d\phi\circ V)=\tilde\nabla_{(\phi\circ\gamma)'(t)}\phi_*\tilde V$, and therefore $d\phi\circ D_t V=\tilde D_t(d\phi\circ V)$.