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.

  • $\begingroup$ Could you post an image of Lee's definition/proposition 4.38? $\endgroup$
    – JoshDH
    Jul 2 '19 at 13:28
  • $\begingroup$ @JoshDH, I have typed Proposition 4.38 part a) exactly as it appear in the book (except that in the book they include the case in which the manifold has boundary). Also, which definition would you like to see? $\endgroup$
    – Lucas Vaca
    Jul 3 '19 at 4:35
  • $\begingroup$ Lee's definition of pullback of a connection, one's I've seen have your question as an immediate consequence! $\endgroup$
    – JoshDH
    Jul 3 '19 at 7:59
  • $\begingroup$ Pullback of a connection are defined in Intro to Riemannian Manifolds as: $(\phi^* \tilde\nabla)_XY = (\phi^{-1})_*(\tilde\nabla_{\phi_*X}(\phi_*Y))$. @JoshDH can you explain why? $\endgroup$
    – Lucas Vaca
    Jul 4 '19 at 11:58
  • $\begingroup$ I'm writing an answer, now, I should post it tomorrow. $\endgroup$
    – JoshDH
    Jul 4 '19 at 21:47

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).


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!


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