I'm looking for a resource regarding the chain rule for covariant derivatives.

The closest version I've seen is the one regarding differentials of maps: If F and G are differential maps, then $d(G \circ F)_p=dG_{F(p)} \circ dF_{p}$

(From deCarmo's Differential Geometry)

I'm not positive this extends to covariant derivatives, although I feel it should, as the differential applied to a vector $v$ and the covariant derivative in the direction $v$ differ only by a projection onto the tangent plane. Could someone please point me towards the right resource? Thanks so much!


Edit: Regarding the comments below by Ryan and Deane:

I'm thinking about a covariant derivative (using the standard Levi-Civita connection) on a surface $S$ embedded in $\mathbb{R}_3$. Let $dN$ be the shape operator (differential of the Gauss map) and let $u \in T_p(S)$. Say $\alpha(t)$ is a curve on the surface with $\alpha'(0) = u$. Let $v(t)$ be a vector field defined on $\alpha(t)$. Then, the equation I am interested in would be: $\nabla_u (dN(v(t))) = (\nabla_u dN) (\nabla_u v(t))$. That is, the R.H.S is the covariant derivative (in the tensor sense) of $dN$ applied to the tangent vector $\nabla_u v$. I would like to know whether equation is true, or if there is a similar type of rule. Thanks again for your time!


closed as not a real question by Ryan Budney, Willie Wong, Deane Yang, Will Jagy, Mariano Suárez-Álvarez Jun 18 '11 at 5:48

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    $\begingroup$ Covariant derivatives usually apply to objects like sections of fibre bundles when you have some kind of connection. So what kind of compositional operation do you want to consider? $\endgroup$ – Ryan Budney Jun 17 '11 at 21:30
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    $\begingroup$ Agree with Ryan. Please cite a specific formula you think might be true. $\endgroup$ – Deane Yang Jun 17 '11 at 22:25
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    $\begingroup$ I'm not sure why this was closed as "not a real question". There actually is a very useful chain rule for the covariant derivative and it has to do with pullback bundles. If $\pi : E \to N$ is a vector bundle (e.g. $TN$), $\nabla^E$ is a covariant derivative on $E$, $\sigma$ is a section of $\pi$, and $\phi : M \to N$ is a map, then $\nabla^{\phi^* E} (\sigma \circ \phi) = (\nabla^E \sigma \circ \phi) \cdot T\phi$. This can be seen trivially using the "connection map" formulation $\nabla^E \sigma := \kappa \cdot T\sigma$. If someone re-opens this question, I'd gladly post a full answer. $\endgroup$ – Victor Dods Dec 16 '11 at 1:18