Proving some basic properties of covariant differentiation I have the following somewhat awkward definition of covariant differentiation along a curve:

Let $S \subseteq \mathbb{R}^N$ be a smoothly and isometrically embedded manifold, and $\alpha : I \to S$ be a smooth curve. The covariant derivative along the curve $\alpha$ of a vector field $V: I \to \mathbb{R}^N$, $V(t) \in T_{\alpha(t)}S$ for each $t \in I$, is the orthogonal projection of $\dot{V}$ onto $T_{\alpha(t)}S$.

I feel like it ought to be easy to show that if I have an isometry $f: S \to \tilde{S}$, then covariant differentiation and the differential $df$ ought to commute in the obvious way, namely
$$ df \left( \nabla_{\dot{\alpha}} V \right) = \nabla_{df(\dot{\alpha})} \left( df(V) \right) $$
but I spent the better part of the afternoon thinking about it without any success, at which point I resorted to taking charts and comparing coordinate expressions. That was easy but it felt like cheating. Is there a way to do this directly from the definition, exploiting the fact that $S$ and $\tilde{S}$ are embedded manifolds?
 A: Thanks for clarifying your question. Given an isometry of only the surface (which say does not extend to the ambient space) it shouldn't be immediately clear that covariant differentiation, given by your formula, commutes with the isometry.  
The reason for this is you're saying that covariant differentiation is intrinsic -- this is a key step in Gauss's Theorema Egregium.   Once you know covariant differentiation is intrinsic, then you have a clear attack at proving the Riemann curvature tensor is intrinsic, so Gauss curvature is intrinsic.  
Usually in a differential geometry course this is where you make a big deal about the fact that a Riemann manifold has a unique Levi-Civita connection and your formula for the covariant derivative is one manifestation of it. This sometimes goes under the name of the fundamental theorem of differential geometry so it's not a trivial thing. 
I think there's a variety of alternative ways of proving what you want but they all require some kind of tool such as the above argument.  I suppose you could reduce it to a linear-algebra argument if you take the next step to view your formula for the covariant derivative as a cute formula for an Ehresmann connection (this starts to relate your covariant derivative to an intrinsic language).  You would then have to ensure that your connection and the pull-back connection (by your isometry) are the same -- in effect you'd be encoding the "fundamental theorem"'s proof but in the language of Ehresmann connections. 
