# 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?

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Why you think coordinate representation is cheating? –  Fabian Feb 28 '11 at 19:34
@Fabian: Well, the particular approach I used felt especially like cheating: pick a chart $\phi$ on $U \subseteq S$, and use the chart $\phi \circ f^{-1}$ on $f(U) \subseteq \tilde{S}$. Then the coordinate expressions for $f$ and $df$ become trivial and it's just a simple matter of noting that the expressions on the left and the expressions on the right have the same form. Then because the chart was arbitrary, this proves the identity abstractly... and throughout this process I have not exploited the fact that the manifolds are embedded isometrically in Euclidean space. –  Zhen Lin Feb 28 '11 at 19:40
I don't see that as an awkward definition at all. Composition with a linear function commutes with differentiation by the chain rule. So the only issue is showing that conjugation by an isometry converts projection to one tangent space into projection into the other but this is tautological. –  Ryan Budney Feb 28 '11 at 20:24
@Ryan: Hmmm, that seems to assume that there is an isometry of the whole Euclidean space extending the isometry which is defined between the two manifolds. Certainly I can see that it's obvious in that scenario. But consider, for example, the case where $S$ is an open disc in the plane, and $\tilde{S}$ is a patch of a cylinder — what happens here? –  Zhen Lin Feb 28 '11 at 22:07