# 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

## 1 Answer

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.

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