# Covariant derivative of a section along a smooth vector field

Suppose $$E$$ is a $$q$$-dimensional real vector bundle on a smooth manifold $$M$$ and $$\Gamma(E)$$ is the set of smooth sections of $$E$$ on $$M$$. A connection on the vector bundle $$E$$ is a map $$D:\Gamma(E) \to\Gamma(T^*(M)\otimes E)\tag{1}$$ which satisfies the following conditions:

• For any $$s_1,s_2\in\Gamma(E)$$, $$D(s_1+s_2)=Ds_1+Ds_2$$.
• For $$s\in\Gamma(E)$$ and any $$\alpha\in C^\infty(M)$$, $$D(\alpha s) = d\alpha\otimes s + \alpha Ds\;.$$

Suppose $$X$$ is a smooth tangent vector field on $$M$$ and $$s\in\Gamma(E)$$. Let $$D_Xs:=\langle X, Ds\rangle\;\tag{2}$$ where $$\langle\;,\rangle$$ represents the pairing between $$T(M)$$ and $$T^*(M)$$. Then $$D_Xs$$ is a section of $$E$$, which is called the covariant derivative of the section $$s$$ along $$X$$. This definition is given in Chern's Lectures on Differential Geometry.

By (1), $$Ds$$ is an element in $$\Gamma(T^*(M)\otimes E)$$, not $$\Gamma(T^*(M))$$. On the other hand, $$X\in\Gamma(T(M))$$. How should I understand the pairing in (2)?

In John Lee's Riemannian Manifolds, a connection in $$E$$ is a map $$\nabla : T(M)\times \Gamma(E)\to \Gamma(E)\tag{3}$$ written $$(X,Y)\mapsto \nabla_XY$$, satisfying

• $$C^\infty(M)$$-linear in the first component;
• $$\mathbb{R}$$-linear in the second component;
• the product rule $$\nabla_X(fY) = f\nabla_XY+(Xf)Y\;.$$

Essentially $$\nabla_XY=D_XY$$ in Chern's notation; we can show that (2) satisfies all the defining properties for (3).

Are there some reasons we would like to go to the more abstract definition in (1) instead of (3)?

• It's totally the same definition. Finding the covariant derivative of $s$ in the direction of $X$ is evaluating the $1$-form in $Ds$ on $X$. Chern does (did) his mathematics with differential forms — as do many of us — and so this approach was more natural. He wants to think of curvature as a $\text{Hom}(E,E)$-valued $2$-form, not as a repeated covariant derivative operator. But forms are beautiful to compute with. Jun 7, 2020 at 22:31
• @TedShifrin: Thanks. Could you say a bit more on Chern's approach? So instead of considering a linear map, say, $T: V^*\times V\to\mathbb{R}$, it may be more convenient to deal with the linear map $T:V^*\to V^*$? Does this have anything to do with "currying"? Also, to see the equivalency between (1) and (3), we need $\Gamma(T^*M\otimes E)=\Gamma(T^*M)\otimes\Gamma(E)$, right?
– user9464
Jun 8, 2020 at 19:14
• See my answer here for your latter question. The real question is this: Do you understand and work with differential forms as differential forms, or do you try to evaluate on vector fields immediately and avoid the forms? :D I don't really see your question with $T$; it's really about whether you .work with $V^*\otimes E$ or with $\text{Hom}(V,E)$. Currying seems to be a computer science issue, but it certainly shows up naturally in analysis and differential geometry. Jun 8, 2020 at 19:29

The pairing $$TM \times (T^* M \otimes E) \to E$$ is really just the canonical pairing $$\operatorname{tr}: TM \times T^* M \to \Bbb R$$ with the tensorial factor $$E$$ coming along (inertly) for the ride: More precisely, by definition $$\langle \,\cdot\, , \,\cdot\, \rangle$$ is the composition $$TM \times (T^* M \otimes E) \stackrel{\otimes}{\longrightarrow} TM \otimes T^* M \otimes E \stackrel{\operatorname{tr} \otimes \operatorname{id}_E}{\longrightarrow} E .$$ On decomposable elements, $$\langle X, \alpha \otimes \xi \rangle = \alpha(X) \xi .$$