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)?

  • $\begingroup$ 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. $\endgroup$ Jun 7, 2020 at 22:31
  • $\begingroup$ @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? $\endgroup$
    – user9464
    Jun 8, 2020 at 19:14
  • $\begingroup$ 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. $\endgroup$ Jun 8, 2020 at 19:29

1 Answer 1


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, $$\langle X, \alpha \otimes \xi \rangle = \alpha(X) \xi .$$

As for comparing the definitions, only a little unwinding is required to show that the two are coincident; I cannot improve on Ted Shifrin's comment about Chern's form approach to geometry.


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