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Let $\pi:E\rightarrow M$ be a vector bundle, and $D$ a connection on it. Suppose $\sigma_1,\sigma_2\in\Gamma(E)$, $p\in M$, $V\in T_pM$ such that $\sigma_1(p)=\sigma_2(p)$. Are the following two conditions equivalent?

  1. $d(\sigma_1)_p(V)=d(\sigma_2)_p(V)$
  2. $D_V\sigma_1=D_V\sigma_2$
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3 Answers

Yes: just write it out in a local basis: $\sigma_i = \sum f_{ij} E_j$ in some local choice of basis $E_j$, and $\sigma_1(p)=\sigma_2(p)$ means that $f_{1j}(p)=f_{2j}(p)$. Then $d(\sigma_1-\sigma_2)$ is defined at $p$ and represented by $d(f_{1j}-f_{2j}) \otimes E_j$, while $D(\sigma_1-\sigma_2)=\sum d(f_{1j}-f_{2j}) \otimes E_j + (f_{1j}-f_{2j}) \otimes D E_j$ so that at $p$, $D(\sigma_1-\sigma_2)(p)=\sum d(f_{1j}-f_{2j}) \otimes E_j$ represented by the same stuff.

To be careful, $d\sigma_1$ and $d\sigma_2$ are not really defined, even at $p$, because $d\sigma(p)$ is defined for a section $\sigma$ just at the points where $\sigma=0$.

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Thanks for your answer. –  user45859 Feb 3 at 7:40
I don't understand the last comment: the differential of $\sigma:M\rightarrow E$ is well-defined as $d\sigma (p): T_p(M)\rightarrow T_{\sigma (p)}(E)$. –  abx Feb 3 at 7:42
When $\sigma\in\Gamma(E)$ and $\sigma(p)\neq 0$, why can't we define $d\sigma_p$? Since $\sigma$ is a smooth map from $M$ to $E$, $d\sigma_p$ should be a linear map from $T_pM$ to $T_{\sigma(p)}E$. –  user45859 Feb 3 at 7:43
Sorry, I should have indicated more clearly: if $\sigma(p)=0$ we can define $d\sigma(p) \in T^*_p M \otimes E_p$, rather than in $T^*_p M \otimes T_{\sigma(p)} E$. –  Ben McKay Feb 3 at 8:39
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The two ways are connected by $D_V\sigma = (K\circ d\sigma)(V)$ where $K:TE\to E$ is the connector for the covariant derivative. See 19.12 of

  • Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)
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Let $D:V(M)\times \Gamma(E)\rightarrow\Gamma(E)$ be the connection on the vector bundle $E$. Now we consider the local representation of $D$.

Let $V|_U=\sum V_ie_i\in V(M)$, $\sigma|_U=\sum \sigma_is_i\in\Gamma(E)$ and $\{e_i\}$, $\{s_i\}$ be local vector basis on $U$. Then we have $$D_V\sigma|_U$$ $$=\sum_k(\sum_i\sum_jV_i\sigma_j\Gamma_{ij}^k+V(\sigma_k))s_k$$

First, from $\sigma_1=\sigma_2$, we have $\sigma_{1k}=\sigma_{2k}$.

Second, $$d(\sigma_1)(V)=d(\sigma_2)(V)$$ $$\Longleftrightarrow V(\sigma_{1k})=V(\sigma_{2k})$$ $$\Longleftrightarrow D_V(\sigma_1)=D_V(\sigma_2)$$

So they are equal.

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