---- Einstein convention is in effect. - $B$ is [[smooth manifold]] of dimension $n$ - $E \xrightarrow{\pi} B$ is a rank-$m$ smooth [[vector bundle]] with typical fiber $V$ > [!definition] Definition. ([[covariant derivative on a vector bundle]]) > A **covariant (or invariant) derivative on $E$** is an $\mathbb{R}$-[[linear map|linear]] map $\nabla^{E}:\Gamma(E)\to \Gamma(T^{*}B \otimes E)$ satisfying the [[derivation|Leibniz rule]] $\nabla^{E}(fs)=df \otimes s+ f \nabla^{E}s$ for any [[section]] $s \in \Gamma(E)$ and function $f \in C^{\infty}(B)$. > Here, $\Gamma(T^{*}B \otimes E)$ denotes the space of sections on the [[tensor product of vector bundles|tensor product bundle]] $T^{*}B \otimes E$, i.e., the space $\Omega_{B}^{1}(E)$ of $E$[[differential form with values in a vector bundle|-valued differential forms]] (likewise $\Gamma(E)=\Omega_{B}^{0}(E)$). With this notation, $\nabla^{E}:\Omega_{B}^{0}(E) \to \Omega_{B}^{1}(E)$, suggestive of the [[exterior derivative]] [[differential of a 0-form|of a]] $0$-form.[^2] > The data specified by $\nabla^{E}:\Omega_{B}^{0}(E) \to \Omega_{B}^{1}(E)$ [[three views on connections|is equivalent to]] that of a [[connection on a vector bundle|connection]]. Sometimes the notation $s'=\nabla^{E}s$ is [[derivative|used]]. > [!basicexample] The most important example. > Let $A$ be a [[connection on a vector bundle|connection]] on $E$, viewed locally in the usual way as a matrix $(A^{i}_{j})_{i,j=1}^{m}$ of $1$-forms. Put $\nabla^{E}=d_{A}$ defined in a [[vector bundle|local trivialization]] $U$ $d_{A}s |_{U} = (ds + As) |_{U},$ where $ds$ should be viewed as taking the [[exterior derivative]][^3] of $s$ — identified with a smooth function $U \subset B \to \mathbb{R}^{m}$ — componentwise. Explicitly, the output is a [[differential form with values in a vector bundle|vector of differential 1-forms]] $\begin{bmatrix} ds ^{1} + A^{1}_{j} s ^{j} \\ \vdots \\ ds ^{m} + A^{m}_{j} s ^{j} \end{bmatrix}=(ds ^{i} + A^{i}_{j} s ^{j})_{i=1}^{m}.$ > Explicitly in terms of the [[vector bundle|local frame of sections]] $(e_{i})$ determined by $\Phi$, $d_{A}(s ^{i}e_{i})=(ds ^{i}+A ^{i}_{j} s ^{j})e_{i}.$ > Should $U=(x^{1},\dots,x^{n})$ furthermore be a [[coordinate chart|coordinate neighborhood]], even more explicitly we may write $d_{A}s |_{U}=\begin{bmatrix} \left( \frac{ \partial s ^{1} }{ \partial x^{k} } + \Gamma_{jk} ^{1} s ^{j} \right) \ dx ^{k} \\ \vdots \\ \left( \frac{ \partial s ^{m} }{ \partial x^{k} } + \Gamma_{jk} ^{m} s ^{j} \right) \ dx ^{k} \end{bmatrix}=\left( (\frac{ \partial s ^{i} }{ \partial x^{k} } + \Gamma^{i}_{jk} s ^{j}) \ dx^{k} \right)_{i=1}^{m},$ where $\frac{ \partial s ^{i} }{ \partial x^{k} }$ and $\Gamma_{jk}^{i}$ are smooth functions $U \subset B \to \mathbb{R}$. Explicitly in terms of the [[vector bundle|local frame of sections]] $(e_{i})$ determined by $\Phi$, $d_{A}(s ^{i} e_{i})=\big( (\frac{ \partial s ^{i} }{ \partial x^{k} } + \Gamma^{i}_{jk}s ^{j} )dx ^{k} \big)e_{i}.$ > (Is $d_{A}$ invariantly defined? Yes.[^1]) > > > > Conversely, every covariant derivative arises as $d_{A}$ for some connection $A$! [[three views on connections|See here.]] ^basic-example [^1]: Here is the verification. Let $s \in \Gamma(E)$. In the local trivialization $(U, \Phi)$, $s$ may be viewed as a vector-valued function $s:U \to \mathbb{R}^{m}$. Suppose $(U', \Phi')$ is another [[vector bundle|local trivialization]]. Let $\psi:U \cap U' \to \text{GL}(n, \mathbb{R})$ be the transition function from $\Phi'$ to $\Phi$. As a matter of definition, $\textcolor{Thistle}{s=\psi s'}$. Recall also the [[transformation law]] for [[connection on a vector bundle|connections]] $\textcolor{Skyblue}{A=\psi A' \psi ^{-1} - (d \psi) \psi ^{-1}}$. Now we get $\begin{align} d_{A} s& = ds + As \\ &= d (\textcolor{Thistle}{\psi s '}) + (\textcolor{Skyblue}{\psi A' \psi ^{-1} - (d \psi) \psi ^{-1}})\textcolor{Thistle}{\psi s '} \\ &= \psi d s' + (d \psi) s'+ \psi A' s' - (d \psi) s' \\ &= \psi (ds' + A's') \\ &= \psi (d_{A'} s') \end{align}$which is the correct transformation law $E$-valued $1$-forms $\omega \in \Gamma(T^{*}B \otimes E)$. ^47f224 > [!basicproperties] > 1. $\nabla^{E}$ is a **local operator**, in the sense that if $s_{1},s_{2}$ are two [[section|sections]] agreeing on an open set $U \subset B$, then $\nabla^{E}s_{1} |_{U}=\nabla^{E}s_{2} |_{U}$. ^properties > [!proof] Proof of Basic Properties. > 1. We show agreement for all $b \in U$. Wrap a small neighborhood $U_{0} \ni b$ such that $\overline{U_{0}} \subset U$. Let $\alpha \in C^{\infty}(B)$ be a cutoff function, satisfying $0 \leq \alpha \leq 1$, $\alpha |_{U_{0}}\equiv 1$, $\alpha |_{B\setminus U} \equiv 0$. Then $\alpha(s_{1}-s_{2})\equiv 0$ (this is multiplication). Evidently $\nabla^{E}\big( \alpha(s_{1} - s_{2}) \big)\equiv 0$ (it is the covariant derivative of the zero section!), so applying the product rule we get $\nabla^{E}\big( \alpha(s_{1}-s_{2}) \big)=d\alpha \otimes (s_{1}-s_{2}) + \alpha \nabla^{E} (s_{1}- s_{2}) .$ Around $b$, $d\alpha=0$ because $\alpha$ is constant. But $\alpha$ is nonzero around $b$, hence it must be the case that $\nabla^{E}(s_{1} - s_{2})\equiv 0$ around $b$; by linearity the desired equality follows. ^proof [^2]: Notably, recall that the [[exterior derivative]] of a $0$-form in particular has relation to the [[directional derivative|directional derivative]]. [^3]: In this case the definitions of [[exterior derivative]] and [[differential of a smooth map between smooth manifolds|pushforward]] align, so ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```