Let $E\\to M$ be a smooth complex vector bundle over a smooth compact manifold $M$. Denote $\\Omega^\\cdot(M;E):=\\Gamma(\\Lambda^\\cdot(T^\\ast M)\\otimes E)$ be the space of smooth sections of the tensor product vector bundle $\\Lambda^\\cdot(T^\\ast M)\\otimes E$.<\/p>\n
A connection on $E$ is an extension of exterior differential operator $\\rd$ to include the coefficient $E$.
\n\\begin{defn}
\n A \\emph{connection} $\\nabla^E$ on $E$ is a $C^\\infty(M)$-linear operator from $\\Gamma(E)$ to $\\Omega^1(M;E)$ such that for any $f\\in C^\\infty(M)$ and $s\\in\\Gamma(E)$, satisfy the \\emph{Leibniz rule}
\n \\[
\n \\nabla^E(fs)=(\\rd f)\\otimes s+f\\nabla^Es.
\n \\]
\n\\end{defn}
\n\\begin{rem}
\n \\begin{itemize}
\n \\item By the \\emph{partitions of unity}, we can construct may connections on a vector bundle.
\n \\item Given a vector field $X\\in\\Gamma(TM)$, we can define the \\emph{direction derivative} of $s$ among $X$ as a map $\\nabla^E_X\\mathpunct{:}\\Gamma(E)\\to\\Gamma(E)$ by
\n \\[
\n \\nabla_X^E s=< X,\\nabla^E s >=i_X(\\nabla^E s).
\n \\]
\n \\item We can extended a connection $\\nabla^E$ to be a map between $\\Omega^\\cdot(M;E)$ and $\\Omega^{\\cdot+1}(M;E)$ such that for any $\\omega\\in\\Omega^\\cdot(M)=\\Gamma(\\Lambda^\\cdot( T^\\ast M))$ and $s\\in\\Omega^\\cdot(M;E)$, we have
\n \\[
\n \\nabla^E\\mathpunct{:}\\omega\\otimes s\\mapsto(\\rd\\omega)\\otimes s+(-1)^{|\\omega|}\\omega\\wedge\\nabla^Es.
\n \\]
\n \\end{itemize}
\n\\end{rem}
\nIn contrast to the exterior differential operator $\\rd$, the connection $\\nabla^E$ is not satisfied $(\\nabla^E)^2=0$ in general, which induce the concept of curvature.
\n\\begin{defn}
\n The \\emph{curvature} $R^E$ of a connection $\\nabla^E$ is defined by
\n \\[
\n R^E=\\nabla^E\\circ\\nabla^E\\mathpunct{:}
\n \\Omega^\\cdot(M;E)\\to\\Omega^{\\cdot+2}(M;E),
\n \\]
\n for brevity, we also write $R^E=(\\nabla^E)^2$.
\n\\end{defn}
\nBefore we go to the properties of curvature, let us first give a lemma of $C^\\infty$-linear maps between two sections of vector bundle.
\n\\begin{lem}
\n Suppose $M$ is a smooth manifold, $E$, $F$ are two vector bundle on $M$. Let $\\Gamma(E)$, $\\Gamma(F)$ be the infinity dimensional vector spaces of smooth sections of $E$ and $F$, respectively. Let $A\\mathpunct{:}\\Gamma(E)\\to \\Gamma(F)$ be a linear map, if for any $f\\in C^\\infty(M)$ and $s\\in\\Gamma(E)$, we have $A(fs)=fA(s)$, then prove that $A\\in\\Gamma(\\mathrm{Hom}(E,F))$.
\n\\end{lem}
\n\\begin{proof}
\n Firstly, we can proof that $A$ is a local operator. In fact, since $A$ is linear, thus we only need to show that if for some section $s\\in\\Gamma(E)$, which vanish at a neighborhood $U$ of $M$ (i.e., for any $p\\in U$, $s_p=0$), then $A(s)$ vanish on $U$ also. Without loss of generality, we can assume that $U$ is a neighborhood such that $E|_U$ is trivial (i.e., $E|_U\\simeq U\\times R^n$, $n=\\mathrm{rank} E$). Take a local basis of sections $s_1,s_2,\\ldots,s_n$ of $E|_U$, then
\n \\[
\n s=\\sum_{k=1}^nf^ks_k.
\n \\]
\n Since $s_p=0$, thus $f^k(p)=0$ for any $p\\in U$, therefore $(As)_p=(A\\sum_kf^ks_k)_p=(\\sum_kf^kAs_k)_p=0$. This proved that $A$ is a local operator, and it is $C^\\infty(M)$-linear, thus it can be defined pointwise.<\/p>\n
Now, we define a map $\\tilde A\\in\\Gamma(\\mathrm{Hom}(E,F))$ (i.e., for any $p\\in M$, $\\tilde A_p\\in\\mathrm{Hom}(E_p,F_q)$) as
\n \\begin{align*}
\n \\tilde A_p\\mathpunct{:}E_p&\\to F_q\\\\
\n v&\\mapsto(\\tilde Av)=Av,
\n \\end{align*}
\n then $\\tilde A_p\\in\\mathrm{Hom}(E_p,F_q)$. Since for any $s\\in\\Gamma(E)$ such that $s_p=v$, we have $\\tilde A_p(v)=(As)_p$, thus $A\\in\\Gamma(\\mathrm{Hom}(E,F))$.
\n\\end{proof}<\/p>\n
Let $\\mathrm{End}(E)$ denote the vector bundle over $M$ formed by the fiberwise endomorphisms of $E$. Then we have the following important properties of curvature
\n\\begin{prop}
\n \\begin{enumerate}
\n \\item The curvature $R^E$ is $C^\\infty(M)$-linear. That is, for any $f\\in C^\\infty(M)$ and $s\\in\\Omega^\\cdot(M;E)$, one has
\n \\[
\n R^E(fs)=fR^Es.
\n \\]
\n Thus $R^E$ can be viewed as an element of $\\Gamma(\\mathrm{End}(E))$ with coefficients in $\\Omega^{\\cdot+2}(M;E)$, in other words, $R^E\\in\\Omega^{\\cdot+2}(M;\\mathrm{End}(E))\\equiv\\Omega^{\\cdot+2}(M;E^\\ast\\otimes E)$.
\n \\item For two smooth section $X,Y\\in\\Gamma(TM)$, then $R^E(X,Y)\\in\\Omega^\\cdot(M;\\mathrm{End}(E))$, and
\n \\[
\n R^E(X,Y)=\\nabla_X^E\\nabla_Y^E-\\nabla_Y^E\\nabla_X^E-\\nabla^E_{[X,Y]}.
\n \\]
\n \\item We have the second type \\emph{Bianchi identity}
\n \\[
\n \\nabla^E(R^E)=[\\nabla^E,R^E]=0.
\n \\]
\n \\end{enumerate}
\n\\end{prop}
\n\\begin{proof}
\n\\begin{enumerate}
\n\\item From the Lemma, We only need to show that
\n\\begin{align*}
\n R^E(f s)&=\\nabla^E((\\rd f)\\otimes s+f\\nabla^E s)\\\\
\n &=(\\rd^2f)\\otimes s+(-1)^{|\\rd f|}\\rd f\\wedge\\nabla^E s+\\rd f\\wedge\\nabla^E s+f(\\nabla^E)^2 s\\\\
\n &=-\\rd f\\wedge \\nabla^E s+\\rd f\\wedge\\nabla^E s+fR^E s\\\\
\n &=fR^E s.
\n \\end{align*}
\n\\item for any $s\\in\\Omega^\\cdot(M;E)$, we have
\n\\begin{align*}
\n R^E(X,Y)s&=(R^Es)(X,Y)=(\\nabla^E(\\nabla^Es))(X,Y)\\\\
\n &=\\nabla^E_X((\\nabla^Es)(Y))-\\nabla^E_X((\\nabla^Es)(X))-(\\nabla^Es)([X,Y])\\\\
\n &=\\nabla^E_X\\nabla^E_Y s-\\nabla^E_X\\nabla^E_Y s-\\nabla^E_{[X,Y]}s.
\n\\end{align*}
\n\\item Note that the connection $\\nabla^E$ on $E$ can naturally induce a connection on $\\mathrm{End}(E)$, which is still denoted by $\\nabla^E$, as
\n \\[
\n \\nabla^E A=[\\nabla^E,A]=\\nabla^EA-A\\nabla^E.
\n \\]
\n Then,
\n \\[
\n \\nabla^E(As)=(\\nabla^EA-A\\nabla^E)s+A(\\nabla^E s)=(\\nabla^EA)s+A(\\nabla^E s),
\n \\]
\n from which, the third item is trivial.
\n\\end{enumerate}
\n\\end{proof}
\n\\begin{prob}
\n \\begin{itemize}
\n \\item Please rewrite the proof of the Lemma.
\n \\item Please explain why $R^E(X,Y)s=(R^Es)(X,Y)$ in the proof of the Proposition (the second item).
\n \\end{itemize}
\n\\end{prob} <\/p>\n","protected":false},"excerpt":{"rendered":"
Let $E\\to M$ be a smooth complex vector …<\/p>\n