Connections and Curvatures on Vector Bundles

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 $\Lambda^\cdot(T^\ast M)\otimes E$.

A connection on $E$ is an extension of exterior differential operator $\rd$ to include the coefficient $E$.

定义 1. A 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 Leibniz rule
\[
\nabla^E(fs)=(\rd f)\otimes s+f\nabla^Es.
\]

注记 1.
  • By the partitions of unity, we can construct may connections on a vector bundle.
  • Given a vector field $X\in\Gamma(TM)$, we can define the direction derivative of $s$ among $X$ as a map $\nabla^E_X\mathpunct{:}\Gamma(E)\to\Gamma(E)$ by
    \[
    \nabla_X^E s=< X,\nabla^E s >=i_X(\nabla^E s).
    \]
  • 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
    \[
    \nabla^E\mathpunct{:}\omega\otimes s\mapsto(\rd\omega)\otimes s+(-1)^{|\omega|}\omega\wedge\nabla^Es.
    \]

In 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.
定义 2. The curvature $R^E$ of a connection $\nabla^E$ is defined by
\[
R^E=\nabla^E\circ\nabla^E\mathpunct{:}
\Omega^\cdot(M;E)\to\Omega^{\cdot+2}(M;E),
\]
for brevity, we also write $R^E=(\nabla^E)^2$.

Before we go to the properties of curvature, let us first give a lemma of $C^\infty$-linear maps between two sections of vector bundle.
引理 3. 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 $A\in\Gamma(\mathrm{Hom}(E,F))$.

证明 . 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
\[
s=\sum_{k=1}^nf^ks_k.
\]
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.

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
\begin{align*}
\tilde A_p\mathpunct{:}E_p&\to F_q\\
v&\mapsto(\tilde Av)=Av,
\end{align*}
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 $A\in\Gamma(\mathrm{Hom}(E,F))$.

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

命题 4.
  1. 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
    \[
    R^E(fs)=fR^Es.
    \]
    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, $R^E\in\Omega^{\cdot+2}(M;\mathrm{End}(E))\equiv\Omega^{\cdot+2}(M;E^\ast\otimes E)$.
  2. For two smooth section $X,Y\in\Gamma(TM)$, then $R^E(X,Y)\in\Omega^\cdot(M;\mathrm{End}(E))$, and
    \[
    R^E(X,Y)=\nabla_X^E\nabla_Y^E-\nabla_Y^E\nabla_X^E-\nabla^E_{[X,Y]}.
    \]
  3. We have the second type Bianchi identity
    \[
    \nabla^E(R^E)=[\nabla^E,R^E]=0.
    \]

证明 .
  1. From the Lemma, We only need to show that
    \begin{align*}
    R^E(f s)&=\nabla^E((\rd f)\otimes s+f\nabla^E s)\\
    &=(\rd^2f)\otimes s+(-1)^{|\rd f|}\rd f\wedge\nabla^E s+\rd f\wedge\nabla^E s+f(\nabla^E)^2 s\\
    &=-\rd f\wedge \nabla^E s+\rd f\wedge\nabla^E s+fR^E s\\
    &=fR^E s.
    \end{align*}
  2. for any $s\in\Omega^\cdot(M;E)$, we have
    \begin{align*}
    R^E(X,Y)s&=(R^Es)(X,Y)=(\nabla^E(\nabla^Es))(X,Y)\\
    &=\nabla^E_X((\nabla^Es)(Y))-\nabla^E_X((\nabla^Es)(X))-(\nabla^Es)([X,Y])\\
    &=\nabla^E_X\nabla^E_Y s-\nabla^E_X\nabla^E_Y s-\nabla^E_{[X,Y]}s.
    \end{align*}
  3. 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
    \[
    \nabla^E A=[\nabla^E,A]=\nabla^EA-A\nabla^E.
    \]
    Then,
    \[
    \nabla^E(As)=(\nabla^EA-A\nabla^E)s+A(\nabla^E s)=(\nabla^EA)s+A(\nabla^E s),
    \]
    from which, the third item is trivial.

问题 1.
  • Please rewrite the proof of the Lemma.
  • Please explain why $R^E(X,Y)s=(R^Es)(X,Y)$ in the proof of the Proposition (the second item).

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