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\begin{document}

\thispagestyle{fancy}
\lhead{Differential Geometry 2017--2018}
\rhead{ENS de Lyon}
\begin{center}
\Large{Tensor fields}
\end{center}
\vspace{-7mm}
\rule{\linewidth}{0.5pt}


\begin{exo}
Let $E$ and $E'$ be two smooth vector bundles over the same base space $M$. Let $F:\Gamma(E) \to \Gamma(E')$ be a $\mathcal{C}^\infty(M)$-linear map.
\begin{enumerate}
\item Prove that there exists a unique $f:E \to E'$ which is a bundle map over $M$ (that is, for any $x \in M$, $f_{|E_x} \in \End(E_x,E'_x)$) and satisfies:
\begin{equation}
\label{eq}
\forall s \in \Gamma(E), \forall x \in M, \qquad F(s)(x) = f(s(x)).
\end{equation}
\item Check that $f$ can be seen as an element of $\Gamma(E^* \otimes E')$.
\item Conversely, check that a section $f \in \Gamma(E^* \otimes E')$ defines a unique $\mathcal{C}^\infty(M)$-linear map $F: \Gamma(E) \to \Gamma(E')$ satisfying~\eqref{eq}.
\end{enumerate}
\end{exo}

\begin{rem}
In other terms we have defined a canonical isomorphism of $\mathcal{C}^\infty(M)$-modules between $\Hom(\Gamma(E),\Gamma(E'))$ and $\Gamma(E^* \otimes_{\R} E')$.

Similarly, we can show that $s \otimes s' \mapsto (x \mapsto s(x) \otimes s'(x))$ defines a canonical isomorphism of $\mathcal{C}^\infty(M)$-modules from $\Gamma(E) \otimes_{\mathcal{C}^\infty(M)} \Gamma(E')$ to $\Gamma(E \otimes_{\R} E')$. For example, $\Gamma(\bigwedge^k T^*M \otimes_{\R} E)$ is isomorphic to $\Omega^k(M,E) := \Omega^k(M) \otimes_{\mathcal{C}^\infty(M)} \Gamma(E)$, the space of $k$-forms with values in $E$.
\end{rem}

\begin{exo}
Let $\nabla$ be a connection on a smooth vector bundle $E \to M$. We can see $\nabla$ as a map from $\Omega^0(M,E)$ to $\Omega^1(M,E)$. We extend $\nabla : \Omega^k(M,E) \to \Omega^{k+1}(M,E)$ by forcing the Leibniz rule:
\begin{equation*}
\forall \alpha \in \Omega^k(M), \forall s \in \Gamma(E), \forall x \in M \qquad \nabla_x (\alpha \otimes s) = d_x\alpha \otimes s(x) + (-1)^k \alpha_x \wedge \nabla_x s
\end{equation*}
and $\R$-linearity.
\begin{enumerate}
\item Prove that $\nabla \circ \nabla$ is $\mathcal{C}^\infty(M)$-linear from $\Omega^0(M,E)$ to $\Omega^2(M,E)$.
\item Using the previous exercise, show that $\nabla \circ \nabla$ defines a section $R \in \Omega^2(M,\End(E))$. We call $R$ the \emph{curvature} of $(E,\nabla)$.
\item Let $(x_1,\dots,x_n)$ be local coordinates on $M$ and $(e_1,\dots,e_r)$ be a local frame for $E$ defined on the same open set $U$. We denote by $(\Gamma_{ij}^k)$ the Christoffel symbols of $\nabla$ in these coordinates. We also define $(R_{ijk}^l)$ by:
\begin{equation*}
\forall i,j,k \in \{1,\dots,n\}, \qquad R\left(\deron{}{x_i},\deron{}{x_j}\right)e_k = \sum_{l=1}^n R_{ijk}^l e_l.
\end{equation*}
Check that for any $i,j \in \{1,\dots,n\}$ and any $k,l\in \{1,\dots,r\}$ we have:
\begin{equation}
\label{Riemann}
R_{ijk}^l = \deron{\Gamma_{jk}^l}{x_i} - \deron{\Gamma_{ik}^l}{x_j} + \sum_{m=1}^r \Gamma_{jk}^m \Gamma_{im}^l - \sum_{m=1}^r \Gamma_{ik}^m \Gamma_{jm}^l.
\end{equation}
\item Check that for any $X,Y \in \Gamma(TM)$ and any $s \in \Gamma(E)$ we have:
\begin{equation*}
\nabla_X(\nabla_Ys) - \nabla_Y(\nabla_Xs) - \nabla_{[X,Y]}s = R(X,Y)s.
\end{equation*}
\end{enumerate}
\end{exo}

\begin{rem}
In particular, if $\nabla$ is the Levi--Civita connection of $(M,g)$ then $R$ is its Riemann curvature. Equation~\eqref{Riemann} is of course valid in this case.
\end{rem}
\end{document}