Let \(M\) be a smooth manifold with or without boundary, and let \(\nabla\) be a connection in \(TM\). Then \(\nabla\) can be uniquely extended to a connection in each tensor bundle \(T^{(k,l)}TM\), which satisfies:
- In \(T^{(1,0)}TM=TM\), \(\nabla\) agrees with the given connection.
- In \(T^{(0,0)}TM=M\times \mathbb{R}\), \(\nabla\) is given by the ordinary differentiation of functions: \begin{equation*} \nabla_Xf=Xf. \end{equation*}
- \(\nabla\) obeys the following product rule with respect to tensor products: \begin{equation*} \nabla_X(F\otimes G)=(\nabla_XF)\otimes G + F\otimes (\nabla_XG). \end{equation*}
- \(\nabla\) commutes with the trace operator: \begin{equation*} \nabla_X(\tr F)=\tr (\nabla_X F). \end{equation*}
Remarks
- The connection obeys a product rule with respect to the natural pairing between a covector field \(\omega\) and a vector field \(Y\): \begin{equation*} \nabla_X\langle \omega, Y\rangle=\langle \nabla_X\omega, Y\rangle+ \langle \omega, \nabla_XY\rangle \end{equation*}
- The remark above implies the covariant derivative of a covector field \(\omega\) is given by \begin{equation*} (\nabla_X \omega)(Y)=X(\omega(Y))-\omega(\nabla_XY). \end{equation*}
- For all \(F\in \Gamma(T^{(k,l)}TM)\), smooth 1-forms \(\omega^1,\ldots, \omega^k\), and smooth vector fields \(Y_1,\ldots, Y_l\), we have \begin{align} \nabla_XF(\omega_1,&\ldots,\omega_k,Y_1,\ldots,Y_l)=X(F(\omega_1,\ldots,\omega_k,Y_1,\ldots,Y_l)\\ &\quad - \sum_{i=1}^{k} F(\omega_1,\ldots,\nabla_X\omega_i,\ldots,\omega_k,Y_1,\ldots,Y_l)\\ &\quad - \sum_{j=1}^{l} F(\omega_1,\ldots,\omega_k,Y_1,\ldots, \nabla_XY_j,\ldots, Y_l). \end{align}