I'm working on proving Lemma 4.6 in John Lee's "Riemannian Manifolds":
Lemma. Let $\nabla$ be a linear connection on a Riemannian manifold $M$. There is a unique connection in each tensor bundle $T_l^kM$, also denoted $\nabla$, such that the following conditions are satisfied:
- On $TM$, $\nabla$ agrees with the given connection.
- On $T^0M$, $\nabla$ is given by ordinary differentiation of functions: $$\nabla_X f = Xf.$$
- $\nabla$ obeys the following product rule with respect to tensor products: $$\nabla_X(F \otimes G) = (\nabla_X F)\otimes G + F \otimes (\nabla_X G).$$
- $\nabla$ commutes with all contractions: if "$\mathrm{tr}$" denotes the trace of any pair of indices, $$\nabla_X(\mathrm{tr}Y) = \mathrm{tr}(\nabla_X Y).$$
- $\nabla$ obeys the following product rule with respect to the natural pairing between a covector field $\omega$ and a vector field $Y$: $$\nabla_X\langle \omega, Y\rangle = \langle \nabla_X \omega, Y \rangle + \langle \omega, \nabla_X Y \rangle.$$
- For any $F \in \mathcal T_l^k(M)$, where $\mathcal T_l^k(M)$ is the space of smooth $(k,l)$-tensor fields on $M$, and for any vector fields $Y_i$ and 1-forms $\omega^j$, \begin{align} (\nabla_X F)(\omega^1, \ldots, \omega^l, Y_1, \ldots, Y_k) = &X(F(\omega^1, \ldots, \omega^l, Y_1, \ldots, Y_k)) \\ &- \sum_{j=1}^l F(\omega^1, \ldots, \nabla_X \omega^j, \ldots, \omega^l, Y_1, \ldots, Y_k) \\ &- \sum_{i=1}^k F(\omega^1, \ldots, \omega^l, Y_1, \ldots, \nabla_X Y_i, \ldots, Y_k). \end{align}
Lee suggests using 1-4 to prove 5 and 6 (and thus to prove uniqueness), and prove existence using 5 and 6 (ie using 5 and 6 to prove 1-4). The one part of this I'm having trouble with is proving 5 from 1-4 (I can prove 2-4 from 6 and 1 (note 5 follows from 6 also), and 6 follows from 3). Certainly 5 bears resemblance to 3, but 3 involves explicit tensor products, and 5 does not; besides, $\langle \omega, Y \rangle \in C^\infty(M) = \mathcal T^0(M)$, which seems to suggest 2 should be used, but again, the exact computation is lost on me. Any insights?