I am reading the book Riemannian Manifolds, written by John M. Lee. On page $53$, the author defines a connection on the tensor bundle $\text{T}_l^k(M)$ and says it satisfies
(d) $\nabla$ commutes with all contractions: if "$\operatorname{tr}$" denotes the trace on any pair of indices, $$\nabla_X(\operatorname{tr}Y) = \operatorname{tr}(\nabla_XY).$$
What does the trace in the definition mean? Or what does the definition mean? Any advice is helpful. Thank you.
This is defined on page $13$ of the same book.
A tensor field $Y \in T^k_l(M)$ is a $C^{\infty}(M)$-linear map $Y : (\Omega^1(M))^l\times (\mathfrak{X}(M))^k \to C^{\infty}(M)$. The trace of $Y$ is the tensor field $\operatorname{tr}Y \in T^{k-1}_{l-1}(M)$ satisfying
$$\operatorname{tr}Y(\omega_1, \dots, \omega_{l-1}, V_1, \dots, V_{k-1}) = \operatorname{trace}(Y(\omega_1, \dots, \omega_{l-1}, \cdot, V_1, \dots, V_{k-1}, \cdot))$$
for all $\omega_1, \dots, \omega_{l-1} \in \Omega^1(M)$ and $V_1, \dots, V_{k-1} \in \mathfrak{X}(M)$ where the right hand side is the trace of the endomorphism corresponding to $Y(\omega_1, \dots, \omega_{l-1}, \cdot, V_1, \dots, V_{k-1}, \cdot) \in T^1_1(M)$.
The pair of indices refers to which two slots we leave blank (note, we must leave one slot of each type blank). There are many different traces ($kl$ of them in fact). The statement in the book indicates that the equality holds for all of these different traces.
