Covariant derivative commutes with trace (contraction)

Question

I'm trying to solve Exercise 4.3 in J. Lee's Riemannian Manifolds book. The setting is as follows: let $\nabla$ be a linear connection on $M$. I want to show property (b) of Lemma 4.6, which says that $\nabla$ commutes with all contractions. To make the notation easier, I want to show it first for $F\in\mathcal T^1_1(M)$, i.e., I want to show that $$ \nabla_X(\text{tr}(F))=\text{tr}(\nabla_X F). $$ We can work locally, so let $E_j$ and $\omega^i$ be frames of $TM$ and $T^*M$ respectively. We can then write $F=\sum F^j_i E_j\otimes\omega^i$, and we know then that $$ \text{tr}F=\sum_m F^m_m, $$ so $\nabla_X(\text{tr}F)=\sum_m X(F^m_m)$. By property (c) we know that $$ \nabla_X(\sum F^j_i E_j\otimes \omega^i)=\sum X(F^i_j)E_j\otimes\omega^i+\sum F^i_j(\nabla_X E_j\otimes\omega^i+E_j\otimes\nabla_X\omega^i). $$ Since we need $\text{tr}(\nabla_X F)=\sum_m X(F^m_m)$, it seems that we would need $$ \text{tr}\left(\sum F^i_j(\nabla_X E_j\otimes\omega^i+E_j\otimes\nabla_X\omega^i)\right)=0, $$ and I don't see why that would be the case. Could someone point me into the right direction?

Edit

Thanks to everyone's help I've realised why I got stuck: I should have worked with a coordinate frame $\partial_i$ along with its dual coframe $dx^i$, and by taking $X=\partial_m$, the expression $\nabla_X\omega^i$ is then $\nabla_{\partial_m}(dx^i)$, whose coefficients can be determined using property (i) of the Lemma (in terms of the Christoffel symbols).

Answer 1

Ok, let $X\in \mathfrak{X}(M)$, $F \in \mathscr{T}^1_1(M)$, and $\nabla$ be any affine connection on $TM$. I'll use Einstein's summation convention and local coordinates.

The idea here is that $$\begin{split}(\nabla_XF)^i_j &= {\rm d}x^{i}((\nabla_XF)(\partial_j)) \\ &= {\rm d}x^{i}(\nabla_X(F(\partial_j)) - F(\nabla_X\partial_j)) \\ &= {\rm d}x^{i}(\nabla_X(F_j^k\partial_k) - F(\nabla_X\partial_j)) \\ &= {\rm d}x^i(X(F^k_j)\partial_k + F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j)) \\ &= X(F^i_j) + {\color{red}{{\rm d}x^{i}(F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j))}}.\end{split}$$We are done once we show that the term in red vanishes when subjected to contraction. On one hand, we have that $$F^k_j\nabla_X\partial_k = F^k_jX^\ell\nabla_{\partial_\ell}\partial_k = F^k_jX^\ell \varGamma_{\ell k}^i\partial_i,$$and on the other $$F(\nabla_X\partial_j) = F(X^\ell\nabla_{\partial_\ell}\partial_j )=F(X^\ell \varGamma_{\ell j}^{k}\partial_k) = F_k^iX^\ell \varGamma_{\ell j}^k\partial_i,$$so that $${\color{red}{{\rm d}x^{i}(F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j))}} = F^k_jX^\ell\varGamma_{\ell k}^i - F^i_{k}X^\ell\varGamma_{\ell j}^k = X^\ell(F^k_j\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell j}^k).$$The factor $X^\ell$ is irrelevant, so make $i=j$ on $F^k_j\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell j}^k$ to obtain $$F^k_i\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell i}^k = 0,$$since the second term is obtained from the first one by renaming the dummy indices $i\leftrightarrow k$.