The components $W_{ij}^k$ of the difference $W=\nabla-\overline{\nabla}$ between two Levi-Civita connections The following text comes from the book Geometric Relativity written by Dan A. Lee.



The expression $(\nabla_i-\overline{\nabla}_i)(v_j)$ may be a little perplexing, but according to what I've learned so far, that means
$$\nabla_{v_i}v_j-\overline{\nabla}_{v_i}v_j.$$
Then we would have
$$W_{ij}^k=\Gamma_{ij}^k-\overline{\Gamma}_{ij}^k=\frac{1}{2}g^{k\ell}(g_{\ell i,j}+g_{\ell j,i}-g_{ij,\ell})-\frac{1}{2}\overline{g}^{k\ell}(\overline{g}_{\ell i,j}+\overline{g}_{\ell j,i}-\overline{g}_{ij,\ell}).\tag{*}$$
Now I'd like to show that
$$W_{ij}^k=\frac{1}{2}g^{k\ell}(\overline{\nabla}_i g_{\ell j}+\overline{\nabla}_j g_{i\ell}-\overline{\nabla}_\ell g_{ij}),\tag{**}$$
where $\overline{\nabla}_i g_{\ell j}$ is used to denote the components of $\overline{\nabla}g$. My strategy is to expand the RHS of ($**$) and try to arrive at ($*$). The expansion is done by recalling that
$$\overline{\nabla}_k g_{ij}=g_{ij,k}-\overline{\Gamma}_{ki}^\ell g_{\ell j}-\overline{\Gamma}_{kj}^\ell g_{\ell i}.$$
Then we see
$$\overline{\nabla}_i g_{\ell j}+\overline{\nabla}_j g_{\ell i}-\overline{\nabla}_\ell g_{ij}=g_{\ell j,i}-\overline{\Gamma}_{i\ell}^k g_{kj}-\overline{\Gamma}_{ij}^k g_{k\ell}
+g_{\ell i,j}-\overline{\Gamma}_{j\ell}^k g_{ki}-\overline{\Gamma}_{ji}^k g_{k\ell}
-g_{ij,\ell}+\overline{\Gamma}_{\ell i}^k g_{kj}+\overline{\Gamma}_{\ell j}^k g_{ki}.$$
This is as far as I can go because I don't know how to get rid of the Christoffel symbols of $\overline{g}$. Can someone tell me what to do next? Thank you.
Edit 1. I'm very sorry. It seems that I have mixed up coordinate frames and the frame $\{v_1,\ldots,v_n\}$. Let me think about it and fix it later. I apologize.
Edit 2. As mentioned in Edit 1, I failed to rightly tell two different frames apart, so I will redo the problem here. Let me start from ($*$). The connection coefficients $\Gamma_{ij}^k$ and $\overline{\Gamma}_{ij}^k$ are not necessarily the Christoffel symbols because $\{v_1,\ldots,v_n\}$ may not be a coordinate frame. Write
$$g=g_{ij}e^i\otimes e^j\text{ and }\overline{g}=\overline{g}_{ij}e^i\otimes e^j$$
with $\{e^1,\ldots,e^n\}$ denoting the coframe dual to $\{v_1,\ldots,v_n\}$.
Then we have
$$\begin{align}
W_{ij}^k&=\Gamma_{ij}^k-\overline{\Gamma}_{ij}^k\\
&=\frac{1}{2}g^{k\ell}(v_i g_{j\ell}+v_j g_{i\ell}-v_\ell g_{ij}-g_{jm}c_{i\ell}^m-g_{\ell m}c_{ji}^m+g_{im}c_{\ell j}^m)\\
&\quad-\frac{1}{2}\overline{g}^{k\ell}(v_i \overline{g}_{j\ell}+v_j \overline{g}_{i\ell}-v_\ell \overline{g}_{ij}-\overline{g}_{jm}c_{i\ell}^m-\overline{g}_{\ell m}c_{ji}^m+\overline{g}_{im}c_{\ell j}^m),\tag{***}
\end{align}$$
where $c_{ij}^m$ are the functions defined by
$$[v_i,v_j]=c_{ij}^m v_m.$$
For more information, one can see the book Introduction to Riemannian manifolds by John M. Lee. Now, on the other hand, we have
$$\overline{\nabla}_k g_{ij}=v_k g_{ij}-\overline{\Gamma}_{ki}^\ell g_{\ell j}-\overline{\Gamma}_{kj}^\ell g_{i\ell}.$$
This gives
$$\frac{1}{2}g^{k\ell}(\overline{\nabla}_i g_{\ell j}+\overline{\nabla}_j g_{i\ell}-\overline{\nabla}_\ell g_{ij})=\frac{1}{2}(g^{k\ell}v_i g_{\ell j}-\overline{\Gamma}_{ij}^k-n\overline{\Gamma}_{ij}^k+g^{k\ell}v_j g_{i\ell}-n\overline{\Gamma}_{ji}^k-\overline{\Gamma}_{ji}^k-g^{k\ell}v_\ell g_{ij}+\overline{\Gamma}_{ji}^k+\overline{\Gamma}_{ij}^k).$$
Now the thing is, how do I dispense with $\overline{g}^{k\ell}$ in ($***$)? Thank you.
 A: Original answer.

Group the three terms like $g_{lj,i}$ in order to form the
$\Gamma^k_{lj}g_{ki}$. For the rest of the terms, observe that four of
them cancel by pairs because $\bar\Gamma$ is symmetric in its two
lower indices. The two remaining terms involving $\bar\Gamma$ can be
dealt with using the identity $$ \bar g_{ij,k}
 = \bar g_{rj}\bar\Gamma^r_{ik}
 + \bar g_{ri}\bar\Gamma^r_{jk} $$ or expanding the $\bar\Gamma$ in terms of the partial derivatives of $\bar g$.

This version was written for a coordinate frame. Now we do it for the general case.
Edit for an arbitrary frame.
Considering an arbitrary frame $\{v_1,\dots,v_n\}$, you still have
\begin{align}
\overline{\nabla}_i g_{\ell j}
+\overline{\nabla}_j g_{\ell i}
-\overline{\nabla}_\ell g_{ij}
&=v_ig_{\ell j}
-\overline{\Gamma}_{i\ell}^k g_{kj}
-\overline{\Gamma}_{ij}^k g_{k\ell} \\
&+v_jg_{\ell i}
-\overline{\Gamma}_{j\ell}^k g_{ki}
-\overline{\Gamma}_{ji}^k g_{k\ell} \\
&-v_\ell g_{ij}
+\overline{\Gamma}_{\ell i}^k g_{kj}
+\overline{\Gamma}_{\ell j}^k g_{ki}. \tag{S}
\end{align}
where the $g_{\bullet\bullet}$ are still symmetric.
How to simplify this?
First observe that
\begin{align}
v_ig_{jk} &= \nabla_{v_i}(g(v_j,v_k)) \\
&= g(\nabla_{v_i}v_j,v_k) + g(v_j,\nabla_{v_i}v_k) \\
&= g(\Gamma_{ij}^rv_r,v_k) + g(v_j,\Gamma_{ik}^rv_r) \\
&= \Gamma_{ij}^rg_{rk} + \Gamma_{ik}^rg_{jr}. \tag A
\end{align}
Applying $(A)$ to $(S)$, you get six terms of the form $\Gamma^\bullet_{\bullet\bullet} g_{\bullet\bullet}$ and six of the form $\bar\Gamma^\bullet_{\bullet\bullet} g_{\bullet\bullet}$
Now, since $\nabla$ has no torsion, you have $[v_i,v_j]=\nabla_{v_i}v_j - \nabla_{v_j}v_i$. Applying $e^k$ on both sides you get $e^k([v_i,v_j])=e^k(\nabla_{v_i}v_j) - e^k(\nabla_{v_j}v_i)$. This is
$c^k_{ij} = \Gamma^k_{ij} - \Gamma^k_{ji}$, so that
$$\Gamma^k_{ij} = c^k_{ij} + \Gamma^k_{ji}. \tag B$$
Analogously, you obtain
$$\bar\Gamma^k_{ij} = c^k_{ij} + \bar\Gamma^k_{ji}. \tag C$$
Substituting with $(B)$ and $(C)$, you will get six $c^\bullet_{\bullet\bullet}g_{\bullet\bullet}$ terms that cancel in pairs, leaving you with the equation you wanted:
$$
\overline{\nabla}_i g_{\ell j}
+\overline{\nabla}_j g_{\ell i}
-\overline{\nabla}_\ell g_{ij}
=
2(\Gamma_{ij}^k-\bar\Gamma_{ij}^k)g_{k\ell}.
$$
