Index-free tensor expressions and what makes the metric tensor different There are two doubts, but all are from the same section and closely related so I thought I'll ask in one post.
I'm studying a section that introduces Christoffel symbols [Core Principles of Special and General Relativity by Luscombe] and came across a quote. Just for context, it references this equation relating any two torsion-free covariant derivative operators on a manifold: $$(\nabla_{\mu}-\nabla'_{\mu})T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_l}=\sum_{i=1}^k C^{\alpha_i}_{\ \ \mu\nu}T^{\alpha_1\ldots\alpha_{i-1}\nu\alpha_{i+1}\ldots\alpha_k}_{\beta_1\ldots\beta_l}-\sum_{j=1}^lC^{\nu}_{\ \ \mu\beta_j}T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_{j-1}\nu\beta_{j+1}\ldots\beta_l}$$

We show that for $g_{\alpha\beta}$ a given metric field on (a smooth manifold $M$), there is a unique torsion-free derivative satisfying $\nabla_{\mu}g_{\alpha\beta}=0$. We start by noting (from the above equation) that $$(\nabla_{\mu}-\nabla'_{\mu})g_{\alpha\beta}=-C^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}-C^{\nu}_{\ \ \mu\beta}g_{\nu\alpha}=-(C_{\ \beta\mu\alpha}+C_{\ \alpha\mu\beta})=-2C_{(\alpha\beta)\mu}$$ where we've lowered an index (This equation applies to the metric tensor and not a general second-rank covariant tensor) and used the symmetry of $C_{\alpha\beta\gamma}$ in the second and third indices (no-torsion requirement).


*

*I'm not sure how to interpret the contraction of the first term on the LHS. $\mathbf{C}$ is supposed to eat one covector and two vectors. How can it "partially eat" one covariant part of the tensor $\mathbf{g}$ (the index $i$)? Maybe my confusion could be resolved if I knew how to express the equation $$-C^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}=-C_{\ \beta\mu\alpha}$$ in index-free notation. For example, I know that $f_iX^i$ can be written in index-free notation as $\mathbf{f}(\mathbf{X})$, where $\mathbf{f}$ is a covector and $\mathbf{X}$ is a vector. Similarly $g_{ij}X^iY^j$ means $\mathbf{g}(\mathbf{X},\mathbf{Y})$. I also suspect that something like $C^i_{\ \ jk}\omega_{i}$ can be written as $\mathbf{C}(\mathbf{\omega}, \cdot, \cdot)$, where $\mathbf{\omega}$ has filled up the covector slot and we're left with a $(0,2)$ tensor. But I can't figure out how something like $C^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}$ is supposed to represented in a index-free way.


*Why doesn't the quoted equation work for any other second-rank covariant tensor?
 A: I am not familiar with Luscombe's book, but it seems to use Penrose abstract index notation. This style of notation has been popularized by Wald's General Relativity.
If I understand your question correctly, you do not understand why $$-C^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}=-C_{\ \beta\mu\alpha}$$
But this is simply a definition!
In general if you have a tensor $T^{\nu}_{\ \ \mu\alpha}$ and you contract an upper index with the lower metric $g_{\nu\beta}$, you lower the index and it gets replaced by the other index in the metric. That is:$$T^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}=T_{\ \beta\mu\alpha}$$ This is tensor algebra 101 and I am sure Lascombe explained it in a previous section. It exploits the non-canonical isomorphism between a vector $V$ space and its dual $V^*$.  You can read it in Wikipedia.
The OP would like to interpret or translate these tensor eqs. into index-free notation.
Index-free notation is not very practical and becomes cumbersome when you have several indexes. It is used (in physics) only when you are dealing with differential forms .The problem with Index-free notation is that it is complicated to identify the indexes being contracted. You have to use contraction operators which indicate on which indexes you are contracting. For example: $$T^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}=T_{\ \beta\mu\alpha}$$ could be written as: $$C^{1}_{\ \ 1}(T \otimes g)=T^\flat $$ where I used $C^{1}_{\ \ 1}$ to express the contraction operator on the first index in $T$ and $g$ and the $\flat$ operator to express the lowering operator (see Musical Isomorphism). Actually , to be really precise, I should have written $T^{\flat 1}$ to indicate that I want to lower the first index.
As you see this is very clumsy and nobody relly does it.
