Graduate Index-Free Expressions & Unique Metric-Compatible Connections

[Shirish]

There are three doubts, but all are from the same section and closely related so I thought I'll ask in one post.

I'm studying about connections 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)

1. 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.

2. Why doesn't the quoted equation work for any other second-rank covariant tensor?

3. Then it goes on to say:

Uniqueness of $\nabla_{\mu}g_{\alpha\beta}$ is implied by $C_{(\alpha\beta)\mu}=0$, which combined with the no-torsion requirement implies $C^{\alpha}_{\ \ \beta\gamma}=0$

I didn't understand the above line. Where did $C_{(\alpha\beta)\mu}=0$ come from? The book mentions in a prior paragraph that covariant derivatives can differ by tensors $C^{\alpha}_{\ \ \beta\gamma}$, but that's all it says about the $\gamma$ index. I'm not sure how $C^{\alpha}_{\ \ \beta\gamma}=0$ follows.

 

[strangerep]

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.

I don't know of any neat way to express it. This is one of the reasons I prefer index-full notation.

2. Why doesn't the quoted equation work for any other second-rank covariant tensor?

It does, provided the "other second-rank covariant tensor" is also covariantly constant, like the metric.

3. Where did $C_{(\alpha\beta)\mu}=0$ come from? The book mentions in a prior paragraph that covariant derivatives can differ by tensors $C^{\alpha}_{\ \ \beta\gamma}$, but that's all it says about the $\gamma$ index. I'm not sure how $C^{\alpha}_{\ \ \beta\gamma}=0$ follows.

$C$ was assumed symmetric in its 2nd and 3rd indices. So, inserting an extra step in your earlier equation, it becomes: $$ [\dots ] ~=~ -(C_{\ \beta\mu\alpha}+C_{\ \alpha\mu\beta})~=~ -(C_{\ \beta\alpha\mu}+C_{\ \alpha\beta\mu}) ~=~ -2C_{(\alpha\beta)\mu} ~.$$