- #1

Shirish

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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: $$(\nabla_{\mu}-\nabla'_{\mu})T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_l}=\sum_{i=1}^k C^{\alpha_i}_{\mu\nu}T^{\alpha_1\ldots\alpha_{i-1}\nu\alpha_{i+1}\ldots\alpha_k}_{\beta_1\ldots\beta_l}-\sum_{j=1}^lC^{\nu}_{\mu\beta_j}T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_{j-1}\nu\beta_{j+1}\ldots\beta_l}$$

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

3. Then it goes on to say:

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: $$(\nabla_{\mu}-\nabla'_{\mu})T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_l}=\sum_{i=1}^k C^{\alpha_i}_{\mu\nu}T^{\alpha_1\ldots\alpha_{i-1}\nu\alpha_{i+1}\ldots\alpha_k}_{\beta_1\ldots\beta_l}-\sum_{j=1}^lC^{\nu}_{\mu\beta_j}T^{\alpha_1\ldots\alpha_k}_{\beta_1\ldots\beta_{j-1}\nu\beta_{j+1}\ldots\beta_l}$$

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 $$-C^{\nu}_{\ \ \mu\alpha}g_{\nu\beta}=-C_{\ \beta\mu\alpha}$$ inWe 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)

**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:

I didn't understand the above line.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##

**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.
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