# Index-Free Expressions & Unique Metric-Compatible Connections

• A
• Shirish
In summary, the conversation discusses the interpretation of a contraction involving the covariant tensor and metric on a manifold. It is shown that the quoted equation applies to any covariant tensor, as long as it is covariantly constant. The uniqueness of the covariant derivative of the metric is implied by the no-torsion requirement and the symmetry of the tensor in its second and third indices.
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.

Last edited:
Shirish said:
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: $$[\dots ] ~=~ -(C_{\ \beta\mu\alpha}+C_{\ \alpha\mu\beta})~=~ -(C_{\ \beta\alpha\mu}+C_{\ \alpha\beta\mu}) ~=~ -2C_{(\alpha\beta)\mu} ~.$$

## 1. What are index-free expressions?

Index-free expressions are mathematical expressions that do not rely on the use of indices or coordinate systems. They are written in a coordinate-independent manner, making them applicable in any coordinate system.

## 2. What is the significance of index-free expressions in mathematics?

Index-free expressions allow for a more elegant and concise representation of mathematical concepts, as they are not dependent on the choice of coordinates. They also make it easier to generalize concepts and apply them in different contexts.

## 3. What are unique metric-compatible connections?

Unique metric-compatible connections are mathematical structures that define how vectors and tensors are differentiated in a curved space. They are unique in the sense that they are independent of the choice of coordinates and metric, and they preserve the metric structure of the space.

## 4. How do index-free expressions and unique metric-compatible connections relate?

Index-free expressions are often used in the context of unique metric-compatible connections, as they allow for a coordinate-independent representation of these connections. They also make it easier to understand and manipulate the properties of these connections.

## 5. What are some applications of index-free expressions and unique metric-compatible connections?

Index-free expressions and unique metric-compatible connections have various applications in mathematics, physics, and engineering. They are used in the study of curved spaces, general relativity, and differential geometry, among others.

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