Connexion disagreement tensor

Let $\nabla$ and $˜ \nabla$ denote affine connexions on a $𝐶 𝛼$ -manifold $𝑀$ . The connexion disagreement tensor $𝐶 𝑐 𝑎 𝑏 \in T 12 (𝑀)$ of $\nabla$ with $˜ \nabla$ is a tensor field defined so that for $𝜔 𝑎 \in Ω 1 (𝑀)$ we have¹

𝐶 𝑐 𝑎 𝑏 𝜔 𝑐 : = (˜ \nabla 𝑎 - \nabla 𝑎) 𝜔 𝑏

and thus

\nabla 𝑎 𝜔 𝑏 = ˜ \nabla 𝑎 𝜔 𝑏 - 𝐶 𝑐 𝑎 𝑏 𝜔 𝑐 .

Proof of tensoriality

We need to show that

𝐶 (𝜔) 𝑎 𝑏 : = 𝐶 𝑐 𝑎 𝑏 𝜔 𝑐 = (˜ \nabla 𝑎 - \nabla 𝑎) 𝜔 𝑏

is a $𝐶 𝛼 (𝑀)$ -linear map into $T 02 (𝑀)$ . To this end let $𝑓 \in 𝐶 𝛼 (𝑀)$ . Then by the Leibniz rule,

𝐶 (𝑓 𝜔) 𝑎 𝑏 = (˜ \nabla 𝑎 - \nabla 𝑎) (𝑓 𝜔) 𝑏 = d 𝑓 𝑎 𝜔 𝑏 + 𝑓 ˜ \nabla 𝑎 𝜔 𝑏 - d 𝑓 𝑎 𝜔 𝑏 = 𝑓 (˜ \nabla 𝑎 - \nabla 𝑎) 𝜔 𝑏 = 𝑓 𝐶 (𝜔) 𝑎 𝑏

as required.

A word of warning for physicists

Physicists might be uncomfortable with the assertion that $𝐶 𝑐 𝑎 𝑏$ is a tensor, and most introductory general relativity courses will spend a lot of time stressing that connexion coëfficients such as the Christoffel symbols are not tensors. Depending on perspective this is either a misunderstanding or disagreement. The connexion coëfficients for a coördinate chart are not covariant since it depended on the choice of coördinate chart, but if you consider the partial derivative as a local affine connexion as extra data attached to our manifold which we retain after change of coördinates, they suddenly are tensorial.

In particular, given local coördinates $𝑥 : 𝑈 \to ℝ 𝑚$ and considering partial derivative as a local affine connexion, we typically denote the connexion disagreement of an affine connexion $\nabla$ with $𝜕$ is denoted $Γ 𝑐 𝑎 𝑏$ and we have

\nabla 𝑎 𝜔 𝑏 = 𝜕 𝑎 𝜔 𝑏 - Γ 𝑐 𝑎 𝑏 𝜔 𝑐 .

or in components

\nabla 𝜇 𝜔 𝜈 = 𝜕 𝜇 𝜔 𝜈 - Γ 𝛿 𝜇 𝜈 𝜔 𝛿 .

We call $Γ 𝛿 𝜇 𝜈$ the connexion coëfficients. If $\nabla$ is the Levi-Civita symbol the connexion coëfficients are called the Christoffel symbols.

Covariant derivative disagreement on vector fields

With the same notation as above, let $𝑋 𝑎 \in 𝔛 (𝑀)$ be a vector field. Then

\nabla 𝑎 𝑋 𝑏 = ˜ \nabla 𝑎 𝑋 𝑏 + 𝐶 𝑏 𝑎 𝑐 𝑋 𝑐 .

Proof

Let $𝑋 𝑎 \in 𝔛 (𝑀)$ and $𝜔 𝑎 \in Ω 1 (𝑀)$ . Then since covariant derivatives all agree with the exterior derivative on scalar fields, we have

(˜ \nabla 𝑎 - \nabla 𝑎) (𝜔 𝑏 𝑋 𝑏) = 0 .

Therefore by the Leibniz rule

0 = 𝜔 𝑏 ˜ \nabla 𝑎 𝑋 𝑏 + 𝑋 𝑏 ˜ \nabla 𝑎 𝜔 𝑏 - 𝜔 𝑏 \nabla 𝑎 𝑋 𝑏 - 𝑋 𝑏 \nabla 𝑎 𝜔 𝑏 = 𝜔 𝑏 (˜ \nabla 𝑎 - \nabla 𝑎) 𝑋 𝑏 + 𝑋 𝑏 (˜ \nabla 𝑎 - \nabla 𝑎) 𝜔 𝑏 = 𝜔 𝑏 (˜ \nabla 𝑎 - \nabla 𝑎) 𝑋 𝑏 + 𝑋 𝑏 𝐶 𝑐 𝑎 𝑏 𝜔 𝑐 = 𝜔 𝑏 ((˜ \nabla 𝑎 - \nabla 𝑎) 𝑋 𝑏 + 𝐶 𝑏 𝑎 𝑐 𝑋 𝑐)

for all $𝜔 𝑎 \in Ω 1 (𝑀)$ . The conclusion follows.

Covariant derivative disagreement on tensor fields

With the same notation as above, let $𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞 \in T 𝑝 𝑞 (𝑀)$ be a tensor field, where we will suppress position since no raising or lowering will take place. Then by induction on applications of the Leibniz rule we see

\nabla 𝑐 𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞 = ˜ \nabla 𝑐 𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞 + 𝑝 \sum 𝑖 = 1 𝐶 𝑎 𝑖 𝑐 𝑑 𝑇 𝑎 1 \dots 𝑑 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞 - 𝑞 \sum 𝑖 = 1 𝐶 𝑑 𝑐 𝑏 𝑖 𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑑 \dots 𝑏 𝑞 .

In other words, to convert from one connexion to the other for the covariant derivative of a general tensor field $𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞$ , we must

add a contraction with each upper index of $𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞$ ,
subtract a contraction with each lower index of $𝑇 𝑎 1 \dots 𝑎 𝑝 𝑏 1 \dots 𝑏 𝑞$ .

Other properties

If both $\nabla$ and $˜ \nabla$ are torsion-free, or more generally if they have the same Contorsion tensor, then $𝐶 𝑐 𝑎 𝑏 = 𝐶 𝑐 (𝑎 𝑏)$ is symmetric in its lower indices.

#state/tidy | #lang/en | #SemBr

2009. General relativity, §1.1, pp. 32–33. ↩

Connexion disagreement tensor

Covariant derivative disagreement on vector fields

Covariant derivative disagreement on tensor fields

Other properties

Footnotes