Definition talk:Metric Compatible Connection

Mistake
What happened here is that I left out the notion of general covariant derivative (for personal reasons, I only had time for most basic notions). The current definition at hand is defined only for vector fields. However, this covariant derivative induces a covariant derivative of arbitrary tensor fields including scalars. And for scalars it reduces to exactly what you wrote. So what we really have to do here is to created a theorem page for existence of such a derivative, and then create a definition page for the more general covariant derivative. Referring to that derivative would then solve the issue.--Julius (talk) 21:44, 26 May 2023 (UTC)


 * OK, this will be correct if you extend the definition of $\nabla_X$ so that especially $\nabla_X \innerprod Y Z = \map X {\innerprod Y Z}$. I think, the simplest and best solution is just to write it $\map X {\innerprod Y Z}$, as it actually is. But OK to leave it until you add the extended definition of $\nabla_X$. --Usagiop (talk) 22:12, 26 May 2023 (UTC)


 * The reasons I prefer to keep the $\nabla_X \innerprod Y Z$ are the following: 1) this is what I found in the book I referred to, so I do not want to confuse readers; if other books simplify it, let's have the second definition and then show the equivalence; 2) after I set up the page for tensor fields, readers will see how the case for scalars is handled there rather than here; 3) This form of equation does not just illustrate a calculation. It tells something about covariant derivatives, namely, that under special circumstances we can pass them through the scalar product and act on individual vector fields according to Leibniz rule. This property is more fundamental than the simplified equation.--Julius (talk) 20:38, 28 May 2023 (UTC)


 * Is this Definition:Koszul Connection not the extended definition? I think, you just need to correct the link. --Usagiop (talk) 22:38, 26 May 2023 (UTC)


 * Koszul connection is more general in the sense that it allows arbitrary vector bundles. Tangent bundle is just an example. It does not define a covariant derivative of scalars, because scalars in general are not vectors. What we need is something which involves tensor fields and tensor bundles.--Julius (talk) 20:38, 28 May 2023 (UTC)


 * Can you elaborate your last point? I thought a scalar $\alpha : M \to \R$ can be considered as a section ${\mathbf s} \in \map \Gamma {M \times \R}$ with $\map {\mathbf s} x := \struct {x, \map \alpha x}$ so that $\nabla_X {\mathbf s} = \mathbf t$ with $\map {\mathbf t} x := \struct {x, \map X \alpha}$. Do you mean, there is a better approach to justify $\nabla_X \alpha = \map X \alpha$ without such a re-interpretation of scalar? --Usagiop (talk) 22:14, 28 May 2023 (UTC)


 * Way over my head I'm afraid. I'll let the experts hammer this out. --prime mover (talk) 21:46, 26 May 2023 (UTC)