How to prove that the covariant derivative obeys the product rule [closed]

Question

In General Relativity the covariant derivative of contravariant vectors $A^\mu$ is: \begin{equation} \nabla_\mu A^\nu=\partial_\mu A^\nu+\Gamma^\nu_{\mu\alpha}A^\alpha \end{equation} where $\Gamma^\nu_{\mu\alpha}$ are Christoffel symbols.

My question: how do we prove it verifies the Leibniz product rule?

Answer 1

On doubly contravariant vectors, for example, we have $$ \nabla_\mu T^{\alpha\beta}= \partial_\mu T^{\alpha\beta}+ \Gamma^\alpha_{\lambda \mu}T^{\lambda\beta}+ \Gamma^\beta_{\lambda\mu} T^{\alpha\lambda} $$ and if you take $T^{\alpha\beta}=A^\alpha B^\beta$, you will see how Leibnitz works.

Answer 2

You need to define what the Leibniz product rule means for tensors of rank higher than 0.

One has:

$$\nabla_V ({\bf T}\otimes{\bf U}) := \nabla_V {\bf T} \otimes {\bf U} +{\bf T}\otimes\nabla_V {\bf U} $$

for $V$ a vector field, $\nabla$ a connection, $\bf T,U$ tensor fields.

Choose $V$ the coordinate base vector field, $\bf U, T$ arbitrary vector fields in the coordinate base, then use the expansions of the covariant derivative of coordinate bases in terms of coordinate bases (which return the Levi Civita symbols), then simply the Leibniz rule for real functions you learn in Calculus III (the real functions here of several variables are the components of the vector fields in the coordinate basis which are partially differentiated) and you should arrive at the result. This computation is standard in introductory differential geometry texts and, indeed, is independent of knowledge of GR.

Answer 3

The book I'm currently reading "The Geometry Of Physics: An Introduction" by Theodore Frankel starts first with the notion of an intrinsic derivative of vector fields. Which is defined as the tangential part of the ordinary derivative $\frac{d\mathbf v}{dt}$ of some vector field $\mathbf v$ tangent to some manifold $M $ in some $\mathbb R^d$.

So for example $\nabla_\mathbf T \mathbf v =\frac {\nabla \mathbf v}{dt} = \frac{d\mathbf v}{dt} $ minus "the part normal to the manifold". So from this it should be obvious that $\frac{d(f\mathbf v)}{dt} $ satisifes the Leibniz rule

$\frac {df}{dt}\mathbf v +f\frac{d\mathbf v}{dt} $ minus "the part normal to the manifold" $ = \nabla_\mathbf T (f\mathbf v) = \mathbf T(f)\mathbf v + f \nabla_\mathbf T \mathbf v $