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

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

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

• I adjusted your notation from $D_\mu$ to $\nabla_\mu$ in order to fit with the standard one. The D we keep for gauge covariant derivatives, as for example in the Standard Model Commented Jul 19, 2019 at 16:03
• You need to clarify what you mean by “the Leibnitz product rule”. Product of what with what? Commented Jul 19, 2019 at 16:03
• Would Mathematics be a better home for this question? Commented Jul 19, 2019 at 16:04
• To define the product rule you need to know how the covariant derivative works on higher order tensors and on 'covariant vectors' rather than contravariant (i.e. lower indices not upper). It is basically defined to satisfy the Leibniz product rule, as you can check yourself once you look up what I just said. Commented Jul 19, 2019 at 16:09
• You can’t start with only a formula for differentiating a contravariant vector. Commented Jul 19, 2019 at 16:34

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.

• May I ask a question here? In your excellent answer in this [post][1], I can understand that $\Gamma_{\lambda \mu}^\alpha$ comes from the transformation of the basis. However, in this answer and many textbooks, we don't write out the basis explicitly. Does this mean that the whole exact formula should include the basis, while we omit it for simplicity? [1]: physics.stackexchange.com/questions/505592/… Commented Mar 26, 2023 at 6:54
• @Daren Yes. I'd say so. Physics books usually omit the ${\bf e}_\mu$, giving other ways to think about the derivative, --- but I think it safer to define with the basis explicit. Of course when I am calculating I leave it out. Commented Mar 26, 2023 at 12:07

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.

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" $$= \nabla_\mathbf T (f\mathbf v) = \mathbf T(f)\mathbf v + f \nabla_\mathbf T \mathbf v$$