# Why are there no connection coefficients appearing here?

This is from a set of lecture notes on differential geometry.

We make the notational definition: $$D_{\mu}\equiv D_{\partial_\nu} \tag{1}$$

The covariant derivative is defined by its action on basis vectors:

$$D_\mu(\partial_\nu)=\Gamma^\lambda_{\mu\nu}\partial_\lambda, \tag{2}$$

where I have used the summation convention and suppressed the evaluation at the point $$p$$ in the derivative operators $$(\partial_\lambda|_p\rightarrow\partial_\lambda)$$.

We then proceed to define the covariant derivative of $$Y$$ along $$X$$: \begin{align} D_XY&=X^\mu D_\mu(Y^\nu\partial_\nu) \\ &= X^\mu(\partial_\mu(Y^\nu)\partial_\nu+Y^\nu D_\mu(\partial_\nu)) \\ &= (X^\mu\partial_\mu Y^\lambda+\Gamma^\lambda_{\mu\nu}X^\mu X^\nu)\partial_\lambda \end{align}.

It is in the second line I am getting lost, we apply the covariant derivative to $$Y^\mu$$ and then to $$\partial_\nu$$ as with any derivative operator, but the connection coefficients do not appear in the first term of the second line?

I am a physics student so I have seen the covariant derivative written out, but this is a more mathematical text and I am only now seeing the covariant derivative defined more carefully.

Any help appreciated, thanks.

A connection $$\nabla$$ is indeed asked to check the Leibniz rule $$\nabla_X(fV)=(Xf)V+f(\nabla_XV),$$ where $$f$$ is a scalar function and $$V$$ a vector field. The first term of the sum does not depend of the connection, so there won't be Christoffel coefficients implied there.
• I'm still definitely missing something, probably something simple, why does $\nabla_X(f)=X(f)$ and not $\Gamma^\lambda_{\mu\nu}\partial_\lambda(f)$? I've just straight applied the definition of the covariant derivative from above. Commented Sep 30, 2020 at 18:38
• The above is just a formula you ask your connection to check, also note that $\nabla_X$ takes vector fields as argument and not functions, so writing $\nabla_Xf$ does not immediately have sense (it has later, because a connection on vector fields induces a nice connection on all tensor fields -so also on functions which are (0,0)-tensors- which acts on functions by $\nabla_Xf=Xf$). Commented Sep 30, 2020 at 21:34