# Identity $\nabla_{j}\partial_{i} = \Gamma ^{k}_{ij} \partial_{k}$ involving covariant derivative

I am trying to understand the identity i have came across, but i am not being able to: $$\nabla_{j}\partial_{i} = \Gamma ^{k}_{ij} \partial_{k}$$

I thought that such equality would become obviously if we see how the left operator acts on a scalar:

$$\nabla_{j}\partial_{i} \phi = \partial_{j}\partial_{i} \phi - \Gamma ^{k}_{ij}\partial_{k} \phi$$

Not so evident as i thought it would be. How can i get the equalitty?

• The partial derivative operator $\partial_i$ can act as a basis for whatever space your working in. Recall the identity: $$\frac{\partial e_i}{\partial x^j}=\Gamma_{ij}^ke_k,$$ where $e_k$ is our basis vectors. In this sense the partial derivative $e_i$ because they are equivalent. Does that clear anything up?
– aygx
May 17, 2022 at 23:40
• The first one is the defining equality of the Christoffel's symbols. The second one is not right. There is a difference between $(\nabla_j\partial_i)\phi$ and $\nabla_j(\partial_i\phi)$.
– MBN
May 18, 2022 at 8:07
• Could you share some context about where did you see this equation or how it's used etc. ? Maybe there are some underlying assumptions about the identity? May 18, 2022 at 9:21
• @MBN I am not sure if i got your thinking. How can we know the action of the operator on an object, if not by this way?
– LSS
May 19, 2022 at 0:08
• Your second equations is not the action of $\nabla_j \partial_i$ on $\phi$. It is the action of $\nabla_j$ on the one form with components $\partial_i\phi$.
– MBN
May 19, 2022 at 8:56