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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – aygx
    May 17, 2022 at 23:40
  • $\begingroup$ 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 $\nabla_j(\partial_i\phi)$. $\endgroup$
    – MBN
    May 18, 2022 at 8:07
  • $\begingroup$ 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? $\endgroup$
    – seVenVo1d
    May 18, 2022 at 9:21
  • $\begingroup$ @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? $\endgroup$
    – LSS
    May 19, 2022 at 0:08
  • $\begingroup$ 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 $\partial_i\phi$. $\endgroup$
    – MBN
    May 19, 2022 at 8:56


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