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It's mentioned in this paper that if $\partial^i \partial^j$ applied on an equation like:

$$ x \delta_{ij} + (\nabla^2 \delta_{ij}- \partial_i \partial_j) y =0 $$

It yields a couple of equations:

$$ \nabla^2 y=0, $$ and,

$$ x=0 $$

This means:

$ \partial^i \partial^j \delta_{ij} =1 ~~~ \star$

And

$ \partial^i \partial^j \partial_i \partial_j = 0 ~~~ \star $

Any explanation for that?

Because I think: $ \partial^i \partial^j \delta_{ij} = \partial^i \partial_i = \nabla^2 $

And even if the $\star$ relations are true why the equation splits into a couple of equations.

Any help is appreciated!

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    $\begingroup$ Where exactly in the paper is it used? (page and Eqn. number please...) $\endgroup$
    – Jos Bergervoet
    Commented Mar 28 at 22:14
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/808045/2451 $\endgroup$
    – Qmechanic ♦
    Commented Mar 29 at 1:22
  • $\begingroup$ Hi @JosBergervoet. In the paper, similar to the equation that I have mentioned equ. (3.11), and similar to the star $\star$ relations, equ. (3.13) and (3.14). $\endgroup$
    – Dr. phy
    Commented Mar 29 at 11:28
  • $\begingroup$ @JosBergervoet. As mentioned in the paper in the paragraph before equ. (3.13) that $\partial^i \partial^j$ is transverse to $\nabla^2 \delta_{ij} - \partial_i \partial_j $, so I would like to know what that means $\endgroup$
    – Dr. phy
    Commented Mar 29 at 11:33

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