# Identities with Div, Grad, Curl

In physics there are lots of identities like:

$$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - (\nabla \cdot \nabla) A$$

I'm wondering if there is an algorithmic algebraic method to prove and/or derive these identities (something like using $e^{i\theta}$ to prove trigonometric identities)?

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Let me work through your example to illustrate. In tensor notation, the $i$th component of the curl of a vector field $v_i$ is given by $\varepsilon_{ijk}\partial_j v_k$, where $\varepsilon_{ijk}$ is the Levi-Civita symbol, that takes the values $1$, $-1$ and $0$ depending on the order in which the coordinates appear in the subscripts $ijk$. You can think of this as a notational shorthand for the plus and minus signs in the curl formula. So the left-hand side of the desired identity is $$\varepsilon_{ijk}\partial_j (\varepsilon_{k\ell m}\partial_\ell A_m)$$ $$= \varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$ because derivatives commute with constants and with each other.
At this point, to simplify simplify $\varepsilon_{ijk}\varepsilon_{k\ell m}$, I just looked up the relevant identity of the Levi-Civita symbol, but it should be possible to derive it simply by algebraic manipulation from the purely combinatorial definition of the symbol. It turns out that $\varepsilon_{ijk}\varepsilon_{k\ell m} = \delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell}$, where $\delta_{ij}$ is the Kronecker delta which is $1$ if and only if $i = j$, and $0$ otherwise. This symbol acts like a substitution operator: $\delta_{ij}v_j = v_i$.
So we have $$\varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$ $$= (\delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell})\partial_j\partial_\ell A_m$$ $$= \partial_m\partial_i A_m - \partial_\ell\partial_\ell A_i$$ $$= \partial_i (\partial_m A_m) - (\partial_\ell\partial_\ell) A_i.$$ The first term is $\nabla(\nabla\cdot A)$, while the second is $-(\nabla\cdot\nabla)A$.