# How can I prove that D'Alembert operator is invariant under Lorentz transformations?

I'm currently taking a course on Classical Electrodynamics and I'm trying to prove that the D'Alembert operator ($\square=\eta^{\mu \nu} \partial_{\mu} \partial_{\nu}$) is invariant under Lorentz-like transformations.

My professor does this using an argument to show that the variation of the operator must be zero, but I don't see where he specifically uses the fact that the transformation must be Lorentz.

I thought at first of proving it by applying the transformation and showing that the operator stays the same.

What he does is the following:

He shows that $[\delta, \partial_{\mu}]=\delta \omega_{\mu}^{\nu} \partial_{\nu}$ and then showing that $[\partial, \square]=0$.

Why is the first fact so?

Let $\square = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$ and let $\Lambda$ be any operator leaving the metric invariant, i. e. $\Lambda\eta\Lambda^{-1}=\eta$.
From the above it follows that the components of a $(2,0)$ tensor transform with twice the matrix $\Lambda$, whereas the vector components transform with $\Lambda^{-1}$, namely (I suppress all the indexes for the sake of the notations, but they are obviously summed over): $$\square' = \eta'\partial'\partial' = (\Lambda \Lambda\eta)(\Lambda^{-1}\partial)(\Lambda^{-1}\partial) = \eta\partial\partial = \square$$ summing over the corresponding indeces.