Why is not the D'Alembert operator a scalar? I am taking a course on classical electrodynamics and my professor has defined the D'Alembert operator to me as:
$$\square=\eta^{\mu \nu} \partial_{\mu} \partial_{\nu}$$
I have been operating using this definition because I know that what it means is (in a 4-dimensional Minkowski space):
$$\square = -\partial_0 \partial_0 + \partial_1 \partial_1 + \partial_2 \partial_2 + \partial_3 \partial_3$$
But today I was looking again at this definition and I started having some doubts. The D'Alembert operator is defined as a (2,0) tensor acting on two vectors. Shouldn't it be then just a scalar? Why is it not just a number?
I'm sure it is a fairly elemental question but I can't understand it right now.
 A: You say in the comments that you think of the metric as a bilinear symmetric form acting on two vectors. This is correct, of course. What we're doing here is pretending that the operator $\partial$ is a covector, and acting with the metric (the inverse metric, actually) on two copies of it: $\Box = \eta(\partial, \partial) = \eta^{\mu\nu}\partial_\mu \partial_\nu$. This is not perfectly rigorous since $\partial$ isn't an element of the cotangent space, but its components $\partial_\mu$ behave as if they were the components of a covector. This means that the resulting operator is a scalar: for any scalar function $f$, $\Box f$ is a scalar.
You might be confused because there are two meaning of "acting on" here. The metric acts on vectors (or covectors) because it is a tensor; if you give it two vectors you get a number. The D'Alembertian $\Box$ and the gradient $\partial$ are differential operators and they act on functions, returning another function in the case of $\Box$ or a covector in the case of $\partial$.
