How can we be sure that the tensor identity $\partial_{\mu}\phi^*\partial^{\mu}\phi -\partial_{\mu}\phi\partial^{\mu}\phi^* = 0 $ is true? Consider the complex scaler field $\phi$.
Why is it that the tensor identity $$\partial_{\mu}\phi^*\partial^{\mu}\phi -\partial_{\mu}\phi\partial^{\mu}\phi^*  = 0 $$
is true.
I thought the fact that we specify up and down indices is important and we cannot just assume that swapping them will allow for such an identity to hold.
Furthermore how can we tell that $\partial_{\mu}\phi\partial^{\mu}\phi^* = -\phi^*\partial_{\mu}\partial^{\mu}\phi$
This is probably quite obvious I am just a little confused.
 A: Assuming $\phi$ is a classical field. Recall that $\partial_\mu$ denotes the covariant derivative, meaning $\partial_\mu = (\partial_0, \partial_1,\partial_2, \partial_3)$ and $\partial^\mu = g^{\mu \nu}\partial_\nu$ (contract in the $\nu$ variable) where $g^{\mu\nu}$ is the inverse metric. For simplicity, take $g = \eta$ to be the Minkowski metric with signature $(-, +, +, +)$ then it follows that
$\partial^\mu = (-\partial_0, \partial_1, \partial_2, \partial_3).$ Finally, we see that
\begin{align}
\partial^\mu\phi^\ast \cdot \partial_\mu \phi =&\ (-\partial_0\phi^\ast, \partial_1\phi^\ast, \partial_2\phi^\ast, \partial_3\phi^\ast)\cdot (\partial_0\phi, \partial_1\phi, \partial_2\phi, \partial_3\phi) \\
=&\  (\partial_0\phi, \partial_1\phi, \partial_2\phi, \partial_3\phi)\cdot (-\partial_0\phi^\ast, \partial_1\phi^\ast, \partial_2\phi^\ast, \partial_3\phi^\ast)\\
=&\ \partial_\mu\phi \cdot \partial^\mu\phi^\ast.
\end{align}
Also, it is not true that
\begin{align}
\partial^\mu\phi^\ast \cdot \partial_\mu \phi= -\phi^\ast \partial^\mu\partial_\mu \phi
\end{align}
since the left hand side only involves first order derivatives whereas the right hand side involves second order derivatives. However, it is true that
\begin{align}
\int\ \partial^\mu\phi^\ast \cdot \partial_\mu \phi= -\int\ \phi^\ast \partial^\mu\partial_\mu \phi
\end{align}
because of integration by parts.
