Restrictions on the form of a scalar-valued function imposed by Lorentz invariance Let $f(p,q)$ be a smooth Lorentz-invariant function of 4-vectors $p$ and $q$.  Should $f$ necessarily be of the form $f(p,q) = g(p^2, q^2, p_\mu q^\mu)$, where $g(x,y,z)$ is some scalar-valued function?  That is, is every Lorentz-invariant smooth scalar-valued function of 4-vectors a function of the invariants constructed using the arguments?
 A: This answer rests upon the following fact, which I'm pretty sure is true but for which I can't find a source: the only Lorentz-invariant tensors are $g_{\mu\nu}$, $\epsilon_{\mu\nu\rho\sigma}$, and tensor products of them. A Lorentz invariant tensor is one that doesn't change under Lorentz transformations: $\Lambda^\mu{}_\nu \Lambda^\rho{}_\sigma g_{\mu\rho} = g_{\nu\sigma}$. Tensor products means that you can form combinations like $g_{\mu\nu}g_{\rho\sigma}$.
Accepting this, let's Taylor expand our function:
$$f(p,q) = f(0) + A_\mu p^\mu + B_\mu q^\mu + C_{\mu\nu} p^\mu p^\nu + D_{\mu\nu}p^\mu q^\nu + E_{\mu\nu} q^\mu q^\nu + \cdots,$$
where we have things with more indices contracted with many copies of $p$ and $q$. Lorentz invariance means that $f(\Lambda p, \Lambda q) = f(p,q)$; let's look at what happens to one term:
$$D_{\mu\nu}p^\mu q^\nu \to D_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma p^\rho q^\sigma.$$
For the transformed function to be equal to the original function for all $p$ and $q$, each term in the series must be equal to itself, so we must have
$$D_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma p^\rho q^\sigma = D_{\rho\sigma}p^\rho q^\sigma,$$
so all the coefficients must be invariant tensors, so they must all be combinations of $g$ and $\epsilon$. But we only have two vectors, so any contraction of $\epsilon$ will involve the same vector contracted with at least two different indices, and so will vanish since $\epsilon$ is totally antisymmetric. Therefore, the coefficients will just be made out of many copies of $g_{\mu\nu}$, which will be contracted with the vectors, and there are only three options: $p\cdot p$, $p \cdot q$ and $q \cdot q$. Any coefficient with an odd number of indices must vanish, since it can't be built out of the rank-2 tensor $g$.
