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?

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    $\begingroup$ $\Theta(p^0)$ and $\Theta(q^0)$ are also invariants. $\endgroup$ – AccidentalFourierTransform Jul 9 '18 at 22:24
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    $\begingroup$ I was thinking of sufficiently smooth functions of $p$ and $q$. I suspect what I am asking must hold, but am not sure based on considering a Taylor expansion of $f$. $\endgroup$ – Coriolis1 Jul 9 '18 at 23:00
  • $\begingroup$ Maybe the answer is yes if you change Lorentz to Poincare, even without smoothness. Is there a reason that you want to consider $f(p,q)$ and not just $f(p)$? I don't see what further interest there is in considering two vectors as inputs. $\endgroup$ – Ben Crowell Jul 10 '18 at 2:54
  • $\begingroup$ I included two vectors to allow invariants such as $p_\mu q^\mu$, besides $p^2$. I am not sure if the question is different with just one variable. $\endgroup$ – Coriolis1 Jul 10 '18 at 13:20
  • $\begingroup$ You may want to have a look at Ticciati's QFT, §4.11. $\endgroup$ – AccidentalFourierTransform Jul 10 '18 at 18:33

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 $g$.

  • $\begingroup$ Nice! Are $g_{\mu\nu}$ and $\epsilon_{\mu\nu\rho\sigma}$ the only Lorentz-invariant tensors? If they are, for a scalar-valued Lorentz-invariant function that depends on four or more 4-vectors the only invariants that $g$ can depend on are those constructed using $g_{\mu\nu}$, $\epsilon_{\mu\nu\rho\sigma}$ and the arguments. $\endgroup$ – Coriolis1 Jul 10 '18 at 19:31
  • $\begingroup$ @Coriolis1 they are not the only ones (for example $g_{\mu\nu}g_{\rho\sigma}$ is invariant too), but all the invariants can be constructed as tensor products of them. And yes, if there are four or more vectors and you only care about the proper orthochronous group, you can use $\epsilon$ too. If you want parity invariance, though, $\epsilon$ is out. $\endgroup$ – Javier Jul 10 '18 at 19:34
  • $\begingroup$ Yes, I meant the tensors and their products. Is it known that they are the only Lorentz-invariant tensors though? $\endgroup$ – Coriolis1 Jul 10 '18 at 20:11
  • $\begingroup$ @Coriolis1 Like I said, I'm pretty sure they are but I couldn't find a source. The proof would involve decomposing tensor representations of the Lorentz group into sums of irreducibles and looking for singlets. $\endgroup$ – Javier Jul 10 '18 at 20:13

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