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Let $F(x^\mu)$ is a scalar function; i.e. $F(x^\mu): \mathbb{R}^{1,3} \rightarrow \mathbb{R}$. How the Poincare Group $P(1,3)$ will act on it; i.e., by which formula I can calculate it for a specific function $F(x^\mu)$?

Edit 1: We know that a subgroup of Poincare group $P(1,3)$ leaves the hypersurface $\Sigma: x^+ = 0$ invariant where $x^+ \equiv x^0 + x^3$. To show this a formula is needed.

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    $\begingroup$ You mean "scalar" like $F : \mathbb{R}^{1,3} \rightarrow \mathbb{R}$ not "scalar" like "is invariant under transformations", right? $\endgroup$
    – ACuriousMind ♦
    Commented Jul 16, 2014 at 19:14
  • $\begingroup$ yes, $F: \mathbb{R}^{1,3} \rightarrow \mathbb{R}$. $\endgroup$
    – rainman
    Commented Jul 16, 2014 at 19:18
  • $\begingroup$ Well, your edit has not so much to do with the original question, but I'll bite: The subgroup consisting of the identity always leaves everything invariant, so there's nothing to show here. Are you asking how to find the maximal subgroup that leaves $ x^+ = 0$ invariant? $\endgroup$
    – ACuriousMind ♦
    Commented Jul 16, 2014 at 19:39
  • $\begingroup$ @ACuriousMind: yes, how can I find the maximal subgroup? $\endgroup$
    – rainman
    Commented Jul 20, 2014 at 20:34

1 Answer 1

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If you do not know anything more about $F$, the only way to do this is to simply plug the transform result $\Lambda^\mu_\nu x^\nu$ into $F$ and see what happens. General functions into $\mathbb{R}$ cannot be expected to behave nicely in any way.

However, I cannot really fathom a situation where you would have to use a function in $\mathbb{R}$ where you would not know (or demand) that it is truly scalar, i.e. transform trivially under whatever group you have. A bit more detail from you could perhaps help here.

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  • $\begingroup$ edited the question to clarify it. $\endgroup$
    – rainman
    Commented Jul 16, 2014 at 19:32

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