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The Lorentz groups act on the scalar fields as: $\phi'(x)=\phi(\Lambda^{-1} x)$

The conditions for an action of a group on a set are that the identity does nothing and that $(g_1g_2)s=g_1(g_2s)$. This second condition is not fulfilled because of the inverse on $\Lambda$. What is then the action of the Lorentz group on the scalar fields?

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user inquisitor

Denote by $g_1\phi$ the field transformed by the action of $\Lambda_1$ : $$(g_1\phi)(x) = \phi(\Lambda_1^{-1}(x))$$ Similarly $g_2$ has action $$(g_2\psi)(x) = \psi(\Lambda_2^{-1}(x))$$ Substitute $g_1\phi$ for $\psi$ $$(g_2g_1\phi)(x) = (g_1\phi)(\Lambda_2^{-1}(x)) = \phi(\Lambda_1^{-1}\Lambda_2^{-1}(x)) = \phi((\Lambda_2\Lambda_1)^{-1}(x)) $$ So the group action looks correct.

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