# Action of the Lorentz group on scalar fields

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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

asked Dec 29, 2012
retagged Mar 25, 2014
Would you mind writing out more carefully why the second condition isn't fulfilled?

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user user1504
$(\Lambda_1 \Lambda_2)^{-1}=\Lambda_2^{-1} \Lambda_1^{-1}$ which is not $\Lambda_1^{-1} \Lambda_2^{-1}$

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user inquisitor
It's OK, nothing else is needed.

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

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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.

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user twistor59
answered Dec 29, 2012 by (2,500 points)
+1, you forgot the subscripts on the first two lines, though it's not really unclear :)

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user kηives
@kηives : thanks - duly edited now!

This post imported from StackExchange Physics at 2014-03-22 17:31 (UCT), posted by SE-user twistor59
Thank you, just to write it on the original notation: $(g_2g_1)\phi(x)=\phi((\Lambda_2 \Lambda_1)^{-1} x)=\phi(\Lambda_1^{-1}\Lambda_2^{-1} x)$, and $g_2(g_1\phi)(x)=g_1\phi(\Lambda_2^{-1} x)= \phi(\Lambda_1^{-1}\Lambda_2^{-1} x)$

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

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