Left Invariant Metric: What I Don't Understand

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SUMMARY

The discussion centers on the semi-direct product ##R_+ \ltimes R^4## and its representation as a matrix group with elements ##\begin{pmatrix} \lambda & x^{\mu} \\ 0 & 1 \end{pmatrix}##. Participants explore the implications of left invariance of the metric, specifically questioning whether the relationship ##\langle g_1, g_2 \rangle = \langle g_3 g_1, g_3 g_2 \rangle## holds for arbitrary matrices ##g_3## in the group. The conversation also touches on the connection to the Poincaré group, noting that this structure allows for operations with fewer multiplications compared to the full Lorentz group.

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ergospherical
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I haven't learned about Lie Groups yet, but came across this question.

1626811932297.png


What I don't understand:
- is the semi-direct product ##R_+ \ltimes R^4## here a matrix group with elements ##\begin{pmatrix} \lambda & x^{\mu} \\ 0 & 1 \end{pmatrix}##? And is the group multiplication then matrix multiplication?
- I guessed that because ##R_+ \ltimes R^4 \sim R^5## that the metric acts on matrices ##g_1, g_2## in the group as it would acting on two vectors in ##R^5##, but what does it mean that the metric is left invariant? Is it that for an arbitrary matrix ##g_3## in the group that ##\langle g_1, g_2 \rangle = \langle g_3 g_1, g_3 g_2 \rangle##?

(Not sure if any of that's right and maybe it'd be better to actually learn the theory first, but sometimes a practical example can't hurt?)
 
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At first glance, I think this is the subgroup of the Poincaré group with all 4 translations, but only one dilatation (multiplication with ##M_{\mu\nu}=\lambda ##) as opposed to the entire Lorentz group. You could basically do all the things you do with the Poincaré group, but with far fewer multiplications, since the entire Lorentz subgroup is reduced to ##\mathbb{R}_+.##
 
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