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Let $G, H$ be locally compact, $\sigma$-compact metric groups equipped with left Haar measures $m_G, m_H$ respectively. Let $\Phi: G \times H \to G$ be continuous such that $\phi: H \to Aut(G), \, h \mapsto \Phi( \cdot, h)$ is a group homomorphism. Denote $\Phi_h = \phi(h)$. Then, the semidirect product $G \rtimes_\phi H$ is defined as the topological space $G \times H$ equipped with $$(g_1, h_1) \cdot (g_2, h_2) = (g_1 \Phi_{h_1}(g_2), h_1 h_2)$$ for $g_1, g_2 \in G, \, h_1, h_2 \in H$. This is a topological group.

I would like to show that $$dm_{G \rtimes_\phi H}(g,h) = mod(\Phi_h) \,dm_H(h)\, dm_G(g) $$ defines a left Haar measure on $G \rtimes_\phi H$, where $mod$ is of course the modular function $mod: Aut(G) \to \mathbb{R}_{>0}$ defined by $$m_G(\varphi^{-1} B) = mod(\varphi) m_G(B)$$ for Borel sets $B \subseteq G$.

How can I do this without any explicit form for $\Phi_h$? The definition for the semidirect product on wikipedia uses conjugation instead of a general automorphism. Is it obvious from my definition that $\Phi_h$ must be some conjugation?

user302234
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  • Regarding the conjugation: You can think of $G$ and $H$ as subgroups of $G \rtimes_\phi H$ via $G \times 1$ and $1 \times H$. Then $(1,h)(g,1)(1,h)^{-1} = (\phi(h)(g),1) = (\Phi_h(g),1)$. In this sense $\Phi_h$ corresponds to conjugation in $G \rtimes_\phi H$. I explained this recently here. – Jendrik Stelzner Jan 02 '16 at 13:22
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    you have to show that the measure you wrote is left invariant, you don't need any explicit description of $\Phi_h$ since the terms involving it cancel out – k76u4vkweek547v7 Apr 11 '16 at 21:35
  • I often go along with duplicate, but can it really be a duplicate if the "original" question is newer? – Robert Soupe Mar 17 '18 at 01:36
  • @RobertSoupe Sure, particularly if the other question has an answer while this one does not. – Xander Henderson Mar 17 '18 at 01:52
  • @Xander Thanks for the explanation. I suppose I can still vote to close, but with four already it's not particularly urgent, so I'll just abstain. – Robert Soupe Mar 17 '18 at 05:11

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