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Let $G$ and $H$ be Lie groups with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$. Suppose $G$ acts on $H$ by automorphisms, i.e. there exists a lie algebra homomorphism $\phi:G\to Aut(H)$. I want to find the Lie algebra of $G\ltimes H$. As a manifold, $G\ltimes H$ is diffeomorphic to $G\times H$ hence the Lie algebra of $G\ltimes H$ is the vector space $\mathfrak{g}\oplus\mathfrak{h}$. However, the groups $G\times H$ and $G\ltimes H$ need definitely not be isomorphic, so their Lie algebras need not be either. I want to describe the bracket structure on Lie($G\ltimes H$). I know that $\phi$ induces a Lie algebra homomorphism $\mathfrak{g}\to Lie(Aut(H))$, but how can I use this?

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  • $\begingroup$ I think this is in Helgason's book somewhere. Do you have it? $\endgroup$ Oct 2, 2019 at 18:57
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    $\begingroup$ Do you mean Differential Geometry, Lie groups and Symmetric spaces? $\endgroup$ Oct 2, 2019 at 19:00
  • $\begingroup$ Yes, that's the one. Now that I've gone and thumbed through it, I'm not so sure. But if you have it, look there. $\endgroup$ Oct 2, 2019 at 19:03
  • $\begingroup$ I can't find the book. Do you have an idea of the proof? $\endgroup$ Oct 2, 2019 at 19:36
  • $\begingroup$ The Lie algebra of a semidirect product is a semidirect product (of Lie algebras). You want to differentiate the action of $G$ on $H$ at the identity to get a map $\mathfrak{g} \to \text{Der}(\mathfrak{h})$, then take the semidirect product wrt this map. $\endgroup$ Oct 2, 2019 at 21:21

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There are two possible conventions for semidirect product, but let's suppose you're using the following one $$ (g_1,h_2)\cdot(g_2,h_2) = (g_1g_2,h_1(\phi(g_1)h_2)). $$ Employ the notation $\phi_g:H\to H, \ \phi_g(h) := \phi(g)h$ and $\phi^h:G\to H,\ \phi^h(g) := \phi(g)h$, and define $$ \phi_g':= T_{e_H}\phi_g:\mathfrak{h}\to\mathfrak{h}, \qquad \dot{\phi}^h:=T_{e_G}\phi^h:\mathfrak{g}\to T_hH. $$ So $$ (g_1,h_1)\cdot (g_2,h_2)=(g_1g_2,h_1(\phi_{g_1}h_2)) \quad\textrm{and} \quad (g,h)^{-1} = (g^{-1},\phi_{g^{-1}}h^{-1}). $$ Calculating $(g,h)\cdot(k,l)\cdot(g,h)^{-1}$, and differentiating wrt $(k,l)$, it is not difficult to show that the adjoint action of $G\ltimes H$ on $\mathfrak{g}\ltimes \mathfrak{h}$ is given by $$ \operatorname{Ad}_{(g,h)}(\xi,\eta) = (\operatorname{Ad}_g\xi,\operatorname{Ad}_h(\phi'_g(\eta))+\sigma_h(\operatorname{Ad}_g\xi)), $$ where $$ \sigma_h:\mathfrak{g}\to\mathfrak{h}, \qquad \sigma_h(\xi):= h\cdot(\dot{\phi}^{h^{-1}}\xi). $$ Here $\dot{\phi}^{h^{-1}}\xi\in T_{h^{-1}}H$, and $h\cdot $ denotes the derivative of left multiplication by $h$ (i.e., in general we define $h_1\cdot v_{h_2} := T_{h_2}L_{h_1}(v_{h_2})$, where $L_{h_1}:H\to H$ is left multiplication by $h_1$).

Now taking the derivative of this wrt $(g,h)$, we obtain an expression for the adjoint action of $\mathfrak{g}\ltimes\mathfrak{h}$ on itself (and hence the Lie bracket): $$ [(\xi_1,\eta_1),(\xi_2,\eta_2)] : =\operatorname{ad}_{(\xi_1,\eta_1)}(\xi_2,\eta_2) = ([\xi_1,\xi_2],[\eta_1,\eta_2]+\xi_1\cdot\eta_2 - \xi_2\cdot\eta_1), $$ where $$ \xi\cdot\eta := (\dot{\phi}')_\xi\eta = (\dot{\phi}')^\eta\xi = T_{(e_G,e_H)}\phi(\xi,\eta), $$ (in the final equality thinking of $\phi:G\times H\to H$).

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  • $\begingroup$ Is the map $T_{(e_G,e_H)}\phi(\xi,\eta)$ a derivation of $\mathfrak{h}$? $\endgroup$ Oct 3, 2019 at 13:46
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    $\begingroup$ $T_{(e_G,e_H)}\phi(\xi,\cdot)$ is: $\phi_g \in Aut(H)$, so $\phi_g(hlh^{-1}) = \phi_g(h)\phi_g(l)\phi_g(h)^{-1}$. Differentiating wrt $l$ gives $\phi_g'(\operatorname{Ad}_h\eta) = \operatorname{Ad}_{\phi_g(h)}(\phi_g'(\eta))$. Then differentiating wrt $h$ gives $\phi_g'([\eta_1,\eta_2]) = [\phi_g'(\eta_1),\phi_g'(\eta_2)]$. Finally differentiating wrt $g$ gives the derivation property. $\endgroup$
    – user17945
    Oct 3, 2019 at 18:32
  • $\begingroup$ I was under the impression that $Ad$ is used to denote the adjoint map at the level of Lie groups and $ad$ is used to denote the differential of the adjoint map. You seem to be using $Ad$ for one time derivation, and $ad$ for derivation in another coordinate.. is that so! $\endgroup$ Jun 20, 2021 at 3:57
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    $\begingroup$ I'm late to the party but I just had to understand how that computation worked and this post was extremely useful. I wrote up a little thing (mostly for myself and crediting this answer, with some notations changed just as matter of taste), but maybe it can be useful for whoever else comes across this too. $\endgroup$
    – Ivo Terek
    Oct 30, 2022 at 8:37
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    $\begingroup$ Thank you for sharing, @IvoTerek - I've saved a copy myself :) Another place you can find a proof of this result (with yet again different notation) is Section 2.2.4 of Momentum Maps and Hamiltonian Reduction by Ortega and Ratiu. I'm not sure if I wasn't aware of this reference in my original answer (I definitely knew the book), or just forgot to include it. $\endgroup$
    – user17945
    Oct 30, 2022 at 21:14

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