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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? $\endgroup$ – Matthew Leingang Oct 2 at 18:57
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    $\begingroup$ Do you mean Differential Geometry, Lie groups and Symmetric spaces? $\endgroup$ – Lucas Smits Oct 2 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. $\endgroup$ – Matthew Leingang Oct 2 at 19:03
  • $\begingroup$ I can't find the book. Do you have an idea of the proof? $\endgroup$ – Lucas Smits Oct 2 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. $\endgroup$ – Qiaochu Yuan Oct 2 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 $\mathfrak{h}$? $\endgroup$ – Lucas Smits Oct 3 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 at 18:32

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