# Describing the Lie algebra structure of a semi-direct product of Lie groups

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?

• I think this is in Helgason's book somewhere. Do you have it? – Matthew Leingang Oct 2 at 18:57
• Do you mean Differential Geometry, Lie groups and Symmetric spaces? – Lucas Smits Oct 2 at 19:00
• 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. – Matthew Leingang Oct 2 at 19:03
• I can't find the book. Do you have an idea of the proof? – Lucas Smits Oct 2 at 19:36
• 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. – Qiaochu Yuan Oct 2 at 21:21

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 $$\phi:G\times H\to H$$).
• Is the map $T_{(e_G,e_H)}\phi(\xi,\eta)$ a derivation of $\mathfrak{h}$? – Lucas Smits Oct 3 at 13:46
• $T_{(e_G,e_H)}\phi(\xi,\cdot)$ 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. – user17945 Oct 3 at 18:32