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I'm trying to solve problem 1.12 of chapter 1 from Duistermaat & Kolk' Lie groups.

In the exercise you have a Lie group $G$ and a finite-dimensional vector space $V$, and a homomorphism $\Phi:G\rightarrow GL(V)$ of Lie groups.

With this it's defined the semidirect product $G\ltimes V$ of $G$ and $V$ as the Lie group with underlying manifold $G\times V$ and multiplication given by $$(g_1,v_1)(g_2,v_2)=(g_1g_2,v_1+\Phi(g_1)v_2),$$ for $g_1,g_2\in G, v_1,v_2\in V$.

I want to deduce the identity $$[(X_1,Y_1),(X_2,Y_2)]=([X_1,X_2],[Y_1,Y_2]+\phi(X_1)Y_2-\phi(X_2)Y_1),$$ where $\phi=T_1(\Phi)$ is the derivative of $\Phi$ at the identity of $G$.

I tried to do this using curves at $G$ going throught $1_G$ and having tangents $X_1,X_2$ and passing through $0\in V$ and tangents $Y_1,Y_2$ taking derivatives.

anfibio900
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1 Answers1

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You can obtain the Lie bracket as follows: let $G$ be a Lie group

  1. Find the derivative of the adjunction map $ad_g:G\to G$ sending $h\mapsto ghg^{-1}$ at the identity to get the map denoted by $Ad_g\in Aut(\mathfrak{g})$.
  2. This in particular describes a map $Ad:G\to Aut(\mathfrak{g})$ by $g\mapsto Ad_g$.
  3. Differentiate this map at the identity to get a map $\mathfrak{g}\to TAut(\mathfrak{g})\cong Der(\mathfrak{g})$. This is the map $\xi\mapsto [\xi,-]$.

In your case, \begin{align} ad_{(g,v)}(h,w) & = (ghg^{-1},v+\Phi(g)w + \Phi(gh)v)\\ & = (ad_gh,(1+\Phi(gh))v + \Phi(g)w). \end{align} Differentiating at the identity, we get $$Ad_{(g,v)}(\eta,w) = (Ad_g\eta,\Phi(g)\phi(\eta)v+\Phi(g)w).$$ Differentiating again: $$[(\xi,v),(\eta,w)] = ([\xi,\eta],\phi(\eta)v+\phi(\xi)w).$$

Daniel Robert-Nicoud
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  • @SKH Are you sure about your edit? The map $\xi\mapsto[\xi,-]$ lands into $Aut(\mathfrak{g})$, not in $\mathfrak{g}$... – Daniel Robert-Nicoud Jun 13 '17 at 06:42
  • SORRY ! I mean not $\mathfrak{g}$ but $gl(\mathfrak{g})=TAut(\mathfrak{g})$ in your notation. In any case $TAut(\mathfrak{g})\cong Aut(\mathfrak{g})$ is not true because RHS is not even a vector space . The Map $\xi\mapsto[\xi,-]$ can never an automorphism in a Lie algebra. – SKH Jun 14 '17 at 07:29
  • Another edit. $T Aut(\mathfrak{g})$ should be $Der(\mathfrak{g})$. – SKH Jun 14 '17 at 07:56
  • @SKH Ok, this makes much more sense! Edited. – Daniel Robert-Nicoud Jun 14 '17 at 08:48