I have the following problem.
Given the following homomorphism $\phi:\mathbb{R}\rightarrow$Aut($\mathbb{R}^2$), $\phi(t)$= $\left(\begin{matrix} e^{kt} & 0\\ 0&e^{-kt}\\ \end{matrix}\right)$, where $k\in\mathbb{N}$ and $e^{kt}+e^{-kt}\neq2$. Then, what is the Lie algebra $\mathfrak{g}$ of $\mathbb{R}\rtimes_\phi\mathbb{R}^2$?
According to "Symplectic Manifolds with no Kähler structures" from Tralle and Oprea, then the solution is $\mathfrak{g}=<X,Y,Z;[X,Y]=kY,[X,Z]=-kZ,[Y,Z]=0, k\in\mathbb{R}>$
I am however clueless as to how this is achieved
Note: To my understanding the product on $\mathbb{R}\rtimes_\phi\mathbb{R}^2$ is $(x,(y,z))\cdot(x',(y',z'))=(x+x',(y+y'e^{kt},z+z'e^{-kt}))$ and $X =(1,0,0),Y=(0,1,0),Z=(0,0,1)$. However if this were correct then $[X,Y]=(e^{k}-1)Y$ which does not agree with the afore mentioned book. Thanks to anyone who is willing to help
You seem to be using the group product to calculate the commutator - you should actually be using the linearisation of it ($X, Y$ are elements of the Lie algebra, not the Lie group). A general derivation of the Lie algebra of the semidirect product is posted here. In the notation there, $$ [(X,0),(0,Y)] = (0,X\cdot Y) = (0,kY), $$ and similarly for other brackets.