Determining the Lie algebra $\frak g$ to a given Lie group $G$

Question

In an old exercise sheet there was the following Lie group:

$$G = \left\{ \begin{pmatrix} a & b \\ 0 & a^{-1}\end{pmatrix} \Big|\ a> 0\ \text{and}\ b\in \mathbb R\right\}$$

and in order to see whether i understood things correctly, I wanted to determine its corresponding Lie algebra $\frak g$. However, I don't exactly know how to approach this problem.

What I've tried:

My first naive approach was to consider $G$ as a Lie subgroup of $\operatorname{GL}_2(\mathbb R)$ and use the one-parameter subgroup $$\gamma(t) = e^{tA}$$ since $$A \in {\frak{g}} \iff e^{tA} \in G\quad \forall t\in \mathbb R$$

However, I'm not sure how to proceed from here since, unlike in the examples of the common Lie subgroups $O(n), \operatorname{SO}(n)$ or $\operatorname{GL}_n(\mathbb R)$, I don't see a straight forward property of elements of $G$ that would be of help in order to use the exponential map accordingly.

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My second thought was choosing a one-parameter subgroup $$\gamma\colon \mathbb R\to G$$ s.t. $\gamma(0) = E$, one possible one-parameter subgroup would be $$\gamma(t) = \begin{pmatrix} 1+t & 0 \\ 0 & (1+t)^{-1}\end{pmatrix}$$

Now clearly $\gamma(0) = E$. But now I'm not entirely sure. I thought about considering $\gamma'(0)$ in order to obtain a tangent vector at $E$ and then determine the unique corresponding left-invariant vector field $\widetilde X$ via $$\widetilde X(H) = dL_H(\gamma'(0))$$ where $H$ denotes a fixed but arbitrary element of $G$ and $dL_H$ is the push-forward (the differential) of the left-translation map $$L_H\colon G\to G,\ A\mapsto HA$$

My question:

Would my second attempt somehow be reasonable? Or do I confuse some things?

Additionally, I'd like to ask whether someone could give me a hint regarding my first approach (embedding $G\hookrightarrow \operatorname{GL_2}(\mathbb R)$)

Thanks for any help.

Answer 1

You can just think of $\mathfrak{g}$ as the tangent space to $G$ at the identity. What do you do then? You consider $$\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$$with $a>0$ and $b\in \Bbb R$, pretend that $a$ and $b$ depend on a parameter $t$, assume that $a(0)=1$ and $b(0)=0$, write (say) $\dot{a}(0)= r$ and $\dot{b}(0)=s$ with $r,s\in \Bbb R$, and take the derivative with respect to $t$ at $t=0$ to obtain a generic element of $\mathfrak{g}$. Since $(a^{-1})^{\boldsymbol \cdot} = -\dot{a}a^{-2}$, it follows that $$\mathfrak{g} \cong \left\{ \begin{pmatrix} r & s \\ 0 & -r \end{pmatrix} \mid r,s\in \Bbb R\right\}.$$

Your $G$ is a subgroup of ${\rm GL}(2,\Bbb R)$, so $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}(2,\Bbb R)$. In particular, $\exp^G = \exp^{{\rm GL(2,\Bbb R)}}\big|_{\mathfrak{g}}$.