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

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.

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.

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, $$\exp^G = \exp^{{\rm GL(2,\Bbb R)}}\big|_{\mathfrak{g}}$$.
• Yes, exactly, but $r$ and $s$ are the derivatives evaluated at $t=0$. Each Lie group has its own exponential map. I am saying that the exponential map of the group $G$ equals the exponential map of the ${\rm GL}(2,\Bbb R)$ restricted to $\mathfrak{g}$. Your second approach requires a lot more of work, because you'd have to do it for one-parameter subgroups realizing all initial velocities in $T_{{\rm Id}_2}G$, but you chose a single one which has no reason to be special at all. Commented Feb 5, 2022 at 17:31
• To elaborate further, what you tried to do was what I did with $a(t)=1+rt$ and $b(t)=st$. But the curves $a$ and $b$ don't need to define together a one-parameter subgroup. We're computing a single tangent space to $G$ and the value of the differential of a function at a single tangent vector doesn't depend on the curve chosen to realize such vector. Commented Feb 5, 2022 at 17:33
• I don't see any quick solution with exp right now. But if you know what a generic element of $G$ looks like, it doesn't matter, the strategy from the answer should always work. Commented Feb 6, 2022 at 4:15