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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.

Zest
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1 Answers1

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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, $\exp^G = \exp^{{\rm GL(2,\Bbb R)}}\big|_{\mathfrak{g}}$.

Ivo Terek
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  • Thanks for your help Ivo Terek. Did i understand you correctly that you take the derivative of $\begin{pmatrix} a(t) & b(t) \ 0 & a^{-1}(t)\end{pmatrix}$ with respect to $t$ and denote $\dot{a}(t)$ by $r$ and $\dot{b}(t)$ by $s$?

    Furthermore, what does $\exp^G = \exp^{\operatorname{GL(2,\mathbb R)}}\Big|_{\frak g}$ mean?

    Would my second approach in my question somehow work?

     – Zest Feb 05 '22 at 17:22
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    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. – Ivo Terek Feb 05 '22 at 17:31
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    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. – Ivo Terek Feb 05 '22 at 17:33
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    Wow, thanks so much. Your help is greatly appreciated. I will try to work it out appropriately and maybe return to your answer in the next couple of hours or at latest tomorrow. Please give me some additional time to fully appreciate what you did but i will gladly accept your answer eventually. Thanks so much! – Zest Feb 05 '22 at 17:40
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    Good luck! Let me know if you get stuck on anything. – Ivo Terek Feb 05 '22 at 18:00
  • Thanks again Ivo Terek, your solution provides great insights, i worked it out and gained some new perspective. I was just wondering, would the approach via the exponential map (the first attempt above) be reasonable for this problem, too? Do you happen to have a hint regarding how to tackle a problem like this via the exponential map if I don't see a particular defining property like for, say orthogonal matrices which satisy $A^TA = E$ which we can throw into $\exp(\cdot)$ as usual? – Zest Feb 06 '22 at 03:58
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    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. – Ivo Terek Feb 06 '22 at 04:15
  • Beautiful, thanks so much! – Zest Feb 06 '22 at 04:29
  • I am trying to follow your answer, but totally lost in the general process. I presume this is a bit of a template on an easy example of how to find the Lie algebra from the Lie group in more general cases. The step "assume that $a(0)=1$ and $b(0)=0$ is something you can always do with the parameters? Is there some rule-of-thumb? Thank you! – JAP Nov 01 '22 at 20:25
  • $a(0) = 1$ and $b(0)=0$ make the whole initial matrix (i.e., at time zero) equal to the identity matrix. And $\mathfrak{g}$ is the tangent space to $G$ at the identity. – Ivo Terek Nov 01 '22 at 23:44
  • Thank you very much! I get it now! However, this doesn't seem to play any role in the actual process of getting the Lie algebra matrix, which starts at the definition of $r$ and $s$ as the derivatives, am I right? – JAP Nov 02 '22 at 12:57
  • (I will erase at least one or two of these last comments, no worries about clutter) - I think I get it! Without $a(0)=1$ the expression $(a^{-1})^{\boldsymbol \cdot} = -\dot{a}a^{-2}$ would become a problem with a $0$ in the denominator. Is that it? It act like a check that the expression derived is well defined? – JAP Nov 02 '22 at 13:40
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    $a(0)=1$ is crucial to say that $(a^{-1})^{\cdot}(0)= -r$. In this case, $b(0)=0$ doesn't matter (but if you don't have it, your original curve doesn't start at the identity matrix). – Ivo Terek Nov 02 '22 at 14:05