# Lie algebra of $\left(\begin{smallmatrix}a & b\\ & a^2\end{smallmatrix}\right )$ in $GL_2(\mathbb{R})$

I'm working on the question

Let $$G$$ be the group of invertible real matrices of the form $$\left [\begin{array}{c c}a & b\\ & a^2\end{array}\right ]$$. Determine the Lie algebra $$L$$ of $$G$$, and compute the bracket on $$L$$.

I'm familiar with how to derive the Lie algebra for a linear group like $$U_n$$, $$SU_n$$, etc. but I'm not sure what to do in a more explicit case like this.

• What definition of a Lie algebra are you working with? Do you define it as the tangent space at the identity, or do you have a definition involving the exponential map? May 3, 2019 at 4:53
• From Artin's Algebra, "The space of tangent vectors to a matrix group G at the identity is called the Lie algebra of the group." So it's the tangent space at the identity. May 3, 2019 at 5:57

Here is another answer which may be slightly more in line with first principles and/or what is learned at introductory level.

Consider the path in $$G$$ given by $$\gamma(t) = \left(\begin{array}{cc} a(t) & b(t) \\ 0 & a(t)^2 \end{array} \right),$$ which is differentiable, with $$\gamma(0) = I$$ (which implies $$a(0)=1$$ and $$b(0)=0$$). Since $$\gamma$$ is in $$GL_2(\mathbb R)$$, we also have $$a(t)\neq 0$$ for all $$t$$. We see that $$\gamma'(t) = \left(\begin{array}{cc} a'(t) & b'(t) \\ 0 & 2a(t)a'(t) \end{array} \right),$$ which implies that $$\gamma'(0) = \left(\begin{array}{cc} a'(0) & b'(0) \\ 0 & 2a'(0) \end{array} \right),$$ since $$a(0)=1$$. This shows that the Lie algebra $$L$$ consists of matrices of the form $$\left(\begin{array}{cc} a & b \\ 0 & 2a \end{array} \right)$$, where $$a, b \in \mathbb R$$. In fact, the above computation shows one inclusion, namely $$\mathfrak g(G)\subset L$$. Proving the other inclusion means specifically constructing a path in $$G$$ in the direction of any $$A \in L$$. So let $$A = \left(\begin{array}{cc} a & b \\ 0 & 2a \end{array} \right),$$ and consider the exponential mapping. We have

$$\begin{equation} \begin{split} e^A &= \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) + \left(\begin{array}{cc} a & b \\ 0 & 2a \end{array} \right) + \frac{1}{2!}\left(\begin{array}{cc} a^2 & 3ab \\ 0 & 4a^2 \end{array} \right) + \frac{1}{3!}\left(\begin{array}{cc} a^3 & 7a^2b \\ 0 & 8a^3 \end{array} \right) + ... \\ &= \left(\begin{array}{cc} 1+a+\frac{a^2}{2!}+\frac{a^3}{3!}+... & b(1+\frac{3a}{2!}+\frac{7a^2}{3!}+...) \\ 0 & 1+2a+\frac{4a^2}{2!}+\frac{8a^3}{3!}+... \end{array} \right) \\ &= \left(\begin{array}{cc} e^a & b(1+\frac{3a}{2!}+\frac{7a^2}{3!}+...) \\ 0 & (e^a)^2 \end{array} \right). \end{split} \end{equation}$$

After convincing yourself that the series in the top-right matrix entry converges (it is not difficult), we see in fact that $$e^A \in G$$. Since $$A \in L$$, then $$\alpha(t)= e^{tA}$$ is a differentiable path in $$G$$ with $$\alpha'(0)=A$$. This shows that $$L \subset \mathfrak g(G).$$

• @user1445709 By the way, you should also consider the 'Lie groups' and 'Lie algebras' tags for further questions like this. May 3, 2019 at 7:05
• This is very clear, thanks. May 4, 2019 at 15:23

At the identity $$a=1$$ and $$b=0$$. Near the identity, the group is parameterised as $$\pmatrix{1+t&u\\0&(1+t)^2}=I+t\pmatrix{1&0\\0&2}+u\pmatrix{0&1\\0&0}+O(\text{higher powers of t and u}).$$ The tangent space at $$I$$ is spanned by $$A=\pmatrix{1&0\\0&2}\qquad\text{and}\qquad B=\pmatrix{0&1\\0&0}$$ and these matrices span the Lie algebra. To find the bracket, all you need is to compute $$AB-BA$$.

• I don't entirely understand what's happening here. How are you finding that parameterization? May 3, 2019 at 6:27
• @user1445709 By setting $a=1+t$ and $b=u$. May 3, 2019 at 6:28