I am trying to increase my understanding of Lie Algebra, but I got stuck on a example. I have a Lie group corresponding to the set of matrices
\begin{bmatrix}
x & y \\[0.3em]
0 & 1 \\
\end{bmatrix}
and I am seeking to determine the basis for the left invariant vector fields. I know that the equation for the fields is
$$ X_{|g}=L_{g*}V=\frac{d}{dt}(g\gamma(t)) $$
where $V=\frac{d}{dt}(\gamma(t)) $ is a vetor on $TG$. Now the matrix is the element $g\in G$. What I have been doing is that I use that $X_{|g}\in TG$ and just take a derivative of $g$ with respect to $t$ and then plug into the equation, and then solve for $\gamma$. But I know what the solution is suppose to be and I don't get the corret result by doing that. So I don't know what I am doing wrong here. I would really appriciate some guidance.
Best Answer
The Lie algebra of your Lie group is$$\left\{\begin{bmatrix}x&y\\0&0\end{bmatrix}\,\middle|\,x,y\in\mathbb{R}\right\},$$with the usual Lie bracket. A basis of this Lie algebra is $\{e_1,e_2\}$, with$$e_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}\text{ and }e_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}.$$ Let us concentrate on $e_1$. What is the correspondent left invariant vector field? Fix$$g=\begin{bmatrix}x&y\\0&1\end{bmatrix}$$in your Lie group $G$ and consider the map$$\begin{array}{rccc}L_g\colon&G&\longrightarrow&G\\&h&\mapsto&gh.\end{array}$$Also, consider the curve $\gamma\colon\mathbb{R}\longrightarrow G$ defined by$$\gamma(t)=\exp(te_1)=\begin{bmatrix}e^t&0\\0&1\end{bmatrix}.$$Then$$L_g\bigl(\gamma(t)\bigr)=\begin{pmatrix}e^tx&y\\0&1\end{pmatrix}.$$Differentiating this expression (with respect to $t$) and putting $t=0$ leads to$$\begin{pmatrix}x&0\\0&0\end{pmatrix}.$$So, now you have the left-invariant vector field whose value at the identity element of $G$ is $e_1$. Now, do the same thing with $e_2$ and you will have a basis of the Lie algebra $\frak g$ of the left invariant vector fields.