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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.

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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.

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  • $\begingroup$ I followed the path you laid out and I got a result, (0,x; 0,0). But I don't get the result my teacher got, he got the two matrices: (1+t,1; 0,1) and (1,t; 0,1). The answer you got agree more with what I got by myself, might he have messed up? @JoséCarlosSantos $\endgroup$ – Orvar Nov 19 '17 at 16:01
  • $\begingroup$ @Orvar I don't know. Perhaps. By the way: the left invariant vector field that I got whose value at the identity element of $G$ is $f$ is the vector field which, at the same $g$ as above, takes the value$$\begin{pmatrix}0&x\\0&0\end{pmatrix}.$$ $\endgroup$ – José Carlos Santos Nov 19 '17 at 16:07
  • $\begingroup$ How did you find lie algebra of given lie group at the beginning? Can you give a hint? $\endgroup$ – Shreya Nov 30 '18 at 13:07
  • $\begingroup$ I suggest that you ask that as a new question. But if, for some reason, you don't want to do that, tell me so and I will give you a hint. $\endgroup$ – José Carlos Santos Nov 30 '18 at 13:14

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