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Consider the set of matrices $$G:=\left\{ \begin{pmatrix}x_1&0&x_2\\0&x_1&x_3\\0&0&1 \\\end{pmatrix}:x_1>0,x_2,x_3 \in \mathbb{R}\right\}$$ Define the single chart that maps such a matrix into $(x_1,x_2,x_3) \in (0, \infty) \times \mathbb{R}^2$. Justify G is a subgroup of $\text{GL}(3,\mathbb{R})$. Show that vector fields in local coordinates $\zeta_1(x_1,x_2,x_3)=(x_1,0,0),\zeta_2(x_1,x_2,x_3)=(0,x_1,0),\zeta_3(x_1,x_2,x_3)=(0,0,x_1)$ constitute a basis of the space of left invariant vector fields.

How do i even define a chart for this kind of problem ? please provide hints for tackling this problem

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    $\begingroup$ The chart is simply given by $$\begin{pmatrix}x_1&0&x_2\\0&x_1&x_3\\0&0&1 \\\end{pmatrix} \mapsto (x_1, x_2, x_3).$$ $\endgroup$ Mar 10, 2020 at 16:04

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A chart is a map (more precisely, a homeomorphism) $\varphi \colon G \supset U \to V \subset \mathbb{R}^n$ for some $n \in \mathbb{N}$. The subsets $U$ and $V$ are open subsets of $G$ and $\mathbb{R}^n$, respectively. Here, $n = 3$. We can define a chart by$$\varphi \colon G \to \mathbb{R}^3 \colon \begin{pmatrix}x_1 & & x_2 \\ & x_1 & x_3 \\ & & 1\end{pmatrix} \longmapsto \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}.$$One can easily see that this already covers all of $G$, i.e. $U=G$, and that this particular choice of $\varphi$ is indeed a homeomorphism.

Now to proof the fact about the basis of the space of left invariant vector fields, you just have to evaluate both sides of the following equation to check the definition of left invariance:$$D l_g \zeta_k (x_1 , x_2 , x_3) = \zeta_k \big( l_g (x_1 , x_2 , x_3) \big).$$Here $g\in G$ is an arbitrary element, $l_g \colon h \mapsto g \cdot h$ is the left translation, $D$ denotes the push forward, $x_1 >0 , x_2,x_3 \in \mathbb R$, and $k$ is between one and three. How does $\zeta_k (x_1 , x_2 , x_3)$ look like? It's an element of $T_x G, (x_1,x_2,x_3) = x \in G$, therefore it is an equivalence class of differentiable paths $[\gamma]$. So to check above equation, you just need to check it for one of those paths $\gamma \colon (-\varepsilon,\varepsilon) \to G$. For example, a path giving $\zeta_1 (x) = \zeta_1 (x_1 , x_2 , x_3)$ is given by $$\gamma_x \colon t \longmapsto \begin{pmatrix} t x_1 & &x_2 \\ & tx_1 & x_3 \\ & & 1\end{pmatrix}.$$

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  • $\begingroup$ Yes this is makes sense, thank you $\endgroup$
    – Siddhartha
    Mar 12, 2020 at 3:57

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