I am reading some books about Lie group and Lie algebra. Denote the set of all the left invariant vector fields as $\mathfrak{X}_L$, and the tangent space at $e$ of $G$ as $T_eG$. They say that the $\mathfrak{X}_L$ and $T_eG$ are isomorphics. So we can extend a vector $\xi\in T_eG$ to the vector field on $G$ by this way: $$X(g)=dL_g(\xi),\mbox{for any }g\in G$$
where $X(g)$ is a vecotr at $g$, $L_g:G\rightarrow G$ is the left translation, and $dL_g:T_eG\rightarrow T_gG$ is the pushforward. My question is what is $dL_g(\xi)$ exactly?
Thank you.
Your first example may not be so illustrative. Since $$ \begin{pmatrix} 1 & 0 & a_1 \\ 0 & 1 & a_2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & x_1 \\ 0 & 1 & x_2 \\ 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} 1 & 0 & a_1+x_1 \\ 0 & 1 & a_2+x_2 \\ 0 & 0 & 1 \end{pmatrix} $$ your Lie group is isomorphic to $\mathbb{R}^2$ under addition. But if you think of those two positions as coordinates $y_1$ and $y_2$, then \begin{align*} \frac{\partial y_1}{x_1} &= 1 &\frac{\partial y_1}{x_2} &= 0\\ \frac{\partial y_2}{x_1} &= 0 &\frac{\partial y_1}{x_2} &= 1 \end{align*} So for each $(a_1,a_2)$, the map $dL_{(a_1,a_2)}$ is the identity map.
A better example might be \begin{align*} G &= SU(2) = \left\{\begin{bmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{bmatrix} : \alpha,\beta\in \mathbb{C},\ |\alpha^2| + |\beta^2| = 1\right\} \\ \mathfrak{g} &= \mathfrak{su}(2) = \left\{\begin{bmatrix} ix & -\bar\beta \\ \beta & -ix \end{bmatrix} : x\in\mathbb{R},\beta\in \mathbb{C}\right\} \end{align*} Let $\xi = \begin{pmatrix} i & 0 \\ 0 &-i\end{pmatrix} \in \mathfrak{g}$. Then $\xi$ is the derivative at $t=0$ of the path in $SU(2)$ $$ \gamma(t) = \exp\begin{pmatrix} it & 0 \\ 0 &-it\end{pmatrix} = \begin{pmatrix} e^{it} & 0 \\ 0 &e^{-it}\end{pmatrix} $$ So for $g = \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix} \in G$, we have \begin{align*} L_g(\xi) &= \left.\frac{d}{dt} \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix} \begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix}\right|_{t=0} \\ &= \left.\frac{d}{dt} \begin{pmatrix} \alpha e^{it} & -\bar\beta e^{-it} \\ \beta e^{it} & \bar\alpha e^{-it} \end{pmatrix}\right|_{t=0} \\ &= \left.\begin{pmatrix} \alpha e^{it}(i) & -\bar\beta e^{-it}(-i) \\ \beta e^{it}(i) & \bar\alpha e^{-it}(-i) \end{pmatrix}\right|_{t=0} \\ &= \begin{pmatrix} \alpha i & \bar\beta i \\ \beta i & -\bar\alpha i\end{pmatrix}\\ &= \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix}=g\xi\\ \end{align*}
Also, I'm not sure what your pictures are supposed to represent.