how to extend a vector at $e$ of a Lie group to a left invariant vector field? 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.
 A: 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.
