Lie algebra of $\mathrm{SL}(2, \mathbb{C})$ using method of differential geometry I am asking a new question because people did not understand what  I meant and closed the other question.
I know that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ are traceless matrix and I can proof it. What I am interested is proof it using the methods of differential geometry.
Let $\gamma(t)$  be a curve on a manifold $M$ and $f$ a real function. We define the vector $V$ as the operator
$V\cdot f=\frac{d}{dt}f(\gamma )$
Now let $U$ be the set
$$
U:=\left\{\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \in \mathrm{SL}(2, \mathbb{C}) \mid a \neq 0\right\}
$$
and define the map
$$
\begin{array}{l}
x: \quad U \rightarrow x(U) \subseteq \mathbb{C}^{*} \times \mathbb{C} \times \mathbb{C} \\
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \mapsto(a, b, c)
\end{array}
$$
Te Lie algebra $\mathcal{L}(G),$ of a Lie group is defined by
$$
\mathcal{L}(G):=\left\{X \in \Gamma(T G) \mid \forall g, h \in G:\left(\ell_{g}\right)_{*}\left(\left.X\right|_{h}\right)=X_{g h}\right\}
$$
which can be  proved to be isomorphic to the Lie algebra $T_{e} G$ with Lie bracket
$$
[A, B]_{T_{e} G}:=j^{-1}\left([j(A), j(B)]_{\mathcal{L}(G)}\right)
$$
induced by the Lie bracket on $\mathcal{L}(G)$ via the isomorphism $j$
$$
\left.j(A)\right|_{g}:=\left(\ell_{g}\right)_{*}(A)
$$
Any $A \in T_{\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)} \mathrm{SL}(2, \mathbb{C})$ can be written as
$$
A=\alpha\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\beta\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\gamma\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)},
$$
then we can show that $$j\left(\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{1}}-x^{2} \frac{\partial}{\partial x^{2}}+x^{3} \frac{\partial}{\partial x^{3}}$$
$$j\left(\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{2}}$$
$$j\left(\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{2} \frac{\partial}{\partial x^{1}}+\frac{1+x^2x^3}{x^1}\frac{\partial}{\partial x^{3}}$$
and so
$$
 j(A)=(\alpha x^1+\gamma x^2)\left(\frac{\partial}{\partial x^{1}}\right)+(-\alpha x^2+\beta x^1)\left(\frac{\partial}{\partial x^{2}}\right)+\left(\alpha x^3+\gamma\frac{1+x^2x^3}{x^1} \right) \left(\frac{\partial}{\partial x^{3}}\right) \tag 1
$$
On the other hand we can show that by other method  that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ consist of traceless matrices.
My question is how can we identify $(1)$ with a traceless matrix?
 A: The proof is still really the same as the standard one. But here it is using the definition of a tangent vector as a derivation.
$\newcommand{\SL}{\mathrm{SL}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\gl}{\mathrm{gl}}
\renewcommand{\sl}{\mathrm{sl}}$
Let $\sl(2,\C)$ denote the space of trace-free $2$-by-$2$ complex matrices. Consider the smooth map (which really consists of 4 separate scalar functions):
\begin{align*}
  f: \SL(2,\C) &\rightarrow \gl(2,\C)\\
  A &\mapsto \begin{bmatrix} a & b \\ c & d\end{bmatrix}
\end{align*}
Given any tangent vector $V \in T_I\SL(2,\C)$, there exists a curve $\gamma(t)$ such that $\gamma(0) = I$ and $\gamma'(0) = V$. Then
\begin{align*}
  Vf(I) &= \left.\frac{d}{dt}f(\gamma(t))\right|_{t=0}\\
        &= \begin{bmatrix} a'(0) & b'(0) \\ c'(0) & d'(0) \end{bmatrix}
\end{align*}
Differentiating $\det \gamma(t) = ac-bd = 1$ with respect to $t$ at $t = 0$, we get
$$
a'(0) + d'(0) = 0.
$$
This defines a linear map $T_I\SL(2,\C) \rightarrow \sl(2,\C)$.
Conversely, given any $A' \in \sl(2,\C)$, let $\gamma(t) = (I + tA')/(\det (I+tA'))$. Let $V = \gamma'(0)$. Then, a straightforward calculation shows that
\begin{align*}
  Vf(I) &= \left.\frac{d}{dt}f(\gamma(t))\right|_{t=0}\\
        &= A'.
\end{align*}
This shows that linear map is surjective.
A: From the formula above we can prove that
$$\left[\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=2\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$
$$\left[\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=-2\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$
$$\left[\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} ,\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \right]=\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$$
Now we define the linear map $$\rho \left(\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
1 & 0 \\
0 & -1
\end{array}\right)}=X_1$$
$$\rho \left(\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
 0 & 1 \\
0 & 0
\end{array}\right)}=X_2$$
$$\rho \left(\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}\right)={\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right)}=X_3$$
Now $\left[X_1,X_2 \right]=2X_2 \quad$,$\left[X_1,X_3 \right]=-2X_3\quad$,$\left[X_2,X_3 \right]=X_1\quad$
Since $\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)} \quad$,$\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$ and $\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}$ are basis of this lie algebra  $\rho$ is a representation of the lie algebra of $\mathrm{SL}(2, \mathbb{C})$.
So traceless matrix are representation of the lie algebra of $\mathrm{SL}(2, \mathbb{C})$
