I am doing an exercise to find the structures constants of $sl(2,F)$ given the basis elements $x_1 = \begin{bmatrix} 0, 1\\ 0,0\end{bmatrix}, x_2 = \begin{bmatrix} 0 , 0 \\ 1, 0\end{bmatrix}, x_3 = \begin{bmatrix} 1 , 0 \\ 0, -1\end{bmatrix}$.
I thought I should apply the jacobi identity to all possible pairs of the basis elements, but this has been a lot of writing and I do not see how I am supposed to determine the values of the coefficients. I eventually get https://i.stack.imgur.com/LtJul.png I can group by $x_1,x_2,x_3$, but after that I do not know what could be done.
I also tried examining the skew-symm property.
$[x_1, x_2] = a_{12}x_1 + a_{12}^{2}x_2 + a_{12}^{3}x_3$. So then $[x_1, x_2] = -[x_2, x_1]$ yields $a_{12}x_1 + a_{12}^{2}x_2 + a_{12}^{3}x_3 = -(a_{12}x_1 + a_{12}^{2}x_2 + a_{12}^{3}x_3)$, but doesn't that imply that each coefficient $a_{12}^{i}$, for $i =1,2,3$ is just $0$? Because we get $a_{12} = - a_{12}$ and such.
Can I please have a kick in the right direction?
Since $x_1, x_2, x_3$ are a basis of the Lie algebra $sl(2,\mathbb{F})$, every element can be represented by a linear combination of them. In particular the bracket of two elements can be represented this way. $$ [x_i,x_j]=a_{ij}^1x_1+a_{ij}^2x_2+a_{ij}^3x_3, \quad a_{ij}^1,a_{ij}^2,a_{ij}^3 \in \mathbb{F} $$ These numbers $a_{ij}^1,a_{ij}^2,a_{ij}^3$ are called the structure constants. For $i=1,j=2$, we get for example $$ [x_1,x_2]=\left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) - \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) = 1 \cdot x_3, $$ so the structure constants are $a_{12}^1 = 0, a_{12}^2 = 0,a_{12}^3 = 1$. Can you calculate the other ones?