Basis for adjoint representation of $sl(2,F)$ Consider the lie algebra $sl(2,F)$ with standard basis $x=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, $j=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $h=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. 

I want to find the casimir element of the adjoint representation of $sl(2,F)$. How can I go about this? Thanks.

 A: A representation for a Lie algebra $\mathfrak{g}$ is a Lie algebra homomorphism $\varphi:\mathfrak{g} \to \mathfrak{gl}(V)$ for some vector space $V$.
Of course, every representation corresponds to a module action. In the case of this representation the module action would be $g \cdot v = \varphi(g)(v)$.
It is not clear what you mean by "basis for the representation". Do you mean a basis for the linear transformations $\varphi(g)$? That would be a basis for $\varphi(\mathfrak{g})$ (the image of $\mathfrak{g}$ in $\mathfrak{gl}(V)$). Or do you mean a basis for the module $V$?
The adjoint representation is the map $\mathrm{ad}:\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ defined by $\mathrm{ad}(g)=[g,\cdot]$. In the case that $\mathfrak{g}=\mathfrak{sl}_2$, $\mathfrak{g}$ has a trivial center so $\mathrm{ad}$ is injective. Thus a basis for $\mathfrak{g}$ maps directly to a basis for $\mathrm{ad}(\mathfrak{g})$.
Therefore, if by "basis for the representation" you mean a basis for the space of linear transformations $\mathrm{ad}(\mathfrak{sl}_2)$, then "Yes" $\mathrm{ad}_e, \mathrm{ad}_f,$ and $\mathrm{ad}_h$ form a basis for this space.
On the other hand, if you mean "basis for the module" then $e,f,$ and $h$ themselves form a basis for $V=\mathfrak{sl}_2$.
By the way, if you are looking for matrix representations for $ad_e,ad_f,ad_h$ relative to the basis $e,f,h$. Simply compute commutators: $[e,e]=0$, $[e,f]=h$, $[e,h]=-2e$. Thus the coordinate matrix of $ad_e$ is $$\begin{bmatrix} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$
