A semi simple Lie Algebra (LA) $\mathbf{g}$ is usually defined as (I) a direct sum of simple LAs: $\mathbf{g}\,=\,\oplus_i\,\mathbf{g_i}$. An alternative characterization seems to be the statement that (II) $\forall z \in \mathbf{g} \;\exists\,x,y \in \mathbf{g}\,;\,z=[x,y]$.
$sl(2)$ is a simple LA. Thus, I wouldn't expect it to satisfy neither (I) nor (II), unless one considers it as a trivial example of (I). Hence the question, does $sl(2)$ satisfy (II)? To wit, is $sl(2)$ trivially also semi simple?
The following argument seems to prove it so. I copy it below as well. Is it correct?
Thanks for your help.
Is sl(2) trivially also semi simple? Apparently yes, according to the following argument:
$sl(2)$ is a simple Lie Algebra (LA) with base $\{L_1,\,L_2\,,L_3\}$ such that
$\forall z \in sl(2),\, z=z_i\,L^i\;,\;\left[ L^i,L^j \right] =\epsilon^{ij\phantom{k}}_{\phantom{ij}k}\,L^k $
where $\epsilon^{ijk}$ is the Levi-Civitta tensor, $i=1,2,3$ and $z_i\in{\mathbb C}$. Thus the commutator of any two arbitrary elements can be written as
$\left[x,y\right]\,=\,x_i\,y_j\,\left[L^i,L^j\right]\,=\,\epsilon^{\phantom{k}ij}_k\,x_i\,y_j\,L^k\,\equiv\, \left(\overrightarrow{x}\wedge\overrightarrow{y}\right)\cdot\overrightarrow{L} $
where $(\overrightarrow{x}\wedge\overrightarrow{y})^k\equiv\epsilon^{kij}x_iy_j$ (it would be the usual cross-product if $x^i,y^i\in\mathbb{R}^3$. Here we just take it as a concise notation as we are working in $\mathbb{C}$).
Given an arbitray element $z\in sl(2)$, can we find $x,y\in sl(2)$ such that $z=\left[x,y\right]$? Be $z^i=\alpha^i\,+\,j\,\beta^i\,\in\,\mathbb{C}$ and $\overrightarrow{\alpha}=(\alpha^1,\alpha^2,\alpha^3)$ and an analogous definition for $\overrightarrow{\beta}$. Let's first consider the case where $\overrightarrow{\alpha}\nparallel\overrightarrow{\beta}$. Notice that this can only make sense if $z$ is a linear combination of at least two base elements $L_i$. We can then always define $\overrightarrow{x}=\overrightarrow{\alpha}\wedge\overrightarrow{\beta}$, then it is
$ z_ix^i\,=\,\epsilon^{i\phantom{jk}}_{kl}\,z_i\,\alpha^k\,\beta^l\,=\, \epsilon^{i\phantom{jk}}_{kl}\,\alpha_i\,\alpha^k\,\beta^l +\,j\, \,\epsilon^{i\phantom{jk}}_{kl}\,\alpha^k\,\beta^l\beta_i\,=\,0 $
Introducing then $\overrightarrow{y}\equiv\overrightarrow{z}\wedge\overrightarrow{x}$ we have
$ \left(\overrightarrow{x}\wedge\overrightarrow{y}\right)^k\,=\,\epsilon^{k\phantom{ij}}_{\phantom{k}ij}\,x^i\,\epsilon^{j\phantom{lm}}_{\phantom{j}lm}\,z^l\,x^m\, =\,\left(\delta_{kl}\delta_{im}-\delta_{km}\delta_{li}\right)\,x^i\,z^l\,x^m\, =\,z^k\,(x_i\,x^i)\,-\,x^k\,(z_i\,x^i)\, =\,z^k\,(x_i\,x^i)\,=\,(\overrightarrow{z})^k\,(x_i\,x^i) $
Rescaling $x$ and $y$ by $\lambda\equiv\sqrt{x_ix^i}$, we have that $z=[x,y]$.
In case $\overrightarrow{\beta}=\lambda\overrightarrow{\alpha}$, we may choose $\overrightarrow{x}=(\alpha^2,-\alpha^1,0)$, asuming $\alpha^1\neq 0\,,\,\alpha^2\neq 0$, and proceed in the same way. In this case it is also satisfied that
$ z_i\,x^i\,=\,z_1\,\alpha^2\,-\,z_2\,\alpha^1\,=\,\alpha^1\,(1+j\lambda)\,\alpha^2\, -\,\alpha^2\,(1+j\lambda)\,\alpha^1\,=\,0 $
and again it is
$ \left(\overrightarrow{x}\wedge\overrightarrow{y}\right)^k\,=\,z^k\,(x_i\,x^i). $
Finally, let's consider the case where $z$ is a multiple of one base element, e.g., $z=z_1\,L_1\,=\,\alpha_1\,(1+j\lambda)\,L_1$. Be $x=L_2$, then $z_ix^i=0$ and $y=z_1\,L_3$. Indeed, it is $[x,y]=z$.
Hence, for $sl(2)$, every element $z$ can be written as the commutator of two given elements $x,y\in sl(2)$. $\qed$
Yes, it is true. It is basically the fact that in the exterior square of a $3$-dimensional subspace, every element is a decomposable tensor.
From the multiplication table of $\mathfrak{g} = \text{sl}(2)$ one sees immediately that $\mathfrak{g} = [\mathfrak{g}, \mathfrak{g}]$, so that if $x_1, x_2, x_3$ is a basis of $\mathfrak{g}$, by bilinearity every element of $\mathfrak{g}$ can be written as $$ a_3 [x_1, x_2] + a_2 [x_1, x_3] + a_1 [x_2, x_3]. $$ Rewrite as $$ [x_1, a_3 x_2 + a_2 x_3] + a_1 [x_2, x_3]. $$ If $a_2 = a_3 = 0$, we're done. So suppose WLOG $a_2 \ne 0$. Then rewrite the above as $$ [x_1, a_3 x_2 + a_2 x_3] + a_1 a_2^{-1} [x_2, a_3 x_2 + a_ 2x_3] = [x_1 + a_1 a_2^{-1} x_2, a_3 x_2 + a_2 x_3], $$ done.