Find all finite dimensional simple Lie algebras satisfying certain conditions Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$. Suppose that $\mathfrak{g}$ can be generated by five nonzero elements $x,y,x',y',h\in \mathfrak{g}$, which satisfy the following relations:
$$[x,y]=h, [h,x]=2x, [h,y]=-2y\tag{1}$$
$$[x',y']=h, [h,x']=2x', [h,y']=-2y'\tag{2}$$
$$[x,x']=[y,y']=0\tag{3}$$
Basically, $\{x,y,h\}$ and $\{x',y',h\}$ each form a $\mathfrak{sl}(2)$, sharing the same CSA, and with $[x,x']=[y,y']=0$. We can represent the relations graphically as follows:

Question: what are all possibilities of $\mathfrak{g}$? My conjecture is that $\mathfrak{g}$ can only be $\mathfrak{sl}(2)$, with $x=\lambda x', y'=\lambda y$, for some $\lambda\in \mathbb{C}$, but I have no idea how to prove it.
[A vague idea: maybe we can first prove that there exist $f\in \mathrm{Aut}(\mathfrak{g})$ with $f(x)=x',f(y)=y',f(h)=h$, and then use our knowledge of automorphisms of simple Lie algebra to restrict the possibilities of $\mathfrak{g}$?]
 A: EDIT: [Failed argument follows in the first paragraph, left here for posterity.] I think this is quite straightforward - in any finite-dimensional quotient ${\mathfrak s}$ of your Lie algebra, $h$ is semisimple (since it acts via scalar multiplication of each generator) and $x,y,x',y'$ are nilpotent (since they are eigenvectors for $h$, with non-zero eigenvalues). Now all eigenvalues of $({\rm ad}\, h)$ on ${\mathfrak s}$ are real (since this is true for the generators) so there exists a Cartan subalgebra ${\mathfrak h}$ containing $h$, and (assuming ${\mathfrak s}$ is simple) a positive system $\Phi^+$ in $\Phi({\mathfrak s},{\mathfrak h})$ such that all weights of $h$ on $\Phi^+$ are non-negative. In particular, this means $x,x'$ are sums of positive root vectors and $y,y'$ are sums of negative root vectors. It follows immediately that $h$ spans ${\mathfrak h}$ [this doesn't actually follow], so ${\mathfrak s}$ is of rank 1, so equals $\mathfrak{sl}_2$. Hence $x$ and $x'$ both map to some (non-zero) multiple of $e$, and we get your ideals $x-\lambda x'$, $y'-\lambda y$. (You need $\lambda\neq 0$.)
I couldn't help making an observation. A more natural presentation of your Lie algebra is:
$$[e_1,f_1]=h_1,\;\; [h_1,e_1]=2e_1,\;\; [h_1,f_1]=2f_1,$$
$$[e_2,f_2]=h_2,\;\; [h_2,e_2]=2e_2, \;\; [h_2,f_2]=-2f_2,$$
$$[e_1,f_2]=[e_2,f_1]=0,\;\; h_1+h_2=0$$
where $e_1=x$, $f_1=y$, $e_2=y'$, $f_2=x'$.
In other words, $x$ and $y'$ are the positive root elements, and $x'$ and $y$ are the negative root elements (different from the above). This is certainly an infinite-dimensional Lie algebra. If we quotient further by the conditions $$({\rm ad}\, e_1)^3(e_2)=({\rm ad}\, e_2)^3(e_1)=0$$ and similarly for $f_1, f_2$ then we obtain the loop algebra $\mathfrak{sl}_2\otimes{\mathbb C}[t,t^{-1}]$. (This follows from the Gabber-Kac theorem - these relations without the condition $h_1+h_2=0$ generate the derived subalgebra of the affine Kac-Moody algebra of type $A_1$; then $h_1+h_2$ is in the centre, and taking the quotient by it produces the loop algebra. Going in reverse, one would construct the affine $A_1$ Lie algebra as a central extension of $\mathfrak{sl}_2\otimes{\mathbb C}[t,t^{-1}]$.)
The map from your Lie algebra to $\mathfrak{sl}_2\otimes{\mathbb C}[t,t^{-1}]$ is given by: $$x\mapsto e\otimes 1,\; y\mapsto f\otimes 1,\; h\mapsto h\otimes 1,\; y'\mapsto f\otimes t,\; x'\mapsto e\otimes t^{-1}.$$
FURTHER EDIT: Ok, the loop algebra does actually help. If ${\mathfrak s}$ is a finite-dimensional simple quotient of your Lie algebra, then the images of $x,y,h$ span an $\mathfrak{sl}_2$-triple, and $x'$ is in the 2-weight space for $h$, and we have $[x,x']=0$, so $x'$ is a highest weight vector for the $\mathfrak{sl}_2$. It follows that $({\rm ad}\, y)^3(x')=0$. Exactly the same argument establishes $({\rm ad}\, y')^3(x)=0$ etc. so in fact any finite-dimensional quotient of your Lie algebra factors through the surjection onto the loop algebra. I think the only finite-dimensional simple quotient of the loop algebra is $\mathfrak{sl}_2$.
