In the same spirit of my question here I want to decompose $\mathfrak{sl}_2 \otimes \mathfrak{sl}_2$ into a direct sum of irreducible representations of $\mathfrak{sl}_2$. There is a unique irreducible representation of dimension of $\mathfrak{sl}_2$ $n\in\mathbb N^*$ which is denoted $V(n)$.
I know that $W\otimes W=S^2W\oplus \bigwedge^{2}W$, here if we write the basis $\{e,f,h\}=\{v_0,v_1,v_2\}$ the exterior power would be generated by $\{e\otimes f,e\otimes h,f\otimes h\}$ and the symmetric part by the remaining tensor products between elements of the basis.
We see here that $\mathfrak{sl}_2\otimes \mathfrak{sl}_2\cong V(4)\oplus V(2)\oplus V(0)$. I tried to show that $\bigwedge^2 \mathfrak{sl}_2$ is irreducible by saying that if there is a sub representation of dimension 1, it corresponds to the trivial one so if $\sum_{i,j}a_{ij}v_i\otimes v_j$ spans this space we must have $$[e,u]=\sum_{i,j}a_{ij}([e,v_i]\otimes v_j+v_i\otimes [e,v_j])$$ and from this I get that $a_{ij}=0$ $\forall i,j$.
So I concluded that it's irreducible, so the 1 dimensional subrepresentation must be in $S^2\mathfrak{sl}_2$, but by doing the same process with an element of $S^2\mathfrak{sl}_2$ I get that there is no one dimensional subrepresentation in $S^2\mathfrak{sl}_2$.
So how do we get the decomposition here ?