Let $\mathfrak{g}$ be the complexified Lie algebra of a simple compact real Lie group. Then the adjoint and coadjoint representation are isomorphic (see Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?).
In the representation category of $\mathfrak{g}$, we naturally have an intertwiner $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ (viewing $\mathfrak{g}$ as the adjoint representation) given by the Lie bracket. My question is wether simplicity and compactness are sufficient conditions for the Lie bracket to be the only such intertwiner (up to a constant factor). Phrased differently, when does $F: \mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ with
\begin{align}
F([x,y] \otimes z) + F(y \otimes [x,z]) = [x,F(y \otimes z)]
\end{align}
for all $x,y,z \in \mathfrak{g}$ imply $F(y \otimes z) = c \cdot [y,z]$, with $c \in \mathbb{C}$?
I already tried to show that $[[y,z],F(y \otimes z)]=0$, but I can't even get that done.
I also tried working in the basis mentioned in the page I reference. Then the formula becomes
\begin{align}
\forall r \in \{1 \ldots \dim \mathfrak{g}\}: \sum_{n=1}^{\dim \mathfrak{g}} \langle [e_k,e_i]f^{(nj)}_r + [e_k,e_j]f^{(in)}_r + [e_k,e_r]f^{(ij)}_n , e_n \rangle = 0
\end{align}
where
$F(e_i \otimes e_j) = \sum_{k=1}^{\dim \mathfrak{g}} f^{(ij)}_k e_k$, and the inner product is defined as $\langle x,y \rangle = \text{Tr}(\text{Ad}_x \text{Ad}_y^*)$, where the star is just taking a matrix adjoint in the basis $\{e_i\}$. However, I don't know how to proceed from here.
Any help is welcome.
Best Answer
Turning comments into an answer:
For any semisimple complex Lie algebra $\mathfrak g$, if $W$ is any finite-dimensional representation of $\mathfrak g$, then $W$ decomposes into a finite direct sum $W \simeq \bigoplus_i V_i^{m_i}$ where the $V_i$ are mutually isomorphic irreducible (=simple) representations with multiplicities $m_i \ge 1$. Further, if $V$ is any irreducible representation of $\mathfrak g$, then either $$Hom_{\mathfrak g} (V,W) = Hom_{\mathfrak g} (W,V) = 0,$$ in which case $V \not \simeq V_i$ for all $i$; or, if $V \simeq V_i$, then $$dim(Hom_{\mathfrak g} (V,W)) = dim(Hom_{\mathfrak g} (W,V)) = m_i.$$ (This is essentially the definition of semisimple representation plus Schur's Lemma.)
Now in our case we let $\mathfrak g$ be simple, and $V$ the adjoint representation (well-known to be simple). According to https://mathoverflow.net/q/129857/27465 and https://mathoverflow.net/q/130185/27465 (or case-by-case check in LiE), the multiplicity of $V$ in $V \otimes V$ is $1$, except in the case of $\mathfrak g \simeq \mathfrak sl_{n \ge 3}$, in which it is $2$.
(Funnily, the MO posts leave a bit of a gap for showing the multiplicity is $\ge 1$: which is exactly what your question shows, namely, the Lie bracket is an obvious non-trivial element of $Hom_{\mathfrak g}(V \otimes V, V)$.)
So in all cases except $\mathfrak g \simeq sl_{n \ge 3}$, the result is true, again due to Schur's Lemma.
In the case $\mathfrak g \simeq sl_{n \ge 3}$, the missing dimension is given by the intertwiner
$$F_{extra}: x\otimes y \mapsto xy + yx - \frac1n Tr(xy+yx)$$
(the "traceless part of $xy +yx$", as Allen Knutson writes in the MO post; note that this is the composition of the map $x\otimes y \mapsto xy+yx$, whose codomain is $\mathfrak{gl}_n$, with the projection $\mathfrak{gl}_n \twoheadrightarrow \mathfrak{sl}_n, z \mapsto z - \frac1n Tr(z)$).
A little computation indeed shows that $F_{extra}$ is an element of $Hom_{\mathfrak g}(V \otimes V, V)$, and is linearly independent from the intertwiner given by the Lie bracket for all $n \ge 3$ (it does vanish for $n=2$, matching earlier results). What happens there seems to be a decomposition of $V \otimes V$ into symmetric versus alternating tensors and what's special in this case is that there is an extra copy of $V$ within the symmetric ones. If, on the other hand, one tries to imitate the map $F_{extra}$ e.g. in the case of $\mathfrak{so}_n$, it is not even well defined i.e. the image does not lie in $\mathfrak{so}_n$ etc.