Let $\mathfrak{g}$ be the adjoint representation of a simple Lie algebra (which is not of type $A$). Then the space of intertwiners between the third exterior power of $\mathfrak{g}$ and the third symmetric power of $\mathfrak{g}$ has dimension one. I have checked on a case-by-case basis using LiE that these two representations have a single irreducible component in common.

I am asking for a natural non-zero element in this space; where by natural I mean constructed from the Lie bracket, permutations, the Killing form and its adjoint. If you can give this in diagrammatic notation that would be even better.

The question stands independently of the motivation. However my motivation is that I am trying to understand Drinfeld's "terrific relation" in the definition of the Yangian. This relation can be interpreted as saying that two actions of the common irreducible representation agree.

I am looking at "A guide to quantum groups" by Chari & Pressley where the "terrific relation" is (4) in Theorem 12.1.1 page 376 (Google Books).

A related question is Lie algebra cohomology.

The constructions work perfectly well in type $A$. The reason type $A$ is exceptional (!) is that in type $A$ (and only in this case) $\mathfrak{g}$ appears as a composition factor of the symmetric square $S^2\mathfrak{g}$. The projection is given by $$ x\otimes y \mapsto xy+yx -\frac{2}{n}\mathrm{trace}(xy).1 $$ where $x,y$ are $n\times n$ matrices with zero trace.

This muddies the water.

**Update** I was confused when I asked the original question. The dimension
of the space of intertwiners seems to 2 for the exceptional groups, 3 for classical groups and 4 for $\mathfrak{sl}(n)$.

However the representation that is relevant in this construction of the Yangian is the kernel of the Lie bracket. This is an irreducible representation of $\mathrm{Aut}(\mathfrak{g})$ and is nonzero except in type $A_1$. Since $H^2(\mathfrak{g})=0$ this representation also appears in the third exterior power.

Then the right hand side of the "terrific relation" shows that this representation also appears in the third symmetric power. This is explained in Robert Bryant's answer.

highest weightof the common irreducible constituent, to see whether this exhibits a predictable pattern. $\endgroup$ – Jim Humphreys Aug 6 '15 at 13:21