What is the homomorphism between the third exterior and third symmetric power of the adjoint representation of a simple Lie algebra?

Question

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.

Answer 1

There is a natural homomorphism that I am pretty sure is not zero in general that can be described most easily using the dual language of the Lie algebra $\frak g$:

Recall that there is a differential $\mathrm{d}:{\frak{g}}^\ast\to \Lambda^2({\frak{g}}^\ast)$ that extends uniquely to a degree-one, square-zero derivation of $\frak{g}^\ast$. It is defined by $\mathrm{d}\alpha(x,y) = -\alpha\bigl([x,y]\bigr)$ for all $\alpha\in \frak{g}^\ast$ and $x,y\in \frak{g}$.

Define a mapping $$ P:S^3(\frak{g}^\ast)\to \Lambda^6(\frak{g}^\ast) $$ by the rule $$ P(\alpha_1\circ\alpha_2\circ\alpha_3) = \mathrm{d}\alpha_1\wedge\mathrm{d}\alpha_2\wedge\mathrm{d}\alpha_3\,. $$ Using the Lie bracket one defines the Cartan $3$-form $\phi\in \Lambda^3(\frak{g}^\ast)$ by $$ \phi(x,y,z) = \kappa\bigl(x,[y,z]\bigr), $$ where $\kappa$ is the Killing form of $\frak{g}$. This can be used to define a mapping $$ \Phi: \Lambda^3(\frak{g}^\ast)\to \Lambda^6(\frak{g}^\ast) $$ by $\Phi(\psi) = \phi\wedge\psi$ whose image is a proper subspace of $\Lambda^6(\frak{g}^\ast)$, except for the cases $A_1$ and $A_2$.

Now, using the natural inner product on $\Lambda^6(\frak{g}^\ast)$ induced by the Killing form, define an orthogonal projection $\pi:\Lambda^6(\frak{g}^\ast)\to \Phi\bigl(\Lambda^3(\frak{g}^\ast)\bigr)$. Then $\pi\circ P\bigl(S^3(\frak{g}^\ast)\bigr) =\Phi\bigl(\Lambda^3(\frak{g}^\ast)\bigr)$, and, as long as these two spaces are not zero (and I think that they are not in general), this will define a map of the kind you seek.

$\Phi$ is not injective since $\phi$ is in its kernel, so you will have to use the adjoint of $\Phi$ (using the inner product induced by the Killing form) to construct the map explicitly.

Alternatively, you could just define the pairing $$ \beta: S^3(\frak{g}^\ast)\times \Lambda^3(\frak{g}^\ast)\longrightarrow \mathbb{C} $$ by $$ \beta(p,\psi) = \kappa\bigl(P(p),\Phi(\psi)\bigr) $$ where I have extended $\kappa$ in the usual way to the exterior powers of $\frak{g}^\ast$, and, noting that both spaces are, of course, self-dual, verify that this pairing is not identically zero, which will then give the map.

P.S. I haven't checked all the details of this, but I believe that the following strategy will yield a (highly unsatisfactory) proof that the pairing $\beta$ is nonzero in all cases except $A_1$: First, check, by hand, that the pairing is nonzero for $A_2$, $B_2=C_2$ and $G_2$. (This is not hard to do. $A_2$ is just a few lines, and $B_2$ is only a little longer. I haven't checked $G_2$.) Second, suppose that $\frak{g}$ is simple and that $\frak{h}\subset\frak{g}$ is a simple sub algebra such that the pair $(\frak{g},\frak{h})$ is an irreducible symmetric pair. Using the fact that the Killing form of $\frak{g}$ restricts to be a non-zero multiple of the Killing form of $\frak{h}$ while the Cartan $3$-form of $\frak{g}$ restricts to be a nonzero multiple of the Cartan $3$-form of $\frak{h}$, one shows that the corresponding restriction of $\beta_{\frak{g}}$ is a nonzero multiple of $\beta_{\frak{h}}$. Thus, if $\beta_{\frak{h}}$ is not zero, then $\beta_{\frak{g}}$ is also not zero. Finally, any simple Lie algebra other than $A_1$ and the $C_n$ $(n>2)$ is the end of a chain of symmetric pairs that starts with either $A_2$, $B_2$, or $G_2$, so the result follows for all of these except the $C_n$. However, $C_n$ contains $\mathbb{C}+A_{n-1}$ as a symmetric subalgebra, and I think that the above inductive argument can be extended to this case.