Let $M$ be a closed $6$-dimensional Riemannian manifold with a spin$^{\mathbb{C}}$ structure. It is known that real $4$-forms on $M$ act on the positive-spinors as trace-free hermitian endomorphisms by Clifford multiplication (say, denoted by $\gamma$). Now for a real $3$-form $\beta$ on $M$, one can see that $\gamma(\beta)^2-|\beta|^2\mathrm{Id}$ is trace-free and hence must represent a $4$-form on $M$. Can we identify this $4$-form?



Identifying a $4$-form on a $6$-dimensional manifold
$\newcommand{\R}{\mathbb{R}}$As you state, there is a $SO(6)$-equivariant map $\delta:\operatorname{Sym}^2(\Lambda^3\R^6)\to \Lambda^4 \R^6$ such that $\gamma(\delta(\beta^{\otimes 2})) = \gamma(\beta)^2 - |\beta|^2\operatorname{id}$ on the positive spinors (the central $U(1)\subset\operatorname{Spin}^c(6)$ cancels, so that it suffices to work with the quotient $SO(6)$). By adjunction, this corresponds to an invariant tensor in $\operatorname{Sym}^2(\Lambda^3 \R^6)\otimes\Lambda^4\R^6\subset (\R^6)^{\otimes 10}$. It is a classical fact, known as the first fundamental theorem for the orthogonal group (compare [this answer](https://mathoverflow.net/questions/238788/invariant-polynomials-under-diagonal-action-of-the-orthogonal-group)), that these invariants have a basis spanned by perfect matchings of the ten indices into five pairs, which are then contracted with the metric tensor. The (anti-)symmetry constraints then exhibit the representation $\operatorname{Sym}^2(\Lambda^3 \R^6)\otimes\Lambda^4\R^6$ as a quotient, in which many of these pairings vanish by antisymmetry; with a little work, one finds that the space of invariants is $1$-dimensional, corresponding to the map $\operatorname{Sym}^2(\Lambda^3\R^6)\to \Lambda^4 \R^6$ which contracts two tensor indices of the two inputs with the metric and then antisymmetrizes the remaining 4 indices. So the map $\delta$ must be a constant multiple of this map, and the constant can be easily determined by considering $\beta = e\_1\wedge e\_2\wedge e\_3 + e\_1\wedge e\_4\wedge e\_5$.


