Let $G=\operatorname{SL}\_6$ act on $V=\Lambda^3 \mathbb C^6$. I would like to find the ring of invariants $\mathbb C[V]^G$. ~~There is an obvious invariant
$$Sq: V \to \mathbb C, \quad \omega \mapsto \omega^2 \in \Lambda^6 \mathbb C^6 \simeq \mathbb C,$$
with the last isomorphism $\operatorname{SL}\_6$-invariant.~~ **Update:** $Sq=0$, but there is a unique invariant $\alpha$ of degree $4$, not $2$ (see the comment of Robert Bryant below).


My idea how to prove $\mathbb C[V]^G = \mathbb C[\alpha]$ is to consider
$$W=\mathbb Cx\_1 \wedge x\_2 \wedge x\_3 \oplus \mathbb Cx\_4 \wedge x\_5 \wedge x\_6 \subset V.$$
If a general $G$-orbit intersects $W$, then a map $f: \mathbb C[V]^G \to \mathbb C[W]$ is injective. But the image lies in $\mathbb C[W]^{\mathbb C^\* \rtimes \mathbb Z/2\mathbb Z}$, where the action of $\mathbb C^\*$ on $W$ is given by
$$t(w\_1 \cdot x\_1 \wedge x\_2 \wedge x\_3 + w\_2 \cdot x\_4 \wedge x\_5 \wedge x\_6)=tw\_1 \cdot x\_1 \wedge x\_2 \wedge x\_3 + t^{-1}w\_2 \cdot x\_4 \wedge x\_5 \wedge x\_6,$$
and the action of $\mathbb Z/2\mathbb Z$ is given by
$$1(w\_1 \cdot x\_1 \wedge x\_2 \wedge x\_3 + w\_2 \cdot x\_4 \wedge x\_5 \wedge x\_6)=w\_2 \cdot x\_1 \wedge x\_2 \wedge x\_3 - w\_1 \cdot x\_4 \wedge x\_5 \wedge x\_6.$$
Now it is easy to check that $\mathbb C[W]^{\mathbb C^\* \rtimes \mathbb Z/2\mathbb Z} \simeq \mathbb C[\beta]$, where $\beta=w\_1^2 w\_2^2$ is an invariant of degree $4$. Therefore the question is whether a general $G$-orbit intersects $W$ (and also how to construct $\alpha$ with $f(\alpha)=\beta$), and here I am stuck. Could you help me?



Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$
The principal isotropy group is $H=SL(3)\times SL(3)$: it has the right dimension (namely 16) and occurs as an isotropy group (namely of a general element of $W$). Now it is a general result of Luna-Richardson that the restriction map $\mathbb C[V]^G\to\mathbb C[W]^N$ is an isomorphism where $W=V^H$ and $N=N\_G(H)/H$.


For a concrete construction of $\alpha$ it might be easiest to consider $V$ as a symplectic vector space with symplectic form
$$\langle\omega\_1,\omega\_2\rangle:=\frac{\omega\_1\wedge\omega\_2}{x\_1\wedge\ldots\wedge x\_6}.$$
Then there is a moment map given by $$m:V\to\mathfrak g^\*:\omega\mapsto[\xi\mapsto\langle\omega,\xi\omega\rangle].$$
Then $\alpha$ is the pullback of an invariant quadratic form on $\mathfrak g^\*$. Concretely: Let $\xi\_i$ be a basis of $\mathfrak g$ and let $\xi^i$ be its dual basis (i.e. with ${\rm tr}(\xi\_i\xi^j)=\delta\_{ij}$). Then
$$
\alpha(\omega)=\sum\_i\langle\omega,\xi\_i\omega\rangle\langle\omega,\xi^i\omega\rangle.$$


