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?

isa nonzero invariant polynomial under $\mathrm{SL}(6,\mathbb{C})$ and itdoesgenerate the ring of invariants, but it is irreducible of degree $4$, not degree $2$. $\endgroup$ – Robert Bryant Dec 20 '16 at 21:49