Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$ 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?
 A: Here's an alternate construction of the invariant $\alpha$ that seems a little simpler (and, besides, gets used in proving the normal forms for $3$-forms in $6$-variables). The details may be found in Nigel Hitchin The geometry of three-forms in six and seven dimensions, section 2.1.
Let $S$ be a vector space over $\mathbb{C}$ of dimension $6$, and fix an isomorphism $\Lambda^6(S^*)= \mathbb{C}$ (i.e., choose a volume form). Then there is an induced natural isomorphism 
$$
S = S\otimes \Lambda^6(S^*) = \Lambda^5(S^*)  .\tag1
$$
Given $\phi\in\Lambda^3(S^*)$, define a mapping $J_\phi:S\to S$ by the rule
$$
J_\phi(s) = (\iota_s\phi)\wedge\phi
$$
where $\iota_s\phi\in\Lambda^2(S^*)$ is the interior product of $s\in S$ with $\phi$.  It is easy to see that the trace of $J_\phi\in \mathrm{End}(S,S) = S\otimes S^*$ vanishes identically.  However, if one sets
$$
\alpha(\phi) = \tfrac16\,\mathrm{tr}\bigl((J_\phi)^2\bigr),\tag2
$$
then one finds that $\alpha(\phi)$ does not vanish identically, and it is obviously a quartic polynomial in the coefficients of $\phi$.  In fact, one has the identity
$$
(J_\phi)^2 = \alpha(\phi)\,\mathrm{Id}_S\,,\tag3
$$
and this identity can be used to put $\phi$ in normal form with respect to the eigenspaces of $J_\phi$ when $\alpha(\phi)\not=0$.
A: Another elementary description of the degree four invariant is as
$$
\epsilon_{i_1i_2i_3i_4i_5i_6}\ \epsilon_{i_7i_8i_9i_{10}i_{11}i_{12}}
\ \omega_{i_1i_2i_3\ }\ \omega_{i_4i_5i_7}\ \omega_{i_6i_8i_9}\ \omega_{i_{10}i_{11}i_{12}}
$$
where indices are summed from 1 to 6. The epsilon notation is as in
this MO answer.
A: 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
$$
\alpha(\omega)=\sum_i\langle\omega,\xi_i\omega\rangle\langle\omega,\xi^i\omega\rangle.$$
