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 $$ \alpha(\omega)=\sum_i\langle\omega,\xi_i\omega\rangle\langle\omega,\xi^i\omega\rangle.$$

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 $\alpha(\phi)\not=0$.

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.