First of all, it is not true that there are only two types of nondegenerate $3$-forms in $\Lambda^3(V^\ast)$ when $\dim_\mathbb{R}V=6$. There are actually $3$ such nondegenerate orbits. Here is the complete list of $\mathrm{GL}(V)$-orbit types in this space: Let $e^1,\ldots, e^6$ be a basis of $V$. Then the following $3$-forms are inequivalent under $\mathrm{GL}(V)$ and every element of $\Lambda^3(V^\ast)$ is $\mathrm{GL}(V)$-equivalent to exactly one of these:

- $\phi_1 = e^1\wedge e^2\wedge e^3 + e^4\wedge e^5\wedge e^6$
- $\phi_2 = e^1\wedge e^3\wedge e^5 - e^1\wedge e^4\wedge e^6-e^2\wedge e^3\wedge e^6- e^2\wedge e^4\wedge e^5$
- $\phi_3 = e^1\wedge e^5\wedge e^6 + e^2\wedge e^6\wedge e^4+e^3\wedge e^4\wedge e^5$
- $\phi_4 = e^1\wedge e^2\wedge e^5 + e^3\wedge e^4\wedge e^5$
- $\phi_5 = e^1\wedge e^2\wedge e^3$
- $\phi_6 = 0$

The first three types are nondegenerate, and the first two types have open $\mathrm{GL}(V)$orbits, while the orbit of the third type is a hypersurface in $\Lambda^3(V^\ast)$. For a proof of this classical fact, you can see the Appendix in my article *On the geometry of almost complex $6$-manifolds* (available at http://arxiv.org/abs/math/0508428).

The dimension of the annihilators are $0$, $0$, $0$, $1$, $3$, and $6$, respectively. The stabilizers are

- $\mathrm{Stab}^+(\phi_1) = \mathrm{SL}(3,\mathbb{R})\times\mathrm{SL}(3,\mathbb{R})$
- $\mathrm{Stab}^+(\phi_2) = \mathrm{SL}(3,\mathbb{C})$
- $\mathrm{Stab}^+(\phi_3) = \mathrm{GL}(3,\mathbb{R})\ltimes{\frak{sl}(}3,\mathbb{R})$
- $\mathrm{Stab}^+(\phi_4) = \bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^\ast\times \mathbb{R}^\ast\bigr)\ltimes\bigl(\mathbb{R}^4\oplus \mathbb{R}^5\bigr)$
- $\mathrm{Stab}^+(\phi_5) = \bigl(\mathrm{SL}(3,\mathbb{R})\times\mathrm{GL}(3,\mathbb{R})\bigr)\ltimes\bigl(\mathbb{R}^3\otimes \mathbb{R}^3\bigr)$
- $\mathrm{Stab}^+(\phi_6) = \mathrm{GL}(6,\mathbb{R})$

The proofs that these are stabilizers are all relatively straightforward, but, since you asked, I will put in a sketch of the arguments for Cases 4 and 5:

As I wrote in the comment below, Case 5 is very straightforward: Assume the index ranges $1\le i,j,k\le 3 < a, b, c\le 6$. If $\bar e^1,\ldots,\bar e^6$ were any other basis of $V^\ast$ such that $\bar e^1\wedge \bar e^2\wedge \bar e^3 = e^1\wedge e^2\wedge e^3$, then we'd have to have $\bar e^i = p^i_j\ e^j$ for some $3$-by-$3$ matrix $(p^i_j)$ that satisfies $\det(p^i_j)=1$. Then we'd also have to have $\bar e^a = q^a_j\ e^j + p^a_b\ e^b$ for some $3$-by-$3$-matrix $(p^a_b)$ such that $\det(p^a_b)\not=0$, while the $3$-by-$3$ matrix $(q^a_j)$ is arbitrary. Thus, the changes of basis that leave $\phi_4$ fixed are the semidirect product of $\mathrm{SL}(3,\mathbb{R})\times\mathrm{GL}(3,\mathbb{R})$ with the module of $3$-by-$3$ matrices, i.e., $\mathbb{R}^3\otimes \mathbb{R}^3$, as I claimed.

Case 4 is only a little bit more tricky: Write $\phi_5 = (e^1\wedge e^2 + e^3\wedge e^4)\wedge e^5$. Since $e^5$ is the only linear divisor of $\phi_5$, it follows that any linear transformation $L:V\to V$ that preserves $\phi_5$ must preserve $e^5$ up to a multiple, say $\bar e^5 = L^\ast(e^5) = \lambda\ e^5$ for $\lambda\not=0$. Moreover, such a linear transformation $L$ has to then satisfy
$$
L^\ast(e^1\wedge e^2 + e^3\wedge e^4)
\equiv \lambda^{-1}\bigl(e^1\wedge e^2 + e^3\wedge e^4\bigr)\quad\text{mod}\quad e^5
$$
so $L$ must be a conformal symplectic transformation in the appropriate 4-dimensional subquotient of $V^\ast$. Up to restricting to the identity component (which means assuming that $\lambda>0$), this means that $L$ reduces on this subquotient to an element of $\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+$. Lifting back to the space $V$, one sees that $L$ must act on the $5$-dimensional span of $e^1,\ldots, e^5$ as an element of a group that is isomorphic to $\bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+\bigr)\ltimes \mathbb{R}^4$. Finally, note that $L$ can do anything to $e^6$ since it doesn't appear in the formula for $\phi_4$. Thus, one has the identity component of the stabilizer group of $\phi_4$ in the form
$$
\bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+\times \mathbb{R}^+\bigr)
\ltimes \bigl(\mathbb{R}^4\oplus \mathbb{R}^4\oplus \mathbb{R} \bigr)
$$
I think that, with this sketch, you should be able to write out the matrices that define this group and supply the rest of the details yourself.

sense of Hitchinthe same sense as the one in your second sentence? If so, a rephrasing might be helpful on your readers! – Mariano Suárez-Alvarez♦ Feb 7 '13 at 20:57