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N.Hitchen in his paper about geometry of three forms wrote that "for a Real vector space $V$ of dimension six, the group $GL(6,V)$ has an open orbit and he referenced it to a thesis which was written in 1907 , how can we prove this fact, is there any open access reference?

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This is very well explained in Hitchin's article freely available at – BS. Jan 13 '13 at 12:59
he only referenced it and never proved it, in which page of this paper he very well explained it? – baba ab dad Jan 13 '13 at 13:06
The proof is given in section 2, pages 3 to 6, first the complex case, then in the real case. – BS. Jan 15 '13 at 11:05
Is the thesis available online? – Mariano Suárez-Alvarez Apr 13 '13 at 20:37
Mariano , No,I think this thesis has been written around 1930. But the proof of Robert is very nice and unique. – baba ab dad Apr 13 '13 at 21:16
up vote 12 down vote accepted

In any case, the proof is very simple. Consider the $3$-form $$ \phi_0 = dx^1\wedge dx^2\wedge dx^3 + dx^4\wedge dx^5\wedge dx^6. $$ I claim that the subgroup $G\subset\mathrm{GL}(6,\mathbb{R})$ that stabilizes $\phi_0$ consists of the obvious subgroup $G_0=\mathrm{SL}(3,\mathbb{R})\times\mathrm{SL}(3,\mathbb{R})$ that fixes the two summands plus the discrete part that switches the two summands. Since this subgroup has dimension $8{+}8=16$ while $\mathrm{GL}(6,\mathbb{R})$ has dimension $36$, it follows that the $\mathrm{GL}(6,\mathbb{R})$-orbit of $\phi_0$ has dimension $20= 36-16$, which is the dimension of $\Lambda^3(\mathbb{R}^6)$. Thus, this orbit is open.

To see the claim, just look at the vectors $v\in \mathbb{R}^6$ that satisfy $\bigl(\iota_v(\phi_0)\bigr)^2=0$. Clearly, such a vector must lie in either the 'first' $\mathbb{R}^3$ or the 'second' $\mathbb{R}^3$. Thus, $G$ must carry each of these subspaces into either itself or the other subspace. The subgroup of $G$ that fixes each subspace must be of index $2$ in $G$, and is clearly $G_0$.

What is not so obvious is that there exist exactly $2$ open orbits. One is the one listed above, and the other is $$ \phi_1 = \mathrm{Re}\bigl((dx^1+i\ dx^4)\wedge(dx^2+i\ dx^5)\wedge(dx^3+i\ dx^6)\bigr), $$ whose stabilizer has an index two subgroup that is isomorphic to $\mathrm{SL}(3,\mathbb{C})$.

Remark: By the way, the existence of the open orbit shouldn't be surprising. It's a general fact that $\mathrm{GL}(n,\mathbb{R})$ has an open orbit in $\Lambda^k(\mathbb{R}^n)$ for $k>0$ whenever $n^2$, the dimension of $\mathrm{GL}(n,\mathbb{R})$, is as large as $n\choose k$, the dimension of $\Lambda^k(\mathbb{R}^n)$. This happens for all $n$ when $k\in\lbrace1,2,n{-}2,n{-}1,n\rbrace$ and otherwise only for $$ (n,k)\in\lbrace (6,3), (7,3), (7,4), (8,3), (8,5)\rbrace. $$

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Dear Robert, you beautifully solved it, thanks – baba ab dad Jan 13 '13 at 17:33
To see there are exactly 2 open orbits, we can use algebraic-group methods as follows. Let $G={\rm{GL}}_6$. By considering tangent spaces analytically (over $\mathbf{R}$ and $\mathbf{C}$) and algebraically (over $\mathbf{C}$), and dimensions of stabilizer groups analytically and algebraically, a point $\omega$ has open orbit under $G(\mathbf{R})$ in the analytic sense if and only if it has Zariski-open $G$-orbit in the algebraic sense (over $\mathbf{C}$, or use $\mathbf{R}$-schemes). By the nature of the Zariski topology, the open algebraic orbit through $\phi_0$ is the only one. (cont'd) – user29720 Jan 13 '13 at 19:57
So the open orbit as an $\mathbf{R}$-variety is $G/H$ with $H = ({\rm{SL}}_3 \times {\rm{SL}}_3) \rtimes (\mathbf{Z}/(2))$. By Cor. 1 to Prop. 36 in Ch. I of Serre's "Galois cohomology" book, $G(\mathbf{R})\backslash (G/H)(\mathbf{R})$ is identified with the kernel of the map ${\rm{H}}^1(\mathbf{R},H)\rightarrow {\rm{H}}^1(\mathbf{R},G)$. The target vanishes by Hilbert 90, and the source bijects onto ${\rm{H}}^1(\mathbf{R},H/H^0)$ of size 2 since $H^0$ has only two $\mathbf{R}$-forms (itself and "${\rm{SL}}_3(\mathbf{C})$ as an $\mathbf{R}$-group") and both have trivial degree-1 cohomology. – user29720 Jan 13 '13 at 20:21
@Hassan: If you like the answer (and you probably like it, because you have accepted it), you can also vote it up! – Mikhail Borovoi Jan 13 '13 at 20:23
@kreck: Thanks for outlining this sophisticated proof. However, one doesn't need such tools to classify the orbits (even the non-open ones) in $\Lambda^3(\mathbb{R}^6)$, as it can be done easily using standard exterior algebra facts. It's just not obvious (which is what I wrote). If the reader wants details on this argument, one place they are available is in the Appendix to my article "On the geometry of almost complex $6$-manifolds" ( – Robert Bryant Jan 14 '13 at 0:26

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