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Define $Stab^{+}(\Omega )$={ $\phi \in GL^{+}(V)$ : $\phi^{*}\Omega=\Omega$ }.

we say three-form $\Omega\in\wedge^{3}V^{*}$ is non-degenerate , if $i_X\Omega\neq 0$ for all $X\in V$-{0}

Let $V\cong \mathbb{R}^{6}$ and $\Omega\in\wedge^{3}V^{*}$ be non-degenerate(by sense of Hitchin) . Then we know $Stab^{+}(\Omega )=SL(3,\mathbb{R})\times SL(3,\mathbb{R})$ OR $Stab^{+}(\Omega )=SL(3,\mathbb{C}))$

We call the three-form $\Omega\in\wedge^{3}V^{*}$ is n-fold degenerate if the annihilator of $\Omega$, i.e.,

$Ann(\Omega)$={ $X\in V$ : $ i_X\Omega=0$ } has $dim(Ann(\Omega)=n$ so my question is what can we say about $Stab^{+}(\Omega )$ when $\Omega$ is n-fold degenerate?( $n$ here is 0,1,3,or 6)

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1  
Is the sense of Hitchin the 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
    
Is the sense of Hitchin –  Hassan Jolany Feb 7 '13 at 21:00

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up vote 14 down vote accepted

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:

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

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

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

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.

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Dear @Robert,is it possible for you to explain with more details the cases of 4 and 5(because in your paper I couldn't find a reason for cases 4,5 about stabilizers) ? –  Hassan Jolany Feb 20 '13 at 0:48
    
We know that for the $Sp(V)$-action we have same results of V.Lychagin about stabilizers of degnerate 3-forms in six dimension . but for the general case $GL(v)$ you left the cases of 4 and 5 without reason –  Hassan Jolany Feb 20 '13 at 15:38
    
@Hassan: Unfortunately, I don't have time to put in these details right now. Case 5 should be clear, though. Using the index conventions $1\le i,j,k\le 3 < a,b,c\le 6$, if ${\bar e}^1,\ldots,{\bar e}^6$ is another basis of $V$ such that ${\bar e}^1\wedge{\bar e}^2\wedge{\bar e}^3= e^1\wedge e^2\wedge e^3$, then we'll have ${\bar e}^i = p^i_j\ e^j$ with $\det(p^i_j)=1$ and ${\bar e}^a = q^a_j\ e^j + p^a_b\ e^b$ where $\det(p^a_b)\not=0$ and the $q^a_j$ are arbitrary. This shows that the group of linear transformations that preserve $\phi_5$ must be what I claim it to be. –  Robert Bryant Feb 20 '13 at 15:41

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