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we know Darboux theorem for higher-symplectic geometry is not correct in general, but is there any Darboux like theorem for non-degenerate 3-forms in 6-manifolds?

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What is nondegenerate? aa –  Alexander Chervov Apr 1 '13 at 20:23
    
Dear@ Alexander Chervov you can find this definition here mathoverflow.net/questions/120768/… –  Hassan Jolany Apr 1 '13 at 20:30
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up vote 13 down vote accepted

This depends on what you mean by 'Darboux-like'. It is certainly not true that a closed nondegenerate 3-form on a 6-manifold is necessarily locally equivalent to one of the 'flat' models, so there is no direct analog of the Darboux' theorem in this case.

As I remark in my article "Remarks on the geometry of almost complex $6$-manifolds" (Asian Journal of Mathematics 10 (2006), 561–606, also available on the arXiv as arXiv:math/0508428), the closed $3$-forms of elliptic type in dimension $6$ essentially depend on 4 arbitrary functions of 6 variables (modulo diffeomorphism).

However, there is an analog if you are willing to consider something stronger: If $\phi\in\mathcal{A}^3_+(M^6)$ is a $3$-form of elliptic type on an oriented $6$-manifold $M$, then there is a unique $J(\phi)\in\mathcal{A}^3_+(M^6)$ with the property that the complex $3$-form $\Upsilon = \phi+i \ J(\phi)$ is decomposable and hence of type $(3,0)$ with respect to an almost complex structure $J_\phi$ on $M$ that induces the given orientation of $M$. All this is algebra. However, now, if one adds the hypothesis that $d\Upsilon = 0$, which is, of course, the same as $d\phi = d\bigl(J(\phi)\bigr)=0$, then one has that there always exist local complex functions $z^1,z^2,z^3$ such that $\Upsilon = dz^1\wedge dz^2\wedge dz^3$. In particular, $\phi = \mathrm{Re}(dz^1\wedge dz^2\wedge dz^3)$, so this is a sort of Darboux-like theorem; it's just that you need more hypothesis than the closure of the original form.

There is a similar result for nondegenerate $3$-forms of hyperbolic type. This is the case in which the form $\phi$ can be written locally as $\phi = \phi_+ + \phi_-$ where each of $\phi_\pm$ is decomposable while $\phi_+\wedge\phi_-\not=0$. (These two summands are unique up to permutation.) In this case, the 'Darboux-like' theorem is that $\phi$ can be put in normal form if and only if $d\phi_+=d\phi_-=0$, which is stronger than $d\phi=0$ (which is not sufficient by itself).

Finally, there is the case of nondegenerate $3$-forms of what might be called 'nilpotent type' (which is not a stable type in Hitchin's sense, but is nondegenerate in the sense described by the OP). A $3$-form $\phi$ on $M^6$ is nondegenerate of nilpotent type if and only if each point lies in some open set $U$ on which there exists a coframing $\omega^1,\ldots,\omega^6$ for which $$ \phi = \omega^4\wedge\omega^2\wedge\omega^3 +\omega^5\wedge\omega^3\wedge\omega^1 +\omega^6\wedge\omega^1\wedge\omega^2. $$ For such a $\phi$, the conditions that it can locally be put in this form with $\omega^i = dx^i$ for some coordinates $x = (x^1,\ldots,x^6)$ consist of two things: First, the condition $d\phi=0$, which is clearly necessary; second, the condition that $3$-plane field $D\subset TM$ defined by $\omega^1=\omega^2=\omega^3=0$ (which is well-defined by $\phi$) should be Frobenius. It is not hard to show that these necessary conditions are also sufficient, so this is the 'Darboux-like' normal form theorem in this case.

Note the interesting fact that, in each of these three cases, the 'Darboux-like' conditions are all first order equations on $\phi$. This does not continue in higher dimensions. In dimension $7$, the two stable types of $3$-forms each have examples that are flat to first order but not flat to second order, so the 'Darboux-like' theorems in this case turn out to involve a mixture of first and second order conditions.

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Is there a characterization of types of forms for which a Darboux-type theorem holds? (I don't know how to make this precise...) –  Mariano Suárez-Alvarez Apr 2 '13 at 4:57
    
@Mariano: I don't know how you want to make your question precise. Here's a precise problem, is it one you want solved? Determine the algebraic types of exterior forms $\tau$ with the property that an exterior differential form $\phi$ that is everywhere of algebraic type $\tau$ is flat if and only if $\phi$ is flat to first order at every point. Some alternative endings to the sentence: "is closed.", "is flat to order $k$ at every point" (where $k$ is fixed and may depend on $\tau$), or "is flat to all orders $k\ge 1$ at every point". –  Robert Bryant Apr 3 '13 at 0:48
    
That looks like what I had in mind (adding the "is closed" for good measure) :-) –  Mariano Suárez-Alvarez Apr 3 '13 at 2:00
    
@Mariano: If $\phi$ is flat to first order at every point, then it is closed. The converse is not generally true, though, of course, it is true for $2$-forms of constant rank (as Darboux' Theorem shows) and for nonvanishing decomposable forms of any degree in any dimension. –  Robert Bryant Apr 3 '13 at 4:09
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@Mariano: (Part 2) The first case in which 'closed' is not the same as 'flat to first order' is the case of a closed, nondecomposable $3$-form in dimension $5$. In the 1980s, F. J. Turiel wrote a series of papers on the geometry of such $3$-forms, noting that there were several different homogeneous models and providing a classification. Of course, only one of those models is flat to first order, namely, the one that can be written locally with constant coefficients in some coordinate system. –  Robert Bryant Apr 3 '13 at 13:06
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