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A 3-plane field $D$ on a 6-manifold $M$ is generic if $[D, D] = TM$. I'd like to do some explicit computations with a general local plane field of this type, and so I want to find a local (quasi-)normal form as possible for these plane fields, given, say, as an explicit triple of 1-forms on $M$, which is as simple as possible in the sense that it depends on as few functions as possible. One expects from an easy heuristic argument that an optimal normal form for such plane fields depends on three functions of six variables.

A well-known example of this kind of normal form is given by a theorem of Goursat, which says that for any 2-plane field $E$ on a 5-manifold $N$ that is generic (which this time just means that $[E, [E, E]] = TN$) and a base point $u \in N$, there are local coordinates $(x, y, p, q, z)$ near $u$ and a function $F(x, y, p, q, z)$ such that the restriction of $E$ to the chart is the common kernel of the triple

$\{ dy - p \, dx , dp - q \, dx, dz - F(x, y, p, q, z) \, dx \}$ $(\ast)$.

(In fact, we can furthermore assume that $u = (0, 0, 0, 0, 0)$ in this chart.) Conversely, given a function $F$, the 2-plane field $(\ast)$ is generic iff $F_{qq}$ is nowhere zero.

(This triple in this example is just the plane field on the jet space $J^{0, 2} (\mathbb{R}, \mathbb{R}^2) = \mathbb{R}^5_{xypqz}$, where $p$ and $q$ are placeholder variables for $y$ and $y'$, respectively, corresponding to the differential equation $z' = F(x, y, y', y'', z)$. I've been told, though, that there might well be no normal form for generic 3-plane fields on 6-manifolds that can be interpreted in this way, i.e., as the plane field determined by a system of differential equations with some unspecified functions on a suitable jet space.)

There are various accessible proofs of Goursat's result in the literature (including in a paper of Bryant & Hsu, and in a recent Ph.D. dissertation of Strazzullo), but the arguments in those references are not obviously adaptable to the 3-plane geometry in which I'm interested.

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up vote 4 down vote accepted

You probably can't do much better than this: Given a point $u$, there always exist $u$-centered coordinates $x^1,x^2,x^3,y^1,y^2,y^3$ and functions $F^i(x,y)$ vanishing at $u$ (i.e., at $(x,y)=(0,0)$) such that the plane field $D$ near $u$ is defined by three equations of the form $$ dy^i - F^i(x,y)\ dx^i = 0,\qquad i = 1, 2, 3. $$ (Such coordinates always exist, by a theorem of Schouten and van der Kulk.)

The nondegeneracy condition (i.e., the condition that $[D,D] = TM$) is that $ F^1_3 F^2_1 F^3_2 - F^1_2F^2_3F^3_1\not=0, $ where $$ F^i_j = \frac{\partial F^i}{\partial x^j} + F^j\ \frac{\partial F^i}{\partial y^j}\ . $$

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Thanks, Robert, this is exactly the kind of result I wanted. –  Travis Sep 7 '12 at 23:20
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