Hi, Again:(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to show that, given a 3-manifold M, and a plane field ρ (i.e., a distribution on

TM) on M, there exists an open set U in M, so that ρ can be represented as the kernel of a

differential form w , for W defined on U.

The idea is that the kernel of a linear map from R^{3}--T_{x}M

is either the whole space, or a two-dimensional space, by , e.g., rank-nullity.

My idea is to start by choosing the assigned plane ∏_{m}at any point m in M. Then we use the fact that any subspace can be expressed as the kernel of a linear map.

Specifically, we choose a basis {v1,v2} for ∏_{m}Subset T_{m}M

and define a form w so that:

w(v1)=w(v2)=0 .

Then we extend the basis {v1,v2} into a basis {v1,v2,v3} for T_{m}M , and

then we declare w(v_{3})=1 (any non-zero number will do ), so that

the kernel of w is precisely ∏_{m}, by some linear algebra.

Now, I guess we need to extend this assignment w at the point m, at least into

a neighborhood U_{m}of M . I guess all the planes in a subbundle have

a common orientation, so maybe we can use a manifold chart W_{m}for

m, which is orientable ( being locally-Euclidean), and then use the fact that there is

an orientation-preserving isomorphism between the tangent planes at any two points

p,q in T_{p}U . Does this allow me to define a form in U whose kernel is ρ ?

Edit: I think this should work: please critique: for each q in U , the hyperplane ∏

_{q}can be described as the kernel of a map , as for the case of m. Again,

we find a basis {w1,w2} for ∏_{q}, and then define an orientation-preserving

map (which exists because U is orientable) between T_{m}M and

T_{q}M, sending basis elements to basis elements.

Thanks.

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