Symmetrizing factor in the case of corank 1

(xournal 195)
Key idea to start reasoning:
Lemma
Given a closed 1-form ω on an open set U⊂Rn, a vector field V is a symmetry of the Pfaffian system S({ω}) if and only if V⌟ω is a first integral of S({ω}).
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Proof
The function V⌟ω is a first integral if and only if d(V⌟ω)∧ω=0. Taking into account Cartan formula this is equivalent to

(LVω−V⌟dω)∧ω=0LVω∧ω=0.

And this is true if and only if LVω=βω for some function β, i.e., V is a symmetry of the Pfaffian system
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Pasted image 20221028073831.png

From here we can conclude several interesting facts, like those on @sherring1992geometric:

Proposition 1
Given a Frobenius integrable 1-form ω and a vector field X∉S({ω})⊥ then a function f satisfies fX is a symmetry of S({ω}) (is a symmetrizing factor) if and only if μ=1fX⌟ω is an integrating factor of ω.
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Proof
We know that there exist an integrating factor μ~, that is, a function such that μ~ω is closed. Taking into account that S({μ~ω})=S({ω}), and according to the Lemma above, fX is a symmetry of S({ω}) if and only if fX⌟μ~ω is a first integral. Now observe that

fX⌟μ~ω=μ~1fX⌟ω

and this is a first integral if and only if 1fX⌟ω is an integrating factor.
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Proposition 2
Given a symmetry V, then 1/V⌟ω is an integrating factor.
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Proof
Take f=1 in Proposition 1.
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Proposition 3
Given a Frobenius integrable 1-form, a smooth function μ is an integrating factor if there exists a symmetry V such that μ=1/V⌟ω.
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Proof
Take any X such that X⌟ω≠0. Since μ is an integrating factor, by Proposition 1 f=1μX⌟ω is a symmetrizing factor for X, so V=fX is a symmetry. It is easy to check that

μ=1V⌟ω.

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