# Obstruction to the existence of a globally defined integrating factor

Let $$U$$ be an open subset of $$\Bbb{R}^n$$ and take $$\omega$$ to be a nowhere-vanishing smooth $$1$$-form on $$U$$. The Frobenius Theorem implies that, near each point of $$U$$, $$\omega$$ may be written as $$g\,{\rm{d}}f$$ for suitable locally defined smooth functions $$f$$ and $$g$$ iff $$\omega\wedge{\rm{d}}\omega=0$$. Here is my question(s): For such a $$1$$-form $$\omega$$ on $$U$$ (never vanishing and $$\omega\wedge{\rm{d}}\omega=0$$) what is the obstruction to the existence of global smooth functions $$f$$ and $$g$$ on $$U$$ with $$\omega=g\,{\rm{d}}f$$? Can that be formulated as the vanishing of any homotopical invariant of $$U$$? What is a good non-example in which such a global presentation of $$\omega$$ fails to exist?

• For non-examples, just take any closed but non-exact $\omega$ on $U$ that fails to be simply connected? – Willie Wong Mar 19 at 3:06
• @WillieWong Well, in that case $\omega$ is not in the form of ${\rm{d}}f$, but why not a multiple of such a thing? For instance, $\omega:=-\frac{y}{x^2+y^2}{\rm{d}}x+\frac{x}{x^2+y^2}{\rm{d}}y$ is famously closed but not exact on $U:=\Bbb{R}^2-\{(0,0)\}$. But it is a multiple of ${\rm{d}}\left(\frac{y^2}{x^2+y^2}\right)$. – KhashF Mar 19 at 3:22
• In your "counter example", your function $g$ is not smooth. (It is singular whenever $x = 0$ or $y = 0$.) – Willie Wong Mar 20 at 18:29
• (More precisely, your function $g = \frac{x^2 + y^2}{2xy}$.) – Willie Wong Mar 20 at 18:34

To expand on my comment: suppose you have $$U$$ not simply connected, and $$\omega$$ closed but not exact, and suppose there exists a closed loop $$\gamma: [0,1]\to U$$ such that $$\omega(\dot{\gamma})$$ is signed. (This is in particular the case with the "example" in your comment, where you can take $$\gamma$$ to be any circle centered at the origin.)

Suppose $$\omega= g df$$ for smooth $$g$$ and $$f$$, then you must have

$$g \nabla_{\dot\gamma} f$$

is signed. This implies both $$g$$ and $$\nabla_{\dot\gamma} f$$ are signed along $$\gamma$$, but this is absurd, since integrating from $$0$$ to $$1$$ you have

$$\int_0^1 \nabla_{\dot\gamma(s)} f(\gamma(s)) ~ds = f(\gamma(1)) - f(\gamma(0)) = 0$$

since $$\gamma$$ is a closed curve.

Simply-connectedness is not enough, however, to ensure that the integrating factor can be globalized. To see this, recall that Frobenius theorem states that the nonvanishing one-form $$\omega$$ satisfies $$\omega\wedge d\omega = 0$$ IFF its kernel is the tangent bundle of a regular foliation.

If $$\omega = g ~df$$ for some function $$f$$, then necessarily $$f$$ will be constant on the leaves of this foliation. Hence a counterexample will be found if you have a regular foliation of a simply connected compact manifold. (As on a compact manifold the function $$f$$ must attain a maximum and $$df = 0$$ there, contradicting the assumption that $$\omega$$ is non-vanishing.)

One such example is given by the Reeb foliation of the 3-sphere.

If we want to examine domains in Euclidean space: embed $$\mathbb{S}^3 \hookrightarrow \mathbb{R}^4$$ and slightly thicken it radially by $$\epsilon$$. Define a foliation on this 4 dimensional (simply-connected) domain by extending the Reeb foliation trivially in the radial direction. The lifted one-form $$\tilde{\omega}$$ is the pull-back of the Reeb $$\omega$$ from $$\mathbb{S}^3$$ by radial projection, and hence $$\tilde{\omega} \wedge d \tilde{\omega} = 0$$. Any $$\tilde{f}$$ that realizes $$\tilde{\omega} = \tilde{g} d\tilde{f}$$ will be constant on the foliation, and hence factors through some $$f$$ on $$\mathbb{S}^3$$, and the argument above shows that this contradicts the fact that $$\omega$$ is non-vanishing.

• You seem to be saying that if the vector field corresponding to $\omega$ has a closed integral curve $\gamma(t)$, then $\omega$ cannot be written as $g\,{\rm{d}f}$: Since $\omega$ is non-zero everywhere, $g$ should be either always positive or always negative. Now $f$ is strictly monotonic along $\gamma(t)$ (a Lyapunov function) due to the fact that $\frac{\rm{d}}{\rm{d}t}f(\gamma(t))=g(\gamma(t))\,\big|\nabla f(\gamma(t))\big|^2$ does not change sign. This cannot be the case since $\gamma$ is closed. – KhashF Mar 23 at 17:58
• Thanks, it works. But here we assume something extra about the $1$-form $\omega$: There is a closed integral curve which of course prevents $\omega$ from being exact. I was looking for a topological property of the domain $U$. For instance, if $U$ is simply connected can we always write a non-vanishing $\omega$ with $\omega\wedge{\rm{d}}\omega=0$ as $g\,{\rm{d}}f$? – KhashF Mar 23 at 18:07
• "Closed integral curve" is not a requirement; that $\omega(\dot{\gamma})$ is signed is a weaker condition. // I don't think simply connected is enough; the Reeb foliation of $\mathbb{S}^3$ should be a counterexample. (The three-sphere is compact so there cannot be global $f$ with non-vanshing $df$.) – Willie Wong Mar 24 at 4:30
• $\dot\gamma$ is a vector field along $\gamma$. You pair it against $\omega$. You get a scalar. Nothing about this requires $\omega$ having a vector field corresponding to it (not using any Riemannian structures at all). In the case where you do have a Riemannian structure, you can imagine the corresponding vector field being perturbed to be spiralling (ever so slightly) instead of having closed integral curves.The circles are no longer integral curves but $\omega(\dot{\gamma})$ can still be signed. – Willie Wong Mar 24 at 12:55
• What if you take a slightly thickened copy of $\mathbb{S}^3 \subset \mathbb{R}^4$ and extend the Reeb foliation trivially in the radial direction? If the leaves were defined as level sets of a function the function projects to a function on $\mathbb{S}^3$ and the same argument goes through, no? – Willie Wong Mar 25 at 19:35