Existence of integrals of 1-forms up to multiples - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:57:59Z http://mathoverflow.net/feeds/question/82701 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82701/existence-of-integrals-of-1-forms-up-to-multiples Existence of integrals of 1-forms up to multiples Florian 2011-12-05T13:33:19Z 2011-12-09T17:45:27Z <p>Let $\Omega\subset \mathbb{R}^n$ an open contractible set (we can assume $n=2$ for a start) and $\omega$ be a 1-form on $\Omega$ which is nowhere zero. Then $\omega=df$ for a function $f$ if and only if $d\omega=0$. If $\omega$ is not closed it might still be possible that it works up to a positive multiple. In other words, do there exist functions $f,g$ on $\Omega$, $g>0$ such that $df=g\omega$?</p> <p>For a contractible $\Omega$ this is equivalent to $dg\wedge \omega+gd\omega=0$, and if $h$ denotes the logarithm of $g$, the question is whether there exists $h$ such that $dh\wedge \omega+d\omega=0$. However, this equation does not tell me much.</p> <p>If the machinery of differential topology has an easy answer to this question I'm also interested what can be said in more general cases (first of all a ring-shaped $\Omega$, or more than 2 dimensions).</p> http://mathoverflow.net/questions/82701/existence-of-integrals-of-1-forms-up-to-multiples/82715#82715 Answer by Robert Bryant for Existence of integrals of 1-forms up to multiples Robert Bryant 2011-12-05T16:47:38Z 2011-12-09T17:45:27Z <p>Well, the local condition that $\omega\not=0$ be a nonzero multiple of an closed $1$-form is that $\omega\wedge d\omega = 0$. This is necessary and sufficient for the local existence of functions $f$ and $g\not=0$ such that $\omega = g\ df$. (This claim is just a special case of the Frobenius Theorem.)</p> <p>For the global question, you are really asking whether a codimension $1$ foliation of a contractible open set is always the level sets of a function without critical points. This is certainly false in dimensions $3$ and higher, and I sort of remember that it's false in dimension $2$ as well, but I can't remember the example. (It <em>is</em> already false in dimension $2$. I looked it up later; see the added remark below.)</p> <p>Added remark: Your question is addressed by Exercises 5 and 6 of Section 16 of Chapter XVIII of Volume IV of Dieudonné's <em>Treatise on Analysis</em>. He gives the above local criterion and a counterexample to its global analog with an example of a $1$-form $\omega$ on $\mathbb{R}^2$ that is nonvanishing and yet cannot be written globally in the form $g\ df$ for two smooth functions on $\mathbb{R}^2$. Just to save you the trouble of looking it up, here is his example $$\omega = y^3(1{-}y)^2\ dx + \big(y^3-2(1{-}y)^2\bigr)\ dy.$$ The point is that , if you could write $\omega = g\ df$, then, away from the lines $y=0$ and $y=1$, the function $f$ would have to be a function of $$F = x + \frac{1}{y^2} + \frac{1}{1-y}$$ and $f$ would have to be constant on the lines $y=0$ and $y=1$. Now, you need to check that you can't rig an $f$ with these properties that is smooth and without critical points on the whole plane.</p>