Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Poincaré's Lemma is often stated as saying that a closed differential form on a star-shaped domain is exact. More generally, it is true that a closed differential form on a contractible domain is exact.

What I am wondering is if there is an easy example of a closed differential form on a simply connected domain which is not exact.

share|cite|improve this question
No. A closed differential form on a simply connected domain is exact. This is stronger than Poincare's lemma, but you have to prove Poincare's lemma to prove it, which is why Poincare's lemma is usually stated your way. – user29743 Jul 1 '12 at 22:24
Every simply connected planar domain is contractible, isn't it? – user31373 Jul 1 '12 at 22:25
It depends on what you mean by "domain". If by domain you mean manifold, then there are simply connected manifolds with non-trivial de Rham cohomology, so yes you can construct such examples. – Matt Jul 1 '12 at 22:26
@VictorBessette If you allow $S^2$, then just take the volume form on it. – Matt Jul 1 '12 at 22:33
Dear @countinghaus, what you write is not correct for differential forms of degree $\geq 2$: my answer gives a counterexample. – Georges Elencwajg Jul 2 '12 at 0:23
up vote 4 down vote accepted

Let $U=\mathbb R^n\setminus \lbrace0 \rbrace\subset \mathbb R^n$, a simply connected domain for $n\geq3$, which we assume from now on.

The $(n-1)$ form $\omega \in \Omega^{n-1}(U)$ defined by

$$\omega (x)=\frac {1}{\mid \mid x\mid \mid^n} \sum _{i=1} ^n (-1)^{i-1}x_idx^1\wedge...\wedge \widehat{dx^i} \wedge dx_n$$ for $x\in \mathbb R^n\setminus \lbrace0 \rbrace)$ is closed but not exact. It is thus an example of what you want.
More precisely, its cohomology class $[\omega ]$ generates the one-dimensional $(n-1)$-th De Rham cohomology vector space of $U$, namely $H^{n-1}_{DR}(U)=\mathbb R\cdot[\omega ]$

a) This form can also be seen as the pull-back $\omega =r^*(vol)$ of the canonical volume form $vol\in \Omega ^{n-1}(S^{n-1})$ of the unit sphere under the map $$r:U\to S^{n-1}:x\mapsto \frac {x}{\mid \mid x\mid \mid}$$
b) To be quite explicit, the value of the alternating form $\omega (x)$ on the $(n-1)$-tuple of vectors $v_1,...,v_{n-1}\in T_x(U)=\mathbb R^n$ is $$\omega (x)(v_1,...,v_{n-1})=\frac {1}{\mid \mid x\mid \mid^n}\cdot det[x\mid v_1\mid ...\mid v_{n-1}]$$
where of course $x, v_1, ...,v_{n-1}$ are seen as column vectors of size $n$.

share|cite|improve this answer

The inverse square vector field $\vec{r}/|\vec{r}|^3$ gives rise to its flux 2-form $$\frac{xdydz+ydzdx+zdxdy}{(x^2+y^2+z^2)^{3/2}}$$ in $R^3$ minus the origin, which is simply connected. This is closed because the field is the gradient of the harmonic function $1/|\vec{r}|$. It is not exact because its integral over the 2-sphere is not zero.

share|cite|improve this answer
"This is closed because the field is the gradient of the harmonic function" I think this is swapping the definition of closed and exact. My understanding is that $\phi$ is closed iff $d\phi = 0$ and $\phi$ is exact iff $\phi = d\psi$. Being the gradient of a function makes a form exact. – Daniel Mahler Jul 20 '14 at 4:53

Because specific examples have already been given, I will make some general statements.

I will assume (without proof) that an $n$ dimensional manifold $M$ is orientable if and only if there is a nonvanishing $n$ form on $M$.

Let $M$ be a compact, orientable, $n$ dimensional manifold without boundary. Then, it is a simple application of Stokes' theorem to show that the nonvanishing form is closed but not exact. To answer your question, we just need to find a simply connected manifold with these properties, for example, $S^n$ for $n \geq 2$.

For an even more general theorem (which is much harder to prove), if $M$ is a connected $n$ dimensional manifold, $H^n_{de}(M) = 0$ if $M$ is non-orientable or non-compact and $H^n_{de}(M) = \mathbb{R}$ otherwise (if it is both orientable and compact.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.