# Simple criteria for “closed $\Longrightarrow$ exact”

In determining whether a closed form is an exact form, there is a lot of differential geometry definitions etc. that come in. I'm interested: what is the dummy, Calc III version of when closed implies exact? Is it sufficient if the domain is a simply connected smooth region in $\mathbb{R}^n$?

I believe this holds in $\mathbb{R}^3$, i.e. for a simply connected smooth region in $\mathbb{R}^3$, a curl-free $C^1$ vector field is the gradient of some function, and a divergence-free vector field is the curl of some function. (Edit: the second statement is not true.)

Is "simply connected" the magic ingredient in all dimensions?

• Maybe this will help: en.wikipedia.org/wiki/Poincar%C3%A9%27s_lemma . But, actually, simple-connectedness is enough in lower dimensions, but you need more conditions to guarantee equality for "holes" in higher dimensions. – user99680 Dec 7 '13 at 20:22
• Basically, when there is a "cycle that does not bound", informally, having a k-manifold that does not bound any (k-1)-manifold (tho the classes are not always manifolds.) on any dimension, you will have non-trivial homology, and, in the finite-dimensional case, you will have non-trivial cohomology, meaning you will have non-zero classes in DeRham cohomology. – user99680 Dec 7 '13 at 20:33

Simply connected will only ensure that closed one forms are exact. If you delete the origin from $R^3$, there should be a 2-form on this space which is closed but not exact (I think you could write it down by pulling back the volume form of the sphere to $R^3-0$).

In general closed forms will always be exact on contractible spaces.

I would recommend reading a book on de Rham cohomology, such as Bott and Tu, or From Calculus to Cohomology.

What you're asking for is exactly what the de Rham cohomology measures. Roughly speaking, if you have $\Omega\subset \mathbb{R}^n$, then

$$H^k_{dR}(\Omega)=\frac{\text{closed k-forms}}{\text{exact k-forms}}$$

Thus if you want some nice easy condition that will ensure the property you want, then you'd need to just come up with a sufficient condition that ensures these are all $0$.

It turns out that simply connected implies the closed $1$-forms are all exact, because simply connected tells us something about the fundamental group which is classically related to $H^1$.

Unfortunately, you need something stronger if you want it to work for all forms. Ah. Looks like Steven just beat me to an answer, but as he points out one sufficient condition would be contractible because these cohomology groups are preserved under homotopy. Thus if your space is homotopic to a point, then you can easily calculate that they are all $0$ for a point to see they are all $0$ for the original space.

I'm not sure if there is a nice way to rigorously define contractible at a Calculus level. One possibility is to note that $H^k$ can also be interpreted as measuring whether or not there are "holes detectable by spheres of dimension $k$."

Thus to check whether or not all closed $2$-forms are exact on $\Omega\subset \mathbb{R}^3$, you think about whether or not every sphere in $\Omega$ can be shrunk to a point staying completely within $\Omega$. In the example of $\Omega = \mathbb{R}^3\setminus \{(0,0,0)\}$ this is not the case because the unit sphere cannot be shrunk to a point without passing through (or ending at) $(0,0,0)$.

• When you write \mathrm{closed \ k-forms} instead of \text{closed $k$-forms}, then it makes the look like a minus sign instead of a hyphen. Thus: $\mathrm{closed \ k-forms}$ versus $\text{closed$k$-forms}$. (I changed it.) – Michael Hardy Dec 7 '13 at 20:51
• Hi Matt -- One easy sufficient condition for contractibility of an open set $\Omega\subset\mathbb R^n$ is that it be "star-shaped," meaning that it contains a point $p_0$ such that for every $q\in\Omega$, the line segment from $p_0$ to $q$ lies in $\Omega$. Every convex domain, for example, is star-shaped. – Jack Lee Dec 7 '13 at 20:51
• Also, for developing intuition about $2$-forms on $\mathbb R^3$, it's better to think of arbitrary closed surfaces rather than spheres. It's possible to define a domain $\Omega\subset\mathbb R^3$ such that every sphere in $\Omega$ bounds a ball, but $\Omega$ contains a torus that does not bound any domain. In such a domain, not every closed $2$-form is exact (and hence not every divergence-free vector field is a curl). – Jack Lee Dec 7 '13 at 20:53
• @JackLee Ah yes. Good point. I've been teaching multivariable calculus (math 324) this quarter and have run into this issue of exactly how much detail to give vs how much to "lie" and just give an intuitive feel. In fact, this exact issue came up when defining simply connected, because for calculus students the idea of a definition involving "for all simple closed curves ..." is really abstract, but the notion "for any circle ..." seemed more manageable. – Matt Dec 7 '13 at 23:25
• Hi, Can you please answer this question? I think you can help me math.stackexchange.com/q/1693969/322103 – M98 Mar 12 '16 at 11:02

This is a somewhat over-simplified answer: in many cases, homology and cohomology coincide; specifically, when both , as vector spaces, are finite-dimensional, they will be equal. Informally, in this case, dealing with homology ( which agrees in this case with cohomology), you're detecting "cycles that do not bound" . Think of cases where the cycles are manifolds: homology detects when these manifolds are not the boundary of a manifold of one-lower dimension ( since the manifold-boundary of an n-manifold is an n-1-manifold.) Think of the case of the torus $T^2=S^1 \times S^1$ , and consider a meridian loop, or a longitude loop. These are cycles ( closed, non-self-intersecting loops.) that are not the boundary of any 2-manifold. But consider a loop in the surface of the torus. This loop is a cycle that does bound, in the sense that removal of the loop would disconnect the torus. Homology detects these cycles that do not bound in all dimensions (up to the dimension of the space, of course).