# Why 'closed differential forms' are called 'closed'?

As is well known a differential form $\omega$ is called closed differential form if it satisfies $\mbox{d} \omega = 0 \, \, (\ast) \,$ where $\mbox {d}$ is the exterior derivative. I think the terminology, 'closed' (which we give the differential forms $\omega$ satisfying the equation $\, \, (\ast) \;$) refers to something that motivated the definition. The question is as follows.

Why 'closed differential forms' are called 'closed' ?

• What I've been told has always been that the exterior derivative $\mathrm{d}$ is the dual to the boundary map $\partial$ by Stoke's theorem. For closed (as in compact with no boundary) manifolds $M$, $\partial M = 0$ and one side of Stoke's theorem vanishes always. Analogously then if $\mathrm{d}\omega = 0$, so the other side of Stoke's theorem vanishes always, we can call the form as "closed". – Willie Wong Feb 11 '14 at 9:39
• In any case, this question seems to be a duplicate of an earlier question on MathOverflow. I am however not sure how satisfactory you will find the answer given there. – Willie Wong Feb 11 '14 at 9:41

## 2 Answers

I believe that this is taken from homology theory in algebraic topology. Simplicical homology studies the maps from simplices (i.e. points, lines, triangles, tetrahedrons and so on in increasing dimensions) into some fixed topological space. An $n$-chain (or just chain if the dimension is implicitly understood) is a formal sum of such maps of $n$-dimensional simplices. For example, a $1$-chain is a set of curves in the space (or more precicely, a set of maps from the unit interval into the space).

A chain is called closed if the total boundary map (with orientation) is zero. I assume the reason is that in the one-dimension case you can take a closed form and stitch together the curves it represents into one or more closed curves. In the two-dimensional case you can stitch together the "triangles" to form spheres ("closed" discs) and so on.

In simplicical cohomology theory you study homomorphisms from chains into some group $G$. It can be shown that the simplicical cohomology theory developed from this is more or less equivalent to the de-Rahm cohomology defined with differential forms on manifolds. It would therefor make sense to transfer much of the terminology, including the word closed.

because we have $$\int_{\partial M}\omega=0$$ for any closed manifold $M$.

• The usual definition of "closed manifold" is "compact, without boundary", so the stated integral vanishes whether or not $d\omega = 0$...? – Andrew D. Hwang Feb 7 '14 at 22:54
• in ancient times the definition was different – user88576 Feb 7 '14 at 23:29
• Adding the relevant definition of "closed" to your answer would be worthwhile for posterity, I think. – Andrew D. Hwang Feb 8 '14 at 1:01
• I confess the thought had occurred to me (particularly since you gave no historical evidence in support of your claim). Since MSE is intended to be a repository of factual information, perhaps deleting your post would be appropriate. – Andrew D. Hwang Feb 9 '14 at 15:11
• "dunno what you are talking about" seems disingenuous, given that someone used your account yesterday to say they say they "...made something up, lol", then edited out the "lol", then removed the comment entirely after I responded to it. But in any case, Willie Wong's comment addresses the original question with mathematical correctness, while your answer -- as it currently stands, in the absence of requested clarifications -- does not. Instead, by your purported rationale, every form is closed. – Andrew D. Hwang Feb 11 '14 at 16:32