# proving the injectivity half of de Rham's Theorem when $p>1$ (if a $p$-form $\omega$ vanishes on all $p$-cycles, then $\omega$ is exact)

Let $M$ be a smooth manifold of dimension $n$, and let $\omega$ be a differential $p$-form on $M$. Then we have the following theorem:

$\omega$ is exact if and only if $\oint_c\omega=0$ for all $p$-cycles $c$ on $M$.

(A $p$-cycle is a closed $p$-chain.)

For lack of a better word, I'll call the second condition, about how $\omega$ behaves on $p$-cycles, "the vanishing condition." (Some people call it conservativity, but others use "conservative" to mean "exact," so I will avoid this term.)

The "only if" part of this theorem is easy: it follows directly from Stokes's theorem. The converse is much more difficult. It is my understanding that the converse is equivalent to the injectivity part of de Rham's theorem. (This is how it is presented in several books, at least, including Frankel's The Geometry of Physics.)

However, the converse is very easy in the special case $p=1$, so easy that it is included in many introductory books on differential geometry, such as O'Neill's. In this case, the vanishing condition implies "path independence": $\int_c\omega=\int_d\omega$ if $c$ and $d$ are 1-chains (i.e. curves) with the same endpoints. This allows us to make a direct integration argument: we write down a $0$-form (i.e. a function) $f$ that will be a potential for $\omega$, namely $$f(x)=\int_c\omega$$ where $c$ is any 1-chain starting at an arbitrarily chosen but fixed point in $M$ and ending at $x$. Path independence guarantees $f$ is well-defined. It is then straightforward to verify that $\omega=df$, and exactness follows.

Here's my question: is there a similar constructive argument to be given when $p>1$, and if not, is there anything interesting to say about why not (besides things get harder in higher dimensions)? The first difficulty we'll face is that when $p>1$, we're no longer after a function, but a $(p-1)$-form. Still, the argument for $p=1$ is so easy that it is surprising to learn that the general theorem is very difficult and highly non-trivial. Why should $p>1$ require such a startling jump in difficulty? When I first started thinking about this issue, I guessed that the most you'd have to do is be a little cleverer about how to do the integration in a "p-chain-independent" (instead of "path independent") way; I didn't anticipate I'd have to give up this direct approach and wade through an incredible amount of complexity.

As far as I know, the two standard approaches to de Rham's theorem are via sheaf theory (as given in Warner's or Morita's books) or a complicated induction argument on open sets (due to Bredon). I haven't read these proofs in detail, and I admit to not really understanding them, but as far as I can tell they aren't constructive: they don't write down a potential for $\omega$ like we do for $p=1$. But maybe I'm wrong about that.

Perhaps the Poincaré Lemma will be helpful here—that is certainly a place where one can find very similar constructive integration arguments, and I know it is used in the proofs of de Rham's theorem—but it is at least not directly relevant, because the theorem above says nothing about contractibility.

• One can give a quite elementary argument for $p=m$. Surely you know it, but I could post a summary of the argument in case it's of any use. I can't think of a reference now... – Jesus RS Mar 12 '15 at 11:22
• @Jesus: post it, then. – symplectomorphic Mar 12 '15 at 14:03

If $p=n$ we must show the injectivity of $\int_M:H^n_c(M)\to\mathbb R$ (forms with compact support). Using a partition of unity, this reduces to the case $M=\mathbb R^n$. Now notice that if we are given a form $\omega=f(x)dx_1\dots dx_n$ and $f$ is $\operatorname{div}(X)$ for some vector field $X=(X_1,\dots,X_n)$, then a primitive of $\omega$ is $\det(X,\cdot,\dots,\cdot)$. The difficulty is to get everything with compact support. In fact, we will prove by induction on $m$:

Prop. Let $x'=(x_1,\dots,x_m), x'=(x_{m+1},\dots,x_n)$. If $f(x,x')$ is a smooth function with compact support and $\int_{\mathbb R^m}f(t,x')dt\equiv0$, then $$f(x,x')=\operatorname{div}_xX(x,x')=\sum_{i=1}^m\frac{\partial X_i}{\partial x_i}(x,x'),$$ where $X(x,x')=(X_1(x,x'),\dots,X_m(x,x'))$ has compact support.

(Here $\operatorname{div}_x$ means we consider derivatives with respect to $x_1,\dots,x_m$.)

Case $m=1$. Just take $$X_1(x_1,x')=\int_{-\infty}^{x_1}f(t_1,x')dt_1,$$ which has compact support. Indeed, for $t_1\le x_1<<0$ we have $f(t_1,x')\equiv0$, and for $x_1>>0$, $$X_1(x_1,x')=\int_{-\infty}^{x_1}f(t_1,x')dt_1= \int_{-\infty}^{+\infty}f(t_1,x')dt_1=0.$$

Suppose $m\ge2$. Set $y=(x_{1},\dots,x_{m-1})$, $y'=(x_{m},x_{m+1},\dots,x_{n})$. Let $\varphi(y)$ be a smooth function woth compact support and $\int_{\mathbb R^{m-1}}\varphi(y)dy=1$, and write $$g(y,y')=f(x,x')-h(y,y'),\quad h(y,y')=\varphi(y)\int_{\mathbb R^{m-1}}f(s,y')ds.$$ This $g(y,y')$ has compact support, as $f$ and $\varphi$ have. By induction, there is a vector field with compact support $$Y(y,y')=(X_1(y,y'),\dots,X_{m-1}(y,y'))$$ such that $$g(y,y')=\operatorname{div}_y Y(y,y')=\sum_{i=1}^{m-1}\frac{\partial X_i}{\partial y_i}(y,y')= \sum_{i=1}^{m-1}\frac{\partial X_i}{\partial x_i}(x,x').$$ On the other hand $$h(y,y')=\frac{\partial X_m}{\partial x_m}(x,x'),\quad\text{with}\quad X_m(x,x')=\varphi(y)\int_{-\infty}^{x_m}\left(\int_{\mathbb R^{m-1}}f(s,t_m,x')ds\right)dt_m.$$ The function $X_m$ has compact support:

• If $\|y\,\|>>0$, $\varphi(y)=0$;

• If $\|x'\|>>0$ or $\,x_m<<0$, then $f(s,t_m,x')=0$; and most important

• If $x_m>>0$, as $f$ has compact support we deduce \begin{equation*}\begin{split} \int_{-\infty}^{x_m}\!&\left(\int_{\mathbb R^{m-1}}f(s,t_m,x')ds\right)dt_m\\ &\qquad\qquad=\int_{-\infty}^{+\infty}\!\!\left(\int_{\mathbb R^{m-1}}f(s,t_m,x')ds\right)dt_m= \int_{\mathbb R^{m}}\!\!f(t,x')dt=0. \end{split}\end{equation*}

All in all $X=(X_1,\dots,X_{m-1},X_m)$ has compact support and $f=\operatorname{div}_xX$.

We are done.

• But without compact supports, one has for instance $\omega=dx\wedge dy$, which is exact: $\omega=d(xdy)$, and $\int_{\mathbb R^2}\omega=+\infty\ne0$. – Jesus RS Mar 12 '15 at 18:03
• yes, thanks; I immediately realized there'd be a problem of that form, which is why I deleted my comment just after I made it. now that I've thought about it, though, my understanding is that the classical statement of de Rham's theorem (i.e. for the general groups, not the compactly supported groups) assumes chains that are compact. that rules out your counterexample. how will that change things? – symplectomorphic Mar 13 '15 at 1:34
• In cohomology terms, the standard thing if the manifold is not compact is compact supported cohomology. This is automatic on compact manifolds. Hence my post! – Jesus RS Mar 13 '15 at 3:06
• Many textbooks directly suppose manifolds are compact. And in general, algebraic topology requires some compactness, which goes to compact support. For instance, Poincaré's duality involves homology and compactly supported cohomology. Concerning my post, I don't think it can be extended to other degrees $p\ne1,n$, and, learnt from a friend some time ago, and know no english reference for the idea. – Jesus RS Mar 13 '15 at 14:11
• Another remark in case of use. Even if one deals with a compact manifold, when using partitions of unity to work in charts, hence in $\mathbb R^n$, one ends up with compactly supported objects in $\mathbb R^n$. – Jesus RS Mar 13 '15 at 14:15