Top deRham cohomology group of a compact orientable manifold is 1-dimensional

Question

Let $M$ be a compact smooth orientable manifold of dimension $n$. I am looking for a simple proof that $H_\mathrm{dR}^n(M) \cong \mathbb R$. Equivalently, an $n$-form which integrates to 0 is exact. I can show this via a rather indirect argument as follows: we know $H_\mathrm{dR}^n(M) \cong H^n(M, \mathbb R)$, where $H^n$ denotes the singular cohomology. By the universal coefficient theorem (and the fact that $\mathbb R$ is a field) this is isomorphic to $\operatorname{Hom}(H_n(M, \mathbb Z) , \mathbb R)$. From the (rather lengthy) proof in Section 3.3 of Hatcher's Algebraic Topology, we find that $H_n(M, \mathbb Z)$ is isomorphic to $\mathbb Z$, and so $\operatorname{Hom}(H_n(M, \mathbb Z) , \mathbb R) \cong \mathbb R$. However, it seems like there should be a simpler way to prove this. Does anyone know of one?

Answer 1

$\def\RR{\mathbb{R}}$The following is what I think of as the standard argument; I don't know whether it counts as simple. I'll be showing that, for $M$ a connected, oriented $n$-manifold, if $\omega$ is a compactly supported $n$-form with $\int_{M} \omega = 0$, then $\omega$ is $d \eta$ for a compactly supported $\eta$. Of course, if $M$ itself is compact, then the condition that $\omega$ is compactly supported is automatic.

Part One: The result is true for $\RR^n$. Fix some compactly supported smooth function $h: \RR \to \RR$ with $\int_{\RR} h(x) dx = 1$. Let $\omega = f(x_1, \ldots, x_n) dx_1 \wedge \cdots \wedge dx_n$; by hypothesis $\int_{(y_1,\ldots,y_n) \in \RR^n} f(y_1, \ldots, y_n) dy_1 \cdots dy_n= 0$. Put $$f_k(x_1, \ldots, x_n) = h(x_1) h(x_2) \cdots h(x_k) \left( \int_{(y_1,\ldots,y_k) \in \RR^k} f(y_1, \ldots, y_k, x_{k+1}, \ldots, x_n) dy_1 \cdots dy_k \right).$$ So $f_0=f$ and $f_n = 0$. We will show that $(f_{k} - f_{k-1}) dx_1 \wedge \cdots \wedge dx_n$ is $d \eta_k$ for a compactly supported $\eta_k$, so $\omega = (f_n - f_0) dx_1 \wedge \cdots \wedge dx_n = d \left( \sum_{k=1}^n \eta_k \right)$.

We have constructed $f_k$ and $f_{k-1}$ to have the same integral on every line parallel to the $x_k$-axis. (Use Fubini and the hypothesis $\int_{z\in \RR} h(x) dz=1$.) So, if we put $$g_k(x_1, \ldots, x_n) = \int_{z=-\infty}^{x_k} \left( f_k(x_1, \ldots, x_{k-1},z,x_{k+1}, \ldots, x_n) - f_{k-1}(x_1, \ldots, x_{k-1},z,x_{k+1}, \ldots, x_n) \right)\,dz,$$ then $g_k$ is compactly supported, and we have $(f_{k} - f_{k-1}) dx_1 \wedge \cdots \wedge dx_n = (-1)^{k-1} d \left( g_k dx_1 \wedge \cdots \wedge \widehat{dx_k} \wedge \cdots \wedge dx_n \right)$.

Part Two: General $M$. Now let $M$ be any connected, oriented $n$-fold. Fix an open cover $U_j$ of $M$ by open sets diffeomorphic to $\mathbb{R}^n$. (For example, first cover it by open sets which embed in $\mathbb{R}^n$, which can be done by the definition of a manifold, and then cover each of those by open cubes of the form $\prod (a_j, b_j)$.) If $\omega$ is any compactly supported $n$-form, then we can cover $\mathrm{Supp}(\omega)$ by finitely many $U_j$. We will prove the result by induction on how many $U_j$ it takes to cover $\mathrm{Supp}(\omega)$. The base case, where $\mathrm{Supp}(\omega)$ is contained in one $U_j$, is the first part.

So, suppose that $\omega$ is supported on $U_1 \cup \cdots \cup U_N$ for $N>1$. Write $\omega = \alpha + \beta$ where $\alpha$ is supported on $U_1 \cup \cdots \cup U_{N-1}$ and $\beta$ is supported on $U_N$. Choose some chain of open sets $V_0 = U_1$, $V_1$, $V_2$, ..., $V_k = U_N$ where $V_j \cap V_{j+1}$ is nonzero. Put $\beta_k = \beta$ and choose forms $\beta_0$, $\beta_1$, ..., $\beta_{k-1}$ with $\beta_j$ supported on $V_{j} \cap V_{j+1}$ so that $\int \beta_0 = \int \beta_1 = \cdots = \int \beta_k$. Then, by Part One on $V_j$, the forms $\beta_{j-1}$ and $\beta_j$ are cohomologous. So $\alpha+\beta = \alpha+\beta_k$ is cohomologous to $\alpha+\beta_0$. Since $\alpha+\beta_0$ is supported on $U_1 \cup \cdots \cup U_{N-1}$, induction shows that it is $d$ of a compactly supported form.