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

Let $M$ be a compact smooth orientable manifold of dimension $n$. I am looking for a simple proof that $H_{dR}^n(M) \cong \mathbb R$. Equivalently, an $n$-form which integrates to 0 is exact. I can show this via a rather indirect argument as follows: we know $H_{dR}^n(M) \cong H^n(M, \mathbb R)$, 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 $Hom(H_n(M, \mathbb Z) , \mathbb R)$. 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 $Hom(H_n(M, \mathbb Z) , \mathbb R) \cong \mathbb R$. However, it seems like there should be a simpler way to prove this. Does anyone know of one?

• You probably want $M$ to be connected. – Zhen Lin Nov 3 '12 at 23:30
• Integrate!! It is fairly straightforward to check that $\omega \mapsto \int_M \omega$ is the desired isomorphism. – Matt Nov 3 '12 at 23:30
• I know that the isomorphism takes this form; the part I am having trouble with is seeing why it is injective. – user15464 Nov 3 '12 at 23:41
• I don't think there is a simple argument. (Standard proof of Poincare duality uses Mayer–Vietoris for induction by covering, I believe.) – Grigory M Nov 5 '12 at 15:02
• @user15464: Maybe too late, but there is a nice elementary inductive proof in Spivak Vol 1, chap. 8. – Luis A. Florit Oct 26 '17 at 23:52

$$\def\RR{\mathbb{R}}$$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 $$f_k(x_1, \ldots, x_n) = h(x_1) x(h_2) \cdots h(x_k) \left( \int_{(y_1,\ldots,y_k) \in \RR^k} f(y_1, \ldots, y_k, x_{k+1}, \ldots, x_n) dy_1 \cdots dy_k \right).$$ 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 $$g_k(x_1, \ldots, x_n) = \int_{z=-\infty}^{x_k} \left( f_k(x_1, \ldots, x_{k-1},z,x_{k+1}, \ldots, x_n) - f_{k-1}(x_1, \ldots, x_{k-1},z,x_{k+1}, \ldots, x_n) \right),$$ then $$g_k$$ is compactly supported, and we have $$(f_{k} - f_{k-1}) dx_1 \wedge \cdots \wedge dx_n = (-1)^{k-1} d \left( g_k dx_1 \cdots \widehat{dx_k} \cdots \wedge dx_n \right)$$.
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 $$\int \beta_1 = \int \beta_2 = \cdots = \int \beta_k$$. 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.