Existence of $k$-form with nonzero integral

Question

Say that $N$ is an oriented, compact, connected manifold without border. If $\operatorname{dim}(N) = k$, does it always exists some $k$-form $\omega$ such that $\int_N \omega \neq 0$?

I know how to proceed in some particular cases (e.g., $S^k$), but I have no idea how to prove the general case, if it is even true. In the literature I've checked (mostly Lee's Introduction to Smooth Manifolds, 2nd edition, and Guillemin/Pollack Differential Topology) there is nothing, also - but I may have missed something.

Answer 1

Let $(U, \varphi)$ be an oriented local chart of $M$. On $\varphi (U) \subset \mathbb R^k$ you have a $k$-form $\omega_0 = dx^1 \wedge \cdots \wedge dx^k$ and a bump function $b : \varphi(U) \to \mathbb R$ (i.e. a smooth function with compact support in $\varphi(U)$), then $$\varphi^*(b \omega_0)$$ is a smooth $k$-form on $M$ (by extending to zero outside of $U$). By definition of the integration,

$$ \int_M \omega = \int_{\varphi (U)} b\omega_0 = \int_{\varphi (U)} b(x) dx^1\cdots dx^k.$$

One can choose this to be non-zero (e.g. choose $b\ge 0$ and $b>0$ on an open subset)

Answer 2

From Lee's Introduction to Smooth Manifolds, 2nd edition:

Theorem 17.31 If $M$ is a compact connected orientable smooth $n$-manifold, then $H_{\text{dR}}^{n}(M)$ is $1$-dimensional, and is spanned by the cohomology class of any smooth orientation form.

Here, $H_{\text{dR}}^{n}(M)$ is the standard de Rham cohomology group, that is, $$H_{\text{dR}}^{n}(M)=\{\text{closed $n$-forms on $M$}\}/\{\text{exact $n$-forms on $M$}\}.$$

If you see the proof of Theorem 17.30, you can see that the isomorphism is given by \begin{align*} H_{\text{dR}}^{n}(M) & \to \mathbb{R}, \\ \omega & \mapsto \int_{M} \omega. \end{align*}

Hence, the answer to your question is affirmative.