How to construct a $k$-form on $S^k$ with nonzero integral ?
I think this can be done by Bump function $\rho$ on $R^k$ and define $w = \rho\, dx_1\,dx_2 \cdots dx_k$ on $R^k$. Now pull it back by the chart map (here stereographic projection map.)Then this pullback form will serve from the required form on $S^k$.
Can anyone help me by giving a rigorous proof of the above question.
Thank you.
Let $\phi: S^k \setminus \{z_0\} \to \mathbb{R}^k$ be the stereographic projection, where $z_0$ denotes the north pole of $S^k$. Then for an arbitrary bump function $\rho: \mathbb{R}^k \to [0,1]$, define $\omega=\phi^* (\rho \, dx_1 \wedge\cdots \wedge dx_k)$ on $S^k\setminus \{z_0\}$. How should we define $\omega$ at $z_0$ so that $\omega$ is a smooth $k$-form on all of $S^k$? What's it equal to at points near $z_0$? I'd bet you can answer that.
In general, to integrate a $k$-form on $S^k$ you need to break $S^k$ up into domains of charts and use a partition of unity. But if $\omega$ is defined properly, its support of lies in the domain of a single chart, namely $\phi$. So integrating over $S^k$ is the same as integrating over the domain of that chart: $$\int_{S^k} \omega = \int_{S^k \setminus \{z_0\}} \omega.$$ It just remains to show that this integral is nonzero.