If $\delta c = 0$, does it follow that $d\xi = 0$?

Question

Let $\mathcal{U} = \{\mathcal{U}_\alpha\}_{\alpha = \infty}^\infty$ be a locally finite open covering of the manifold $M^n$, with smooth functions $\lambda_\alpha$, compactly supported in $U_\alpha$. Let $c = \{c_A = c_{\alpha_0 \alpha_1 \ldots \alpha_k}\}$ be a Cech $k$-cochain on $\mathcal{U}$, with constant real coefficients. Define a $k$-form $\xi$ on $M$ by$$\xi = \sum_A c_{\alpha_0 \alpha_1 \ldots \alpha_k} \lambda_{\alpha_0} d\lambda_{\alpha_1} \wedge \ldots \wedge d\lambda_{\alpha_k}.$$If $\delta c = 0$, does it follow that $d\xi = 0$?

Answer 1

I think this cannot work in general, since the choice of the functions $\lambda_a$ gives you too much freedom. To get an explicit counter-example, consider a covering $\{U_1,U_2,U_3,U_4\}$ of $S^2$ such that all possible triple intersections are non-empty but the intersection of all 4 sets is empty. (This is dual to the triangulation as a tetrahedron.) Then choose $\lambda_1$ and $\lambda_2$ with support contained in $U_{12}$ such that $d\lambda_1\wedge d\lambda_2\neq 0$. (Just take two coordinate functions and multiply by an appropriate bump function.) Further, choose $\lambda_3$ and $\lambda_4$ in such a way that their support is disjoint from $U_{12}$. Now for $k=1$, you will certainly find a Czech cocycle $c$ such that $c_{12}\neq 0$. But among the one-forms $\lambda_id\lambda_j$ with $i<j$, only $\lambda_1d\lambda_2$ is non-zero by construction. So you simply get $\xi=c_{12}\lambda_1d\lambda_2$ and $d\xi=c_{12}d\lambda_1\wedge d\lambda_2\neq 0$.