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Apr 7, 2015 at 19:27 comment added Grimolatto Also, notice that, for the case $\partial =\mathrm{d}$, $H^1(S^1,\mathrm{d})=\mathbb{R}$...
Apr 7, 2015 at 18:52 comment added Grimolatto Sorry for the slopiness in the question. What I mean is that the functions $f_i$ could be different for each open subset $U\subset M$, and for each $p\in U$, $\{\partial f_1(p),\ldots,\partial f_m(p)$ must be a basis of $T^*_p M$.
Apr 7, 2015 at 17:11 comment added Mark Grant What do you mean by local basis here? If you just want a basis of $\Omega^1(M)$ of the given type, then it seems like $H^1(\Omega(M),\partial)=0$ is a necessary and sufficient condition.
Apr 7, 2015 at 9:27 history asked Grimolatto CC BY-SA 3.0