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Mar 30, 2010 at 16:18 comment added Steve Huntsman ...recall, e.g., $\int F' dx = F + C$, where really $C$ means an arbitrary constant and could be rewritten as $\mathbb{R}$.
Mar 30, 2010 at 16:15 comment added Steve Huntsman I should have said "plus the kernel of $d$".
Mar 30, 2010 at 15:58 comment added Mariano Suárez-Álvarez Not really. I am only stating the only sensible "general formulation of the notion of antiderivative that incorporates basic information about de Rham (co)homology" I can think of. As Scott observed, antiderivatives are determined---when they exist---up to cocycle. This is not the same as being determined up to a cohomoloy class (consider, for example, the situation in $\mathbb R^n$ where there are very few cohomology classes...), and I do not know what "a form plus a cohomology group" means.
Mar 30, 2010 at 15:53 comment added Steve Huntsman If I read this correctly you are saying that an antiderivative is a form plus a cohomology group.
Mar 30, 2010 at 15:46 history answered Mariano Suárez-Álvarez CC BY-SA 2.5