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Jun 8, 2019 at 10:33 vote accept Ali Taghavi
Mar 27, 2014 at 17:34 comment added Ali Taghavi In the previous comment I assume $f(n\pi , m\pi )=0$ for all integer $m$ and $n$
Mar 27, 2014 at 14:48 history edited Ben McKay CC BY-SA 3.0
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Mar 27, 2014 at 14:42 comment added Ali Taghavi @BenMcKay In the generic case I think the cohomology could be finite. Identify the torus with $\frac{\mathbb{R}^{2}}{2\pi \mathbb{Z}^{2}}$.Put $\alpha=sinxdx+sinydy$. Assume that $\beta=f(x,y)dx\wedge dy$ is a two form. I guess that there is an analytic one form $\gamma$ on torus with $\alpha \wedge \gamma= \beta$. because (perhaps) we can write $f(x,y)=Psin(x)+Qsin(y)$ for periodic analytic functions $P$ and $Q$, provided $f$ vanishes at point with coordinates=$k\pi$, so a finite dimensional codimension.Am i mistaken?
Mar 27, 2014 at 12:46 comment added Ali Taghavi @BenMcKay see page 9(the second line) of Geometry of foliation by Tondeur for usage of partition of unity
Mar 27, 2014 at 12:29 comment added Ali Taghavi @BenMcKay Thanks, Did you considered the real analytic assumption in my question? Before that I post this question, I was aware of the Cartan lemma, but i think that its proof is based on usage of partition of unity(If I am not mistaken)
Mar 27, 2014 at 10:58 comment added Ben McKay If $\alpha$ vanishes on a hypersurface, you get infinite dimensional cohomology by the same reasoning: write $\alpha=\alpha_0/x$, say, with $\alpha_0$ not vanishing on $x=0$, and look at $\beta$ not vanishing but divisible by $\alpha_0$. The interesting case might be for a generic 1-form $\alpha$, so that $\alpha$ has discrete zeroes and near each zero, in suitable coordinates, $\alpha=\sum A_{ij} x^i dx^j$ with $A$ a bilinear form of full rank. I think you still get infinitely dimensional cohomology.
Mar 27, 2014 at 9:46 comment added Mark Grant What happens in the intermediary case between "nowhere zero" and "zero"?
Mar 27, 2014 at 8:13 history answered Ben McKay CC BY-SA 3.0