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Let $\alpha$ be a non vanishing one form on a manifold which which defines a codimension one foliation. With this $\alpha$ we define the following complex:

$$\phi:\Omega^{i}(M)\to \Omega^{i+2}(M)\;\;\;\phi(\beta)=d(\alpha\wedge \beta)$$ Obviously $\phi$ satisfies $\phi \circ \phi=0$, so we have cohomologies associated with this complex. The total cohomology is denoted by $H^{*}(\alpha)$

Are these cohomologies finite dimensional vector space?

Are there some dynamical information in this cohomology?

Is this cohomology independent of choosing the one form $\alpha$ which kernel is tangent to the foliation? This means that: Is it true to say $H^{*}(\alpha) \simeq H^{*}(f\alpha)$ for a non vanishing smooth function $f$?

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First, it's not finite dimensional, even in the case of a torus. Just let $x,y$ be the $2\pi$-periodic functions on the torus and take $\alpha = \mathrm{d} x$, and you'll see that $H^0$ is all the functions of the form $f(x)$. On the other hand, if you let $\alpha = \mathrm{d} x + \sqrt{2}\,\mathrm{d} y$, then $H^0$ just consists of the constants, so it definitely depends on the foliation.

Second, the answer is 'yes, it is independent of the multiple' because you can simply replace all of your representatives by the inverse of that multiple.

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  • $\begingroup$ Thank you very much for your answer. According to your example, is it natural to ask "what is a dynamical interpretation for finite dimensionality of $H^{*}(\alpha)$ for all dimension $*$, where the manifold is not necessarilly a low dim manifold. Moreover is wedge product well defind in this cohomology? And finally is it easy to compute these cohomology for some higher dimension:ex $S^{3}$ with the Reab foliation? Thanks again for your answer. $\endgroup$ – Ali Taghavi Jan 30 '15 at 12:12
  • $\begingroup$ In the other word, by dynamical interpretation I mean: Is finite dimensionality of these cohomology, a topological invariant: that is: Assume that two foliations are topological equivalents. Do they have the same cohomology? $\endgroup$ – Ali Taghavi Jan 30 '15 at 12:23

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