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Edit: Accoring to the comment of Asura Path I revise the question.

Let $X$ be a vector field on a Riemannian manifold $(M,g)$. So we have a $1$-form $\beta$ with $\beta(Y)=\langle Y,X\rangle_g$.

We consider the following subcomplex of de Rham complex $\Omega^*(M)$: $$\{\alpha\in \Omega^*(M)\mid d\alpha=\alpha \wedge \beta\}$$

This generates a cohomology. For zero vector field we get the standard de Rham cohomology.

Is this cohomology always a finite dimensional space? Does it depend on the Riemannian structure? Does it contain some dynamical information about the vector field $X$?

Is there an appropriate analogy of this cohomology in algebraic topology or other type of cohomologis? (In the context of algebraic topology, we replace $1$-form $\beta$ with a $1$-cochain and the exterior derivative with coboundary and wedge product with cup product)

Improve this question
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    $\begingroup$ "This generates a cohomology with the standard exterior derivation." That sentence sounds somewhat confusing to me. Do you just mean that it is a complex and we can compute its cohomology (note that mentioning exterior derivation seems to be redundant, since you already said that it is a subcomplex, so there is only one possible differential on it)? $\endgroup$
    – user141498
    Commented Jun 8, 2019 at 11:44
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    $\begingroup$ I changed $$\beta(Y)=<Y,X>_g$$ to $$\beta(Y)=\langle Y,X\rangle_g.$$ That is standard usage. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Jun 8, 2019 at 16:27
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    $\begingroup$ Two comments: (1) You don't mention it, but there are other conditions for $\alpha$ to be in the subcomplex: for instance, $\alpha \wedge d\beta = 0$. Hence cocycles in the subcomplex are killed by exterior product with both $\beta$ and $d\beta$. (2) If $X$ is nowhere vanishing, so that neither is $\beta$, then the cohomology of the subcomplex is trivial: since the differential agrees up to a sign with exterior product by $\beta$ and exterior product by a nowhere-vanishing 1-form defines an exact sequence. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jun 8, 2019 at 19:30
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    $\begingroup$ @AliTaghavi Let me know if I am misinterpreting. If $X$ is the zero vector field, then $\beta$ is the zero covector field, so for any $\alpha$ we have $\alpha \wedge \beta=0$. Then the condition for your subcomplex just makes it the subcomplex of cocycles, which has trivial differential. This subcomplex does not give the same cohomology as the de Rham complex. $\endgroup$
    – Connor Malin
    Commented Jun 10, 2019 at 4:24
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    $\begingroup$ @op, u might want to look up "twisted de rham cohom" but nb this is about taking differential $d+\lambda$ for a form $\lambda$. (so u're taking the cycles wrt this differential as ur subcomplex). Eg if the vec field is Hamiltonian u can check u get just dR cohom (multiply by exp(f)). Hopefully this will help, but doesn't answer ur question sadly $\endgroup$
    – user108998
    Commented Jun 13, 2019 at 20:04

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