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In this question all objects are real analytic.(manifolds, differential forms..)

Assume that $M$ is a compact manifold and $\alpha \in \Omega^{1}(M)$ is a one form.

We define a map $\phi:\Omega^{*}(M)\to \Omega^{*+1}(M)$ with wedge product; $\phi(\beta)=\alpha \wedge \beta$. Then $\phi \circ \phi=0$. Then we have a complex of vector spaces. So we naturally obtain a cohomology.

Is each cohomology, a finite dimensional vector space?

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  • $\begingroup$ For an $\alpha$ whose support has nonempty interior, the degree-$0$ cohomology is the space of smooth functions on $M$. $\endgroup$
    – Gunnar Þór Magnússon
    Mar 27, 2014 at 17:07
  • $\begingroup$ @GunnarMagnusson in degree zero the cohomology=$\ker \phi$.. If $\alpha$ is a non vanishing 1-form, the kernel of $\phi$ is trivial, So the 0-cohomology is trvial. Am I correct? $\endgroup$
    – Ali Taghavi
    Mar 27, 2014 at 17:07
  • $\begingroup$ Ah, of course, silly me. I still think you won't get finite-dimensional groups in general. We can maybe see that in your torus example by looking at the Fourier expansions of the coefficient functions of forms. $\endgroup$
    – Gunnar Þór Magnússon
    Mar 27, 2014 at 17:10

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Suppose that $\alpha$ is nowhere zero. A differential form $\beta$ satisfies $\alpha\wedge \beta=0$ just when $\beta=\alpha \wedge \gamma$ for some $\gamma$ by Cartan's lemma. So the cohomology vanishes, finite dimensional. On the other hand, take $\alpha=0$. Then the kernel is everything, the image nothing, so the quotient is everything, infinite dimensional.

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  • $\begingroup$ What happens in the intermediary case between "nowhere zero" and "zero"? $\endgroup$
    – Mark Grant
    Mar 27, 2014 at 9:46
  • $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Mar 27, 2014 at 10:58
  • $\begingroup$ @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) $\endgroup$
    – Ali Taghavi
    Mar 27, 2014 at 12:29
  • $\begingroup$ @BenMcKay see page 9(the second line) of Geometry of foliation by Tondeur for usage of partition of unity $\endgroup$
    – Ali Taghavi
    Mar 27, 2014 at 12:46
  • $\begingroup$ @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? $\endgroup$
    – Ali Taghavi
    Mar 27, 2014 at 14:42
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The cohomology is finite-dimensional for a generic $1$-form. In fact, it has a fairly simple description.

Indeed, as Ben McKay pointed out, it is sufficient to check this locally near a generic singularity, which has the form $\alpha= A_{i,j} x^i dx^j$. We might as well write this as $\sum_j y_j dx^j$, where $y_i = \sum_i A_{i,j}x^i$. Any failure of exactness among analytic functions will occur already in polynomials, since we can just take the leading terms of the power series. So we can write a complex of modules

$$\mathbb R[y_1,\dots,y_n] \to \bigoplus_{i} \mathbb R[y_1,\dots,y_n]dx^i \to \dots \to \mathbb R[y_1,\dots y_n] dx^1\dots dx^n$$

with the arrows coming from wedging with $\alpha$.

This complex is the tensor product of the complexes $\mathbb R[y_i] \to \mathbb R[y_i]$, with the arrow coming from multiplication by $y_i$, for $i$ from $1$ to $n$. This complex has cohomology $0$ except for degree $1$, where it is $\mathbb R$. So the tensor product has cohomology $0$ except for degree $n$, where it is $\mathbb R$. This $\mathbb R$ comes from the fact that all forms produced by wedging with $\alpha$ vanish at $0$, whereas a general $1$-form need not vanish.

So the total cohomology is $0$ except in dimension $n$, where it is $\mathbb R^d$, $d$ the number of singular points.

For a $1$-form with special singularities, the complex is equivalent to the Koszul complex with respect to the coefficients of the $dx^i$.

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