Deformation of foliation Suppose $\kappa$ is a no-where vanishing 1-form, then its kernel is integrable is equivalent to condition $d\kappa \wedge \kappa = 0$.
My question is, can such foliation smoothly deformed such that actually $d\kappa = 0$?
EDIT: Smooth means real $C^\infty$, and I want the $\kappa$ to be nowhere-vanishing along the deformation. Thanks Loïc Teyssier pointing out the lack of context.
Thank you.
 A: If $\mathcal F$ is a smooth codimension one foliation on $S^3$ then $\mathcal F$ cannot be defined by a closed $1$-form. This follows from Novikov's compact leaf theorem which says that one of the leaves of $\mathcal F$ is a torus $T$ bounding a Reeb component. If $\mathcal F$ were defined by a closed $1$-form then every leaf near $T$ would also be compact. Indeed, at a neighborhood of $T$,
 any  closed $1$-form defining $\mathcal F$ would admit a primitive as we can normalize local primitives by setting them to be zero at $T$. Level sets of this primitive would be other compact leaves for $\mathcal F$. Reeb global stability Theorem implies that every leaf of $\mathcal F$ is compact, and that $Sˆ3$ admits a map to $S^1$ with connected fibers. But this is impossible since from it we get a surjective map from $\pi_1(S^3)=0$ to $\pi_1(S^1)=\mathbb Z$.
The same argument applies to codimension one foliations on  compact $3$-folds with finite fundamental group.
A: The answer is "no" in some rather natural sense of "deformation". There is a construction due to G. Hector of an everywhere regular smooth (=real $C^\infty$) foliation of $\mathbb R^3=\{(x,y,z)\}$ of codimension $1$, everywhere transverse to the fibers of the projection $(x,y,z)\mapsto(x,y)$, with a distinguished leaf $\{z=0\}$, such that the holonomy computed on some transverse $\Sigma:=\{(x,y)=cst\}$ by lifting the unit circle of $\{z=0\}$, coincides outside a neighborhood of $\Sigma\cap \{z=0\}$ with some beforehand-given smooth diffeomorphism $h : \mathbb R\to \mathbb R$. Now deforming this foliation in the space of Frobenius-integrable foliations could mean  that the holonomy is preserved up to $C^\infty $ conjugacy (what would you want to deform a foliation for if it means breaking the transverse structure?). But only the identity can be obtained as the monodromy of a regular closed $1$-form on $\mathbb R^3$ since $d\kappa=0$ implies the foliation has a smooth first-integral $f$, i.e. $\kappa=df$. 
A: Although jvp's argument is correct, there's a more elementary argument that a foliation on $S^3$, or any closed integral homology sphere $M$ ($b_1(M)=0$), cannot admit a foliation defined by a closed (nowhere zero) 1-form. The point is that if $d\kappa=0$, then since $H_1(M;\mathbb{R})=0$, there exists a smooth function $f:M\to \mathbb{R}$ so that $\kappa=df$. Let $p\in M$ be a point at which the maximum of $f$ is achieved, then $\kappa_p=df_p = 0$, contradicting that $\kappa$ is nowhere zero. 
In fact, there is a complete characterization of foliations defined by closed nowhere-zero 1-forms. Suppose $\kappa$ is a closed, nowhere zero 1-form. Then $ker(\kappa)$ defines an integrable plane field, which integrates to a foliation $\mathcal{F}$ with a transverse measure given by integration agains $\kappa$. A closed manifold with a measured foliation must fiber over $S^1$, and $\kappa$ may be deformed to a closed nowhere zero 1-form $\alpha$ with integral periods, which defines a submersion to $S^1$, and therefore a fibration $\Sigma \to M \to S^1$, where $\Sigma$ is a closed surface tangent to $ker(\alpha)$.
The deformation argument is simple. Take $k=b_1(M)$ loops $c_1,\ldots,c_k$ which generate $H_1(M;\mathbb{Z})/Torsion$. For any closed 1-form $\beta$, we obtain a $k$-tuple of periods $(\beta(c_1),\ldots,\beta(c_k))\in \mathbb{R}^k$. Choose $k$ closed 1-forms $\alpha_1,\ldots,\alpha_k$ generating $H^1(M;\mathbb{R})$. Then for small $\epsilon$, we have $\kappa+\sum t_i \alpha_i$ is a nowhere zero closed 1-form for $|t_i|< \epsilon$. Moreover, the period map takes these forms to a small open set in $\mathbb{R}^k$. This set must contain a point $\kappa+\sum t_i\alpha_i$ with rational coordinates, so we choose $\alpha=m(\kappa +\sum t_i \alpha_i )$, where $m$ is the $lcm$ of the denominators of the coordinates. Clearly $\alpha$ smoothly deforms through closed nonzero 1-forms to $\kappa$, and $\alpha$ has integral periods.  
The foliations described by closed 1-forms may be characterized  up to isotopy: there is a unique such foliation for each element in the projective cone over the fibered faces of the Thurston norm unit ball. If two nowhere zero closed 1-forms are cohomologous, one can show that the corresponding foliations are isotopic, and the same if one is a multiple of the other (by rescaling).
Associated to each face of the Thurston norm unit ball, Thurston showed that there is a canonical element of $H^2(M)$ which realizes the Thurston norm when evaluated on norm-minimizing surfaces in that face. For a nowhere zero integrable 1-form $\alpha$ defining a foliation (so $\alpha \wedge d\alpha=0$), the Euler class of $ker(\alpha)$ has the property that it gives the Euler characteristic of closed leaves. Deformation of integrable 1-forms preserves the Euler class. Thus, if an integrable 1-form is deformable to a closed 1-form, then it must induce the same Euler class as the class associated to a fibered face of the Thurston norm unit ball. For example, associated to each homology class on the boundary of a fibered face, there is a depth one foliation realizing the Euler class of the fibered face, but which is not itself cohomologous to a fibration. To completely answer your question, one would therefore like to know if integrable 1-forms which define the same Euler class are deformations of each other. This question is partially resolved. Suppose that two foliations of a 3-manifold are deformable, then their tangent bundles are homotopic. Eynard-Bontemps proved a converse, that if two oriented foliations have homotopic tangent bundles, then they are deformable, but only through $C^1$ foliations. If the two foliations are taut, then the deformation may be made smoothly. In your case, at one end the foliation is taut (when defined by a closed 1-form), but the initial foliation defined by $\kappa$ might not be taut. 
