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Is there a $2$- dimensional foliation of $\mathbb{R}^4\setminus \{0\}$ whose tangent space is contained in $\ker \alpha$ where $\alpha$ is the following non integrable $1$-form?

$$\alpha=(x^2+y^2)dx+(xz+yw)dy+(xz+yw)dz+(z^2+w^2)dw$$

In the matrix language, the $1$-form $\alpha$ can be counted as $\alpha=trace(AA^{tr}dA)$ where $$A=\begin{pmatrix}x&y\\ z&w \end{pmatrix}$$

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    $\begingroup$ $\ker\alpha$ is not a $3$-dimensional foliation, but is, instead, a rank $3$ distribution away from the origin in $\mathbb{R}^4$. Are you asking for a codimension $2$ foliation such that $\alpha$ vanishes when pulled back to each leaf? Do you need the foliation to be well-defined at the origin, where $\alpha$ vanishes and hence does not define a smooth $3$-dimensional distribution? $\endgroup$
    – Robert Bryant
    Commented Mar 6, 2018 at 9:55
  • $\begingroup$ @yes I search for a 2 dimensional (hence 2 codimensional) foliation in the punctured space whose tangent space(of leaves) ate contained in the 3 dim dostribution. I remove the origon from $\mathbb{R}^4$. $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2018 at 10:05
  • $\begingroup$ @RobertBryant I think I did not indicated in the question that the distribution defines a foliation. $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2018 at 11:27
  • $\begingroup$ @RobertBryant My apology. I was not realized of my mistake in the title of the previous version. Thanks for your comment. $\endgroup$
    – Ali Taghavi
    Commented Mar 6, 2018 at 14:50
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    $\begingroup$ @AliTaghavi: By the way, I don't know if you have noticed, but Daniele's observation that the $1$-form that you wrote down is degenerate along the whole line $y = z = x-w = 0$ shows that the problem is tricky near there. That is the locus of places where $\alpha\wedge \mathrm{d}\alpha$ vanishes, even though $\alpha$ itself only vanishes at the origin of $\mathbb{R}^4$. Are you sure that the $\alpha$ that you want to ask about is the one you wrote? $\endgroup$
    – Robert Bryant
    Commented Mar 7, 2018 at 20:52

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There is no co-dimension 2 foliation of the desired kind in a neighborhood of any point on the line $L\subset\mathbb{R}^4$ defined by $x-w=y=z=0$. Thus, if one wants to find such a foliation, it will be necessary to remove this entire line from the domain.

Here is the reason: Consider the $3$-form $\Upsilon = \alpha \wedge \mathrm{d}\alpha$. It vanishes along the line $L$. When one expands it in the $wxyz$-coordinates, the coefficients of this $3$-form are cubic polynomials in $wxyz$.

Now, suppose that there were a codimension $2$ foliation of the desired kind in a neighborhood of a point $p\in\mathbb{R}^4$ and let it be defined as the simultaneous level sets of two functions $f$ and $g$ defined in a $p$-neighborhood $U\subset\mathbb{R}^4$ and independent there: $\mathrm{d}f\wedge\mathrm{d}g$ is nonvanishing on $U$. The condition that $\alpha$ vanish when pulled back to the leaves of this foliation is that $\alpha\wedge\mathrm{d}f\wedge\mathrm{d}g$ vanish identically. This also implies that, when $\alpha$ is pulled back to a level set of $\mathrm{d}f$, it must become integrable, i.e., $\Upsilon = \alpha \wedge \mathrm{d}\alpha$ must vanish when pulled back to a level set of $f$, i.e., $\alpha\wedge\mathrm{d}\alpha\wedge\mathrm{d}f$ vanishes identically on $U$. Similarly, $\alpha\wedge\mathrm{d}\alpha\wedge\mathrm{d}g$ must vanish identically on $U$.

Now, using a Taylor series expansion about $p$, it is easy to show that, when $p$ lies on the line $L$, the equation $\alpha \wedge \mathrm{d}\alpha\wedge \mathrm{d}f = 0$ implies that, at $p$, $\mathrm{d}f$ must be a multiple of $\mathrm{d}(x{+}w)$. Similarly, at $p$, $\mathrm{d}g$ must be a multiple of $\mathrm{d}(x{+}w)$. Thus, $\mathrm{d}f\wedge \mathrm{d}g$ must vanish at $p$, contrary to the way $f$ and $g$ were chosen.

Thus, the claim is established.

Remark: When $p$ is the origin, the Taylor series expansion shows that $\mathrm{d}f$ must actually vanish at the origin, so the exclusion is even stronger there.

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