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A linear $1$-form on $\mathbb{R}^n$ is a $1$-form $\alpha=\sum_i P_i(x_1,x_2,\ldots,x_n)dx_i$ such that each $P_i$ is in the linear form $P_i=\sum_j a_{ij}x_j$. A linear foliation of $\mathbb{R}^n \setminus \{0\}$ is a foliation tangent to the kernel of a linear $1$-form $\alpha$ whose corresponding matrix $(\alpha_{ij})$ is a non singular matrix.

As I learned from this answer, there are linear $1$-forms which are Frobenius integrable but are not closed $1$-form.

For $n>2$, is there a complete classification and dynamical description of all linear foliations? In particular, is there a linear foliation of $\mathbb{R}^n \setminus \{0\}$ which has a (compact) leaf with non trivial holonomy?

I think that this situation can not occurs when the corresponding $1$-form $\alpha= \sum_i \sum_j(a_{ij}x_j)dx_i $ is a closed $1$-form.(Equivalently the matrix $(a_{ij})$ is a symmetric matrix)

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    $\begingroup$ Is $\alpha$ assumed to be regular outside $0$ ? $\endgroup$ Oct 10, 2017 at 12:10
  • $\begingroup$ @LoïcTeyssier Yes it is assumed that the matrix $(\alpha_{ij})$ is a non singular matrix. I revise the question to emphasis this regularity. $\endgroup$ Oct 10, 2017 at 12:13
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    $\begingroup$ Actually, I gave the classification of the cases in which the foliation is integrable in the answer that you cite. The answer is that these are exactly either the closed 1-forms or the non-closed ones that can be pulled back by a linear projection to $\mathbb{R}^2$. You should be able to deduce the holonomy of the leaves directly from this. In fact, it follows immediately from the fact that the holonomy will be the same as the holonomy in the $2$-dimensional case, namely that it is trivial because it's trivial there. $\endgroup$ Oct 10, 2017 at 14:49
  • $\begingroup$ @RobertBryant I try to deduce this from your answer to my previous question. Thanks $\endgroup$ Oct 10, 2017 at 15:00

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I don't know offhand the answer of your first question, but I can answer the particular situation you describe afterwards : the holonomy is always trivial.

First, notice that a compact leaf $L$ is everywhere transverse to the radial vector field $R$ (if a ray $x\mathbb R_{>0}$ were tangent to $L$ then it would be included in $L$, breaking compactness). This implies that $L$ meets any such ray exactly once : it is diffeomorphic to a $(n-1)$-sphere.

Now, every rescaling $\lambda L$ for $\lambda\in \mathbb R^\times$ is also a compact leaf, thus $\alpha$ foliates the space with smooth spheres. The foliation is therefore smoothly equivalent to the boring trivial fibration $\mathbb R_{>0}\times\mathbb S^{n-1}$.

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  • $\begingroup$ Thanks for your answer. But I think that some thing is missing in the second paragraph "A compact leaf can exist if the corresponding matrix has no real eigenvalue". For example, let $n=2$ The identity matrix gives $xdx+ydy=0$ which gives foliation by circle. $\endgroup$ Oct 10, 2017 at 14:35
  • $\begingroup$ A real eigenspace of $A$ is invariant under the flow of 1 dimensional foliation defined by $X'=AX$. But I am considering a different object. Right? $\endgroup$ Oct 10, 2017 at 14:59
  • $\begingroup$ Sure, I'm too used to dealing with vector fields ;) I'll remove that part of my answer, which is not relevant anyway (see the edit). $\endgroup$ Oct 10, 2017 at 15:08
  • $\begingroup$ Why does transversality of the compact leaf with radial direction imply that the leaf is diffeomorphic to sphere? Is not possible that we embed a tori in $\mathbb{R}^3$ with transverse intersection with radial direction? $\endgroup$ Oct 11, 2017 at 5:26
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    $\begingroup$ Well, no, it's not possible to embed a torus in such a way. Try a connectedness argument. $\endgroup$ Oct 11, 2017 at 7:06

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