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Inspired by An algebraic Hamiltonian vector field with a finite number of periodic orbits (2) we ask if there is a 1 dimensional analytic foliation of $\mathbb{R}^4$ which has at least 1 compact leaf and the number of compact leaves of the foliation is finite?

Note: If there is no an analytic example what about a smooth example?

LSpice
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Ali Taghavi
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Yes, there is. Conceptually, imagine a flow on $\mathbb{R}^3$ with a single closed orbit on the unit circle in the plane $z=0$, while every other trajectory has an increasing z-coordinate. It's then easy to embed that flow in $\mathbb{R}^4$ to get the foliation you want.

More precisely, consider the vector field $$f(x_1, x_2, x_3, x_4) = (-x_2, x_1, (1 - x_1^2 - x_2^2)^2 + x_3^2, x_4)$$

It has no zeros, because if the third coordinate is zero, then at least one of the first two coordinates is nonzero. So the flow given by $x'= f(x)$ defines a 1D foliation of $\mathbb{R}^4$.

And it has exactly one closed orbit (i.e. compact leaf), the circle of the form $(a, b, 0, 0)$ where $a^2 + b^2 = 1$. To see this, note that the 4th coordinate has to be zero in a closed orbit, and that the quantity $x_1^2 + x_2^2$ is constant on flow trajectories.

Martin M. W.
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    Thank you and my +1 for your answer. Does not it work for every arbitrary $x_3=c$?So we have infinite number of closed orbits. Am I mistaken? – Ali Taghavi Jul 29 '23 at 19:57
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    Now that is very great thank you – Ali Taghavi Jul 29 '23 at 19:59
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    Great point, left out the key $x_3^2$ term, I just edited it to include this. Thank you! – Martin M. W. Jul 29 '23 at 19:59
  • I wonder what is the dynamics and topological behavior of codimension 1 distribution orthogonal to this beautiful example? I would appreciate if you send me an email "alitghv@yahoo.com" for more discussion. – Ali Taghavi Jul 29 '23 at 20:15