2
$\begingroup$

Assume that $X$ is a non-vanishing vector field on $\mathbb{R}^3$.

Is there a $2$-dimensional foliation of space such that every trajectory of $X$ is contained in a leaf of the $2$-dimensional foliation?

As a related question:

Is there a classification of all $1$-dimensional foliations of space tangent to the unit speed vector field $t$ for which the distributions $\{t,n\}$ and $\{t,b\}$ are integrable? Here $\{t,n,b\}$ is the associated Frenet frame.

$\endgroup$
4
+50
$\begingroup$

No. For a counterexample, start with the Hopf map $S^3\to S^2$, a fiber bundle with $S^1$ fibers. Its fibers are the leaves of a $1$-dimensional foliation of $S^3$ in which all leaves are closed and the space of leaves is $S^2$. Choose a vector field tangent to the leaves. Remove one point from $S^3$ to get $\mathbb R^3$. A $2$-dimensional foliation of the kind asked for would give a $1$-dimensional foliation of $S^2$.

$\endgroup$
  • $\begingroup$ Thank you very much for this interesting answer. I understand the tangent vector field to the Hopf foliation is not contained in any 2 dimensional smooth distribution $D$ otherwise since any such distribution would be map to a $1$ dimensional foliation of $S^2$.Am I correct? $\endgroup$ – Ali Taghavi Oct 1 '17 at 11:16
  • $\begingroup$ So a natural question is that "Is the Reeb foliation transverse to the hopf fibration, at all points?* $\endgroup$ – Ali Taghavi Oct 1 '17 at 11:20
  • 1
    $\begingroup$ Yes, that was my reasoning. No, the Reeb foliation has a torus leaf, but no torus in $S^3$ can be transverse to the Hopf fibration because that would make it a covering space of $S^2$. In fact, no $2$-dimensional foliation of $S^3$ that has a compact leaf can be transverse to the Hopf fibration. $\endgroup$ – Tom Goodwillie Oct 1 '17 at 18:28
  • $\begingroup$ My apology if my question is elementary. How the foliation on total space, which is not necessarily $S^1$ invariant, give us a foliation of the bases space? $\endgroup$ – Ali Taghavi Aug 4 at 21:55
  • $\begingroup$ For each point in $S^2$ its preimage in $\mathbb R^3$ would be contained in one of the leaves. Therefore each leaf would be the preimage of some subset of $S^2$. In order for the preimages of these subsets to be the leaves of a foliation, the sets themselves would have to be the leaves of a foliation. $\endgroup$ – Tom Goodwillie Aug 5 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.