# Structure of foliations of codimension 2 on three dimensional torus

Is it possible to have a one-dimensional foliation on three dimensional torus such that the foliation has a trefoil knot as its leaf? Moreover, does a one dimensional foliation on three dimensional torus have both compact and non-compact leaves?

Thank you very much!

[Apologies: the answer I wrote below is for $S^2 \times S^1$, not the 3-torus. Corrections added. The case of a 3-torus, or any 3-manifold, is included at the end.]

1. First, there are different kinds of "trefoil knots" in $M = S^1 \times S^2$. Let's say we are interested in the "split trefoil knot", which lies in a 3-ball in $M$. This does in fact occur as a leaf of a foliation and it is easy to see.

2. To build it, first recall the standard foliation of $S^3$ by circles where the one leaf is the trefoil knot $T$. This can be obtained by taking the orbits of the circle action $(z,w) \mapsto (t^2 z, t^3 w)$, $|t|=1$, on the 3-sphere in $\mathbb C^2$.

3. Next we observe that the trefoil knot $T$ and a "disjoint" (split) unknot $U$ in $S^3$ can arise as the leaves of a foliation of $S^3$. (Split means there is an $S^2$ separating $U$ from $T$). To see this, consider the unknot $U'$ in (2) that arises as the orbit of $(z,w) = (1,0)$. This would almost do, except that $U'$ links $T$, i.e. $U'$ and $T$ are not split.

4. To fix this, we use the fact (mentioned by Zare) that the space of constant slope foliations of a 2-torus is connected (it is just a circle). First adjust the foliation in (2) so it contains a solid torus $D^2 \times S^1$ foliated by $p\times S^1$'s with $U' = 0 \times S^1$. Here $D^2 = D^2(1)$ is the unit disk. By rotating the foliations on the tori $N_r = S^1(r) \times S^1$ as $r$ varies from 0 to 1, we can arrange that for some $r>0$ the torus $N_r$ is foliated by leaves of the form $U = S^1(r)\times p$, $p$ in $S^1$. Each of these leaves is an unknot split from $T$, so we have shown 3.

5. To complete the picture, we now observe that $S^2 \times S^1$ can be obtained from $S^3$ by Dehn surgery along $U$, i.e. by cutting out a solid torus (disjoint from a ball containing $T$) with core curve $U$ and gluing it in again differently. Using the rotating foliation idea again, it is easy to extend the foliation from (3) to a foliation of $M$ with $T$ as a leaf. By construction, $T$ is a split trefoil in $M$.

Case of the 3-torus: It is also true that for any closed 3-manifold $M$ and any link $L$ in $M$, there exists a nowhere zero vector field on $M$ such that $L$ is invariant under the flow. One way to see this is to use Zare's construction, which depends on two facts: (a) $M$ can be obtained by Dehn surgery on a link $L'$ in $S^3$ (Lickorish) and (b) any link $L''$ in $S^3$ can be presented as a closed braid (Alexander). This construction will be a little less explicit than the one above, since it uses (a) and (b).

The main point is to prove the statement when $M=S^3$. For this one can start with any flow (e.g. the Hopf flow) such that there exists a closed trajectory. Thickening this flow line, we obtain an unknotted torus $S^1\times D^2$ with the product foliation. The foliation is obtained by suspending the identity map $D^2 \rightarrow D^2$. Now let us present $L$ as the closure of a braid $B$ with $n$ strands. Then $B$ can be obtained by suspending a diffeomorphism of $D^2$ to itself, fixing the boundary, and permuting $n$ points. Plug this new foliation into the original $D^2 \times S^1$. We then have a flow on $S^3$ such that $L$ is periodic.

For the general case, let $W = L' \cup L''$ such that $M$ is obtained by Dehn surgery on $L'$, and $L''$ becomes the desired link $L$ in $M$ after surgery. Then apply the construction above to $(S^3,W)$, and rotate the foliation near $L'$ as in step (5) above so it is compatible with Dehn surgery.

• $T^3$ cannot be obtained by Dehn surgery on the unknot in $S^3$, since $T^3$ has Heegaard genus 3, which is not a lens space. – Zhiyun Cheng Jul 30 '18 at 5:41

The $1$-dimensional foliations of $3$-manifolds are very flexible. You can have any tame knot $K$ as a closed leaf of a $1$-dimensional foliation of a $3$-torus.

Step 1: Foliate the torus by parallel circles, $S^1 \times \mathbb{T}^2$.

Step 2: Change the foliation near one circle $C_0$ to include a meridional circle $C_1$ with a product foliation around that. You can maintain cylindrical symmetry and that the leaves are tangent to tori around $C_0$. Change the angle with a longitude smoothly from $0$ to $\pi/2$, hold constant for an interval, then change the angle back to a multiple of $\pi$.

Step 3: Perturb the foliation near $C_1$ to include $K$ as a leaf with any representation of $K$ as a closed braid you want.

• The knot leaf (trefoil knot) on 3-torus, I mean is that the bounded ball in 3-torus intersects with the leaf is the knotted 1-tangle. Hence, I think the braid you construct is not the one I wanted. Thank you! – Xifeng Su Aug 5 '16 at 6:29
• What I constructed was a foliation with a trefoil knot as a leaf, which is what you asked. The trefoil is contained in a ball. If you mean that you want the knot to be nontrivial in homology, then I don't think it is normally called a trefoil knot, but it is easy to modify the construction to produce such a knot. – Douglas Zare Aug 6 '16 at 8:28
• Would you please give some more details in step 2 and how is the knot K constructed in step 2? Some references and some pictures about the construction in step 2 are preferred. (I am working on dynamical systems and not quite familiar with your sentences of knot theory in step 2 and so I think I have some misunderstanding. I think that C_0 is a leaf, C_1 is another generator of the 2-torus other than C_0, the neighborhood you consider is the thickened torus C_0*C_1* I where I is the interval you mentioned above. Am I right? What is the cylindrical symmetry, the angle and "hold constant"?) – Xifeng Su Aug 7 '16 at 6:56
• $C_0$ is a leaf of the foliation from step 1. It is not homologically trivial. $C_1$ is a meridional curve on a solid torus whose core is $C_0$. $C_1$ is homologically trivial. In step 2, I replace the foliation in a larger solid torus around $C_0$ so that it agrees with the old foliation on the boundary. Inside, all of the leaves are tangent to the tori of fixed distances from $C_0$. The induced foliations of the tori change in slope from the longitude to the meridian, $C_1$, then back to the longitude next to $C_0$. – Douglas Zare Aug 7 '16 at 12:05

We can foliate a torus (product of circles) by circle factors, with all leaves compact. If the torus has dimension 2, we can foliate it by lines with a given irrational slope quotiented down from the plane, and similarly in any dimension 2 or more, with no leaves compact. I don't know about knots.

Suppose that the three dimensional torus is the quotient of $R^3$ by the group $G$ generated by the translations $t_{e_i},=1,2,3$ of respective directions $e_i, i=1,2,3$. We denote by $p:R^3\rightarrow T^3$ the covering map.

The vector field $X$ of $R^3$ defined by $X_(x,y,z)=e_1+sin(2\pi y)e_2$ is invariant by $G$ and defines a vector field $Y$ on $T^3$, the orbit of $p(0,0,0)$ is compact, but the orbit through $p(0,1/8,0)$ is not compact.