Can we foliate the punctured space by tori?

Is it possible to have a 2 dimensional foliation of $\mathbb{R}^{3}-\{0\}$ such that each leaf is homeomorphic to the torus? what algebraic topological obstruction exist?

Another question: is there a foliation as above with the following additional property : The foliation is stable at the origin. that is for every neighborhood V of origin there is a smaller neighborhood W such that the saturation of W is contained in V?

the motivation for this question is the concept of "Blow up" of singularities of vector field. when we blow up a singularity in R^3, we replace the singularity by a S^2. The blow up processes is based on the fact that R^3-{0} is foliated by a familly of S^2. Now if the answer of the above question is positive we can obtain a new type of torus- blow up. However the blowing up was my main motivation for this question, but ,by this question, I never mean "is R^3-{0} homeomorphic to R x tori?"

• The vanishing of the Euler characteristic is a necessary condition for the existence of a codimension 1 foliation, but of course the Euler characteristic of $R^3-0$ does vanish, so this doesn't put an obstacle in this case. – ThiKu Dec 4 '13 at 17:41
• I am not convinced that the vanishing of the euler characteristic is necessary for open 3-manifolds to support a codimension 1 foliation -- it certainly is not necessary for open surfaces. – J. Martel Dec 5 '13 at 0:17
• @J.Martel, I agree with you. R^{3}-{0} is foliated by a one parameter familly of 2- spheres. Do you have any Idea on the main question:the foliation of R^3-{0} by torus? Thanks – Ali Taghavi Dec 5 '13 at 13:18
• @AliTaghavi: I really like your question and i don't have an answer, but i'm thinking about it. What is obvious (via the long exact homotopy sequence) is that there is no fibration of $R^3-0$ by tori -- but a foliation is rather far from being a fibration. – J. Martel Dec 5 '13 at 13:56
• A related (and possibly easier) question would be to show that R^3-0 is not foliated by R^2-planes, here seeing R^2 as the universal cover of T^2, and likewise can we foliate by higher genus (possibly open) surfaces? I expect no', but not exactly sure why. – J. Martel Dec 5 '13 at 14:09

I would like to offer another explanation of the impossibility of foliating $R^3-0$ by tori (or by higher genus closed surfaces), at least in the $C^\infty$ case.

Previously I commented that "foliations are rather far from fibrations". Closer to the truth, foliations are submersions' onto their (potentially very weird!) leaf spaces. So the distance between foliations and fibrations is, in some respects, comparable with the distance between submersions and fibrations. In the case of smoothly foliating open manifolds by compact submanifolds, the distinction however is very small -- in fact, I want to claim below that foliations in this setting are exactly fibrations.

The starting point is Ehresmann's fibration theorem: if $f:V\to M$ is a proper submersion of smooth manifolds, then $f$ is a locally trivial fibration. A proof can be found in Brocker & Janich's "Introduction to differential topology", section 8.12.

Hence if we have a proper smooth function $f$ on $R^3-0$ having no critical points, then the fibres $f^{-1}(pt)$ foliate $R^3-0$ by compact embedded submanifolds. Ehresmann's theorem tells us $f$ is a locally trivial fibration, and of course as $R$ is contractible, $f$ actually defines a globally trivial fibration. In otherwords, all the fibres are diffeomorphic and we have $R^3-0 \simeq f^{-1}(pt) \times R$. From here we can determine that any fibre must be $\simeq S^2$.

An important point which needs some further justification is this: given a smooth foliation $\mathscr{F}$ of $R^3-0$ by, say, compact tori, how do I know that the quotient map from $R^3-0$ to the leaf space is a smooth submersion of $R^3-0$ onto a smooth $1$-manifold? I see how compactness ensures distinct leafs remain separated (and so the quotient is hausdorff), however i am not clear on pinning down the smooth structure of the quotient map.

Granting the above point, we would know that the quotient is necessarily a noncompact smooth 1-manifold, i.e. the real line $R$, and hence the total space must be a product $T\times R$ -- which we know it isn't.

These arguments, however, suffer the shortcoming of not being able to establish whether or not $R^3-0$ can be foliated by punctured surfaces, a question which seems interesting itself.

• An orientability hypothesis is needed since an open Möbius band is smoothly foliated by circles. In this case the quotient is a half-open interval, so it's a 1-manifold with boundary. – Allen Hatcher Dec 7 '13 at 0:43
• @AllenHatcher: thank you for the comment. The non-orientability of the total space may be a red herring. The issue with the mobius strip being rather that we have a leaf which is non-separating (namely the centre leaf). My argument above is then only valid so long as every leaf is two-sided, i.e. separating $R^3-0$. – J. Martel Dec 7 '13 at 3:41
• Now are there any non-separating embedded tori in $R^3-0 \ldots \ldots$ – J. Martel Dec 7 '13 at 4:11
• @J Martel. Thank you very much for your answer. Regarding your last statement I think that every compact codimension one submanifold of R^n is oriantable and separating.If I remember well I saw this theorem some years ago in Hirsch "Differential Topology" – Ali Taghavi Dec 7 '13 at 14:00
• "smoothness of the quotient map" is not a consequence of "rank theorem"? – Ali Taghavi Dec 7 '13 at 15:45

The answer is "No": it is impossible.

If $\mathbb{R}^3 - \{0\}$ is foliated by tori, then it has to homeomorphic to a torus bundle over a 1-manifold. It is obviously this is impossible.

To explain why it is a torus bundle over a 1-manifold, we only need to prove for each leaf $T$ of the foliation, there exists a neighborhood $U(T)$ such that the foliation restricted in $U(T)$ is trivial ($U(T)= T^2 \times \mathbb{R}$). Essentially, this is the direct consequence of the holonomy group of $T$ is trivial based on two things:

1. the assumption that each leaf is compact;
2. each embedded torus in $\mathbb{R}^3 - \{0\}$ is separating.

More precisely, we can understand the local structure as follows.

• choosing two simple closed curves $m$ and $l$ in $T$ such that they inersect at one point $P$. - choosing a small interval $l_0$ transverse to the foliation passing through $P$ ($l_0$ is called a transversal).
• choosing a small neighborhood $U(T)$ of $T$ and a leave $F$ close to $T$ and $F\cap l_0 \neq \emptyset$.
• along the orientation of $m$, we can lift $(m,P)$ to $(m_0,(P_0,P_1))$. Here $m_0\subset F$, $P_0,P_1 \in l_0$ and $P_0$ and $P_1$ are the starting point and ending point of $m_0$ respectively.
• we claim that $P_0 =P_1$, otherwise there are two possibilites:

• (1) $P_0$ and $P_1$ are in two sides of $T$. This impossible since $T$ is separating in $\mathbb{R}^3 -\{0\}$.

• (2) $P_0$ and $P_1$ are in one side of $T$. This will provide a path $L$ in $F$ approaching to $T$. This conflicts to the fact $F$ is compact ($F$ is a torus). (the path $L$ is the union of many lift $m_0$, $m_1$, $m_2$, ... of $m$: the starting point of $m_{i+1}$ is the ending point of $m_i$ in the transversal $l_0$)

• $T - l \cup m$ is a disk (contractable).

• Finally, "The foliation structure restricted in $U(T)$ is trivial" can be followed by the above statements.
• Thank you for the answer.you proved that the foliation gives us a fibre bundle, since the holonomy is trivial. Why this fibre bundle is globally trivia? – Ali Taghavi Dec 6 '13 at 8:33
• @Taghavi, I don't prove the fiber bundle is trivial. I just say "it is a torus bundle over a 1-manifold". But every torus bundle over a 1-manifold can't be homeomorphic to $R^3 -\{0\}$. – Bin Yu Dec 6 '13 at 9:22
• As a consequence of leary hirsch theorem? – Ali Taghavi Dec 6 '13 at 12:39
• I find this argument to be unclear. As i see it, there should be a uniform argument applicable to tori and closed higher genus surfaces. If indeed any such foliation is actually a locally trivial fibration, then within the smooth category it will be a trivial bundle over (the only noncompact smooth 1-manifold) $R$. – J. Martel Dec 6 '13 at 15:06
• @J.Martel as a combination of your idea and Bin's proof it seems that we can conclude " A codimension one foliation of an open manifold with compact leaves, gives a trivial bundle" Am i write? is it a known theorem in foliation theory? However some part of Bin's proof is not clear for me: How is the open set U(T) constructed? what was the role of contractibility of T-l\cup m? – Ali Taghavi Dec 6 '13 at 16:21