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A straightforward argument reveals that a taut foliation is Reebless, and of course there are many examples of Reebless foliations that are not taut. I guess that there are many examples of three-manifolds having a Reebless foliation but not a taut one.

Looking for an example I found the following one. The 3-manifold obtained by doing $+37/2$-surgery along the $(-2, 3 ,7)$-pretzel knot contains a Reebless foliation (essentially because it contains an incompressible torus whose complement consists of two copies of the complement of the trefoil knot) but does not contain a taut foliation since it is a Heegaard Floer L-space ($+18$-surgery along $K$ gives a lens space, and an L-space is forever!*).

So my main questions are:

  1. Is there a classical proof (i.e. that does not make use of Heegaard Floer homology) of the fact that the $+37/2$ surgery along $K$ does not support a taut foliation?

  2. Can someone give me other (in some sense simpler) examples of three-manifolds having a Reebless foliation but not a taut one?

  3. Are there examples in which the use of Heegaard Floer homology can't be apparently bypassed?

*i.e. if $S^3_n(K)$ is an L-space, then so is $S^3_q(K)$ for every rational number $q\ge 2g(K)-1$.

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    $\begingroup$ A similar result holds for 2-fold branched covers over alternating knots which have Conway spheres. Ozsvath and Szabo proved these are L-spaces. On the other hand, the preimage of a Conway sphere is an incompressible torus, so there is a Reebless foliation. $\endgroup$
    – Ian Agol
    Commented Jun 7, 2015 at 9:49
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    $\begingroup$ Is it true that a 3-manifold containing an incompressible torus always contains a Reebless foliation? Can you give me a reference? $\endgroup$
    – Antonio Alfieri
    Commented Jun 7, 2015 at 11:40
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    $\begingroup$ If the 3-manifold is irreducible, then yes. This follows from a result of Gabai, who showed that an irreducible manifold with incompressible torus boundary admits a taut foliation transverse to the boundary. Then spin these leaves around the boundary and include the boundary torus to get a foliation, then glue. I'll look for a specific reference. $\endgroup$
    – Ian Agol
    Commented Jun 7, 2015 at 20:49
  • $\begingroup$ Ah, ok. This is precisely the argument I've done in the specific case I described above. In that case Gabai's theorem is trivially true since the trefoil knot is fibered. Ok, cool! If you can, give me a reference for Gabai's theorem. $\endgroup$
    – Antonio Alfieri
    Commented Jun 7, 2015 at 22:23
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    $\begingroup$ That's right: projecteuclid.org/euclid.jdg/1214437784 See Theorem 5.1 for the second statement, taking $\partial M=R_+$ in the sutured manifold structure on $M$, when $\partial M$ is a union of incompressible tori and $M$ is irreducible. $\endgroup$
    – Ian Agol
    Commented Jun 8, 2015 at 8:39

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I don't have definitive answers to your questions, but I'll try to make some comments.

For 1), I haven't done a literature search to see if this was known. However, Boyer and Clay propose a related open question Problem 1.11.

For 2), as I mentioned in the comments above, any irreducible manifold $M$ containing an incompressible torus admits a Reebless foliation. This follows from Theorem 5.1 of Gabai's paper. Cut $M$ along an incompressible torus to get manifolds $M_1, M_2$. Then let these be taut sutured manifolds with boundary being $R_+$. By Theorem 5.1, there is a foliation containing $\partial M_i$ as leaves, and which is taut in the interior (hence Reebless). Then glue these two foliations together to get a Reebless foliation of $M$. A caveat here is that the foliation might not be smooth.

Hence, double branched covers of prime alternating knots which have a Conway sphere admit Reebless foliations but not taut ones. The preimage of a Conway sphere is an incompressible torus, but by Ozsvath-Szabo, the double branched cover is an L-space, so does not admit a (smoothe co-oriented) taut foliation.

Simpler examples might come from manifolds admitting taut non-oriented foliations, but no taut oriented foliation, for example sol manifolds that semi-fiber (with trivial first betti number).

For 3), I would suggest that the answer is yes, in the sense that before the work of Ozsvath and Szabo, we knew of very few classes of manifolds not admitting taut oriented foliations (hence, it appears that for most of their examples, Heegaard Floer homology cannot be bypassed). We knew that there was an algorithm to (in principle) detect if a given manifold admits a taut foliation, but it is not practical to implement. Something akin to this was implemented to find infinite collections of manifolds admitting no taut foliation (see also Calegari-Dunfield). The big open question though is whether the converse might hold: if (irreducible orientable) $M$ does not admit a taut (smooth co-oriented) foliation, then is it an L-space? I think Juhasz may have posed this as a question or conjecture.

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  • $\begingroup$ Regarding the attribution of the final question, Boyer--Gordon--Watson attribute the question to Ozsváth and Szabó on p.2 here -- arxiv.org/abs/1107.5016 . But they don't give a reference. $\endgroup$
    – HJRW
    Commented Nov 22, 2022 at 14:55

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