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Specifically in a closed, orientable 3-manifold. I'm not necessarily looking for a complete answer, as I don't expect one. Is there prior literature on this question? Also interested in this question but for elements of $H^1(M, \mathbb{Z})$. A necessary condition is that if $\Sigma$ is a compact leaf of your taut foliation, it must have nonzero algebraic intersection with the homology class of the curve. I can't figure out if this is sufficient, or what other necessary conditions there may be.

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    $\begingroup$ Not really an answer, but I think if a loop can be homotoped to be transverse to the taut foliation, then it ought to act by a non-trivial translation on the leaf space of the pullback of the taut foliation to the universal cover, which is a non-hausdorff simply-connected 1-manifold. I think that the converse might also be true. mathscinet.ams.org/mathscinet-getitem?mr=1969207 $\endgroup$
    – Ian Agol
    Jan 20, 2022 at 4:01
  • $\begingroup$ This is a nice question. I makes me wonder about the possibility of a further "promotion" - once the loop is transverse, might it also have lots of friends, in the form of a flow? $\endgroup$
    – Sam Nead
    Jan 22, 2022 at 22:51

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One perspective on this question is to consider the leaf space of the pullback $\tilde{\mathcal{F}}$ of the taut foliation $\mathcal{F}$ of $M$ to the universal cover ($\tilde{M} \cong \mathbb{R}^3$ foliated by planes in the interesting case) together with the action of the fundamental group on the space of leaves $\Lambda=\tilde{M}/\tilde{\mathcal{F}}$, which is a simply-connected but possibly non-Hausdorff 1-manifold by a result of Palmeira.
I think that a loop will be homotopic to be transverse to the foliation if and only if the corresponding group element acts by a translation on part of this manifold $\Lambda$. This seems quite complicated though since a non-Hausdorff 1-manifold can have branching like a tree, so let’s consider the case in which the leaf space $\Lambda$ is Hausdorff.

In this case, the leaf space $\Lambda$ is homeomorphic to $\mathbb{R}$. An example to keep in mind is a 3-manifold fibering over $S^1$. Assume that the foliation $\mathcal{F}$ is cooriented. Then an element $g\in \pi_1(M)$ acts on $\mathbb{R}$ by a homeomorphism $f: \mathbb{R}\to \mathbb{R}$ which preserves orientation. Suppose that $f(x)\neq x$ for some $x\in \mathbb{R}$. Then consider a point in a leaf $L \subset \tilde{\mathcal{F}}$ of the foliation in the preimage of $x$, and its image under the covering translation $g(x)\subset g(L)$. One may connect $x$ to $g(x)$ by an interval transverse to $\tilde{\mathcal{F}}$, and hence projecting to a closed loop transverse to $\mathcal{F}$ in $M$. Thus the only elements of the fundamental group that cannot be homotoped to be transverse to $\mathcal{F}$ are the elements that fix $\Lambda$ pointwise. In the case of a fibration, these are the elements in the fiber surface subgroup. In general, they will correspond to a normal subgroup of $\pi_1(M)$ which is the intersection of the fundamental groups of the leaves of the foliation.

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  • $\begingroup$ I think that "transverse iff translation" can't be true in the non-Hausdorff case. This is because the failure of transversality is "local" while having positive translational part is "global". For example, suppose that $A, B, C_t$ are all connected distinct leaves, with the Hausdorff limit of the $C_t$ containing $A \sqcup B$. Then an arc from $A$ to $B$ will never be transverse. $\endgroup$
    – Sam Nead
    Jan 22, 2022 at 21:41
  • $\begingroup$ @SamNead Well, what I meant was that there is a point in the 1-manifold whose translate lies in the same chart (interval). So I wouldn’t count your example as a translation. $\endgroup$
    – Ian Agol
    Jan 23, 2022 at 2:41
  • $\begingroup$ Ah, I did not understand your definition of "translation". With this definition I think you are precisely correct. :) $\endgroup$
    – Sam Nead
    Jan 23, 2022 at 9:41

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