Specifically in a closed, orientable 3manifold. 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.

1$\begingroup$ 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 nontrivial translation on the leaf space of the pullback of the taut foliation to the universal cover, which is a nonhausdorff simplyconnected 1manifold. I think that the converse might also be true. mathscinet.ams.org/mathscinetgetitem?mr=1969207 $\endgroup$– Ian AgolJan 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 NeadJan 22, 2022 at 22:51
1 Answer
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 simplyconnected but possibly nonHausdorff 1manifold 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 nonHausdorff 1manifold 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 3manifold 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.

$\begingroup$ I think that "transverse iff translation" can't be true in the nonHausdorff 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 NeadJan 22, 2022 at 21:41

$\begingroup$ @SamNead Well, what I meant was that there is a point in the 1manifold whose translate lies in the same chart (interval). So I wouldn’t count your example as a translation. $\endgroup$– Ian AgolJan 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 NeadJan 23, 2022 at 9:41