Is any closed, orientable 3-manifold $M\neq S^1\times S^2$ admitting a co-orientable taut foliation irreducible?

Question

I have seen the following statement in three places: David Gabai's paper Foliations and the topology of 3-manifolds (in the introduction, for Reebless foliations so slightly more general) and Danny Calegri's book Foliations and the Geometry of 3-Manifolds (page 166, Historical Remark) and Mehdi Yazdi's notes Foliations and Three-Dimensional Manifolds:

Theorem:

Let $ M\neq S^1 \times S^2 $ be a closed, orientable 3-manifold that admits a co-orientable taut foliation. Then $ M $ is irreducible.

However, I believe this statement to be incorrect. The manifold $ S^1\times S^2 $ finitely covers $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ (see the first answer to this question), and since $ S^1\times S^2 $ admits a co-orientable taut foliation, this foliation descends to a co-orientable taut foliation on $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ via the covering map. But $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ is clearly not irreducible, contradicting the theorem.

My Question:

1. Is this theorem actually true, and if so, why is $ \mathbb{R}P^3 \# \mathbb{R}P^3$ not a counterexample as I've argued?

I suspect that the inherited taut foliation on $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ is not co-orientable. However I struggle to see why.

I have also seen the same theorem (Theorem 4.35 in here for example), but instead of stating $M \neq S^1 \times S^2$, it is stated that $M$ is not finitely covered by $S^1 \times S^2$. This makes me tend to think that the statement might be wrong after all.

Thank you for your help.

Answer 1

You are making two mistakes in your argument:

1. If $(M,F)$ is an manifold $M$ equipped with a foliation $F$ and $p: M\to N$ is a 2-fold covering map, then $p$ maps $F$ to a foliation $p(F)$ on $N$. For this to be true you want the group $G$ of covering transformation to preserve the foliation $F$, i.e. map leaves to leaves.

2. If $F$ is a cooriented foliation, then (assuming that $p(F)$ is a foliation) $p(F)$ is also cooriented. This is similar to assuming that if $M$ is an orientable manifold, then every manifold covered by $M$ will be also orientable. For coorientability of $p(F)$ to hold, you want $G$ to preserve the coorientation of $F$.

Now, consider the specific example, $M=S^2\times S^1$ with foliation $F$ whose leaves are spheres $S^2\times \{z\}$, $z\in S^1$. I will regard $M$ as the quotient space of $S^2\times [-\pi, \pi]$, where $(\mathbf x,\pi)\sim (\mathbf x,-\pi)$, $\mathbf x\in S^2$. Consider the following involution $\tau$ of $S^2\times [-\pi, \pi]$: $$ \tau(\mathbf x,t)= (-\mathbf x, -t). $$ Then $\tau$ preserves the equivalence relation and, hence, descends to an involution $\sigma$ of $M$. It's easy to see that $\sigma$ is fixed-point free and preserves exactly two leaves of the foliation $F$: One (let's call it $\Sigma_0$) corresponding to $S^2\times \{0\}\subset S^2\times [-\pi, \pi]$ and one (denoted $\Sigma_1$) corresponding to $S^2\times \{\pm \pi\}\subset S^2\times [-\pi, \pi]$. The quotient $N=p(M)$ of $M$ by the group $G$ generated by $\sigma$ is $P^3\# P^3$: A separating sphere in the connected sum will be the projection of the spheres $S^2\times \{\pm \pi/2\}$. The involution $\sigma$ clearly preserves the foliation $F$ of $M$, hence, $p(F)$ is again a foliation of $N$. Generic leaves of the foliation $p(F)$ will be spheres, with two exceptions: The projections $p(\Sigma_0), p(\Sigma_1)$, which will be projective planes. These will be the leaves along which $p(F)$ does not admit a coorientation. (This is a good exercise to work out.)