Dual surfaces of a first cohomology class of a 3-manifold

Question

Let $M$ be closed 3-manifold and $\alpha\in H^1(M;\mathbb Z_2)$ an arbitrary element. (In my case we know that $M$ is non-orientable and $\alpha^3=0$.) It is well known that there is a closed 2-submanifold $S\subset M$ so that $[S]=\alpha$. But there are a lot of such $S$.

Question: what can we say about these $S$ itself?

E.g., is there any known lower bounds of the genus of such $S$, in terms of some invariants of $M$?

Conjecture. If $\alpha^2=0$, then $S$ can be chosen so that the normal bundle $\mathcal{N}S$ is trivial. (The opposite implication is clear.)

I think I need some references (books/articles) with theorems of this sort.

Answer 1

Some things are known in the non-orientable case (in orientable 3-manifolds). Bredon and Wood work out the genus in lens spaces $L(2n,q)$ (and some other manifolds) in their paper Non-orientable surfaces in orientable 3-manifolds. Invent. Math. 7 (1969), 83–110. There is related work using Heegaard Floer theory in Ni-Wu Correction terms, ℤ/2-Thurston norm, and triangulations. (English summary) Topology Appl. 194 (2015), 409–426.

If you have access to Mathscinet, you can look at papers that cite the Bredon-Wood paper; you'll find several that are relevant to your question.

Answer 2

$L(4,1)$ is a counterexample to your conjecture, taking $\alpha$ to be the nontrivial element of $H^1(L;\Bbb Z/2)$. Notice that this element squares to zero (the square is the same as the Bockstein, and this element lifts to $\Bbb Z/4$, so its Bockstein is zero).

I claim that no oriented surface represents this homology class. If it did, then $PD(\alpha) \in H_2(L;\Bbb Z/2)$ would lift to an integral class, but this is false as $H_2(L;\Bbb Z) = 0$.

An embedded surface $S$ representing this homology class must be non-orientable. Because $L(4,1)$ is oriented, this implies the normal bundle $NS$ is nontrivial, isomorphic to the orientation line bundle of $TS$.