A Seifert manifold $M$ is a $3$-dimensional orientable smooth manifold with an effective circle action with no fixed points.
Closed connected Seifert manifolds are classified up to an equivariant diffeomorphism, by the Seifert invariant that encodes the genus of the orbit space, together with data on the exceptional (non-free) orbits.
Question: Is there anything known about the classification of non-compact Seifert manifolds?
Naively, one may try to take the classification of non-compact surfaces, and define a new invariant by keeping track of the surface-type of the orbit space, and the data on the exceptional orbits (which will now be an infinite list), but it's unclear to me if this approach will work.
This can also be translated to orbifold language, if I'm not mistaken, to a question of classifying circle bundles over non compact $2$-dimensional orbifolds.
Edit: maybe I should've started by asking the same questions for circle bundles over non compact smooth surfaces.
Best Answer
This is copied, in part, from the comments. So I've made it community wiki. Please feel free to edit and improve.
The consensus in the comments is that "the classification in the non-compact case follows from the classification in the compact case". Here are a few details.
Suppose that $M$ is a oriented, connected, smooth three-manifold (possibly with boundary).
Definition: Suppose that $M$ is equipped with a smooth circle action $\sigma \colon S^1 \times M \to M$ so that, for every point $x \in M$, the orbit map $S^1 \to \mathcal{O}_x = S^1 \cdot x$ is a covering map. Suppose also that, away from an isolated collection of exceptional orbits, this degree is one. Then we call $\sigma$ a Seifert fibre structure on $M$, and we say that $M$ is Seifert fibered.
Definition: Suppose that $\sigma$ is a Seifert fibering of $M$. We use $B = M/\sigma$ to denote the base orbifold: that is, the quotient of $M$ by the $S^1$ action, where the classes of exceptional orbits are marked by the degree of their covering.
Definition: We call two Seifert fibre structures $\sigma$ and $\sigma'$ on $M$ orbit equivalent is there is an isotopy of $M$ taking the orbits of $\sigma$ to the orbits of $\sigma'$. If all Seifert fibre structures on $M$ are orbit equivalent then we say that the fibering is unique.
For example, the solid torus $S^1 \times D^2$ has infinitely many inequivalent Seifert fibre structures. Doubling, we find that the same is true of $S^1 \times S^2$. However, if $M$ is compact and the base orbifold $M/\sigma$ is "sufficently complicated", then the fibering is unique.
Classification in non-compact case: Suppose $M$ (oriented, connected, smooth three-manifold) admits a Seifert fibre structure $\sigma$. Then $\sigma$ is unique, unless it is homeomorphic to one of the known counterexamples in the compact case (perhaps minus some boundary components).
Proof: Let $B = M / \sigma$ be the base orbifold of the given fibering. If $B$ has finite topological type then we are (essentially) in the compact case. Suppose instead that $B$ has infinite topology. (So, infinite genus, infinitely many ends, infinitely many classes of exceptional orbits, or some combination of these.)
Suppose now that there is another Seifert fibering $\sigma'$ of $M$. Suppose that the new base orbifold $B' = M / \sigma'$ has finite topological type. Thus $M$ admits a horizontal surface of finite topological type. This covers $B$, a contradiction.
So $B'$ has infinite type. So $\sigma'$ admits (essential) vertical tori. These, again, cannot be horizontal surfaces with respect to $\sigma$. So they are isotopic to vertical surfaces. Taking a maximal (but locally finite...) collection of vertical tori gives the desired isotopy. $\Box$
Hmm. There may be a subtle point here. Why is there a maximal locally finite collection of vertical tori? If no such collection exists, then I do not see how to promote a collection of vertical tori to an isotopy of Seifert fibered structures.