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Let $M$ be an open orientable three-manifold such that $\pi_1 (M)$ is isomorphic to the fundamental group of a closed orientable surface $S\ncong \mathbb{S}^2$. Furthermore, suppose that $\tilde{M} \cong \mathbb{R}^3$. Is it true that $M \cong S \times \mathbb{R}$?

Without making any assumptions about $\tilde{M}$, there are counterexamples to the question, (see for instance... ), but in those cases the universal cover is not very nice.

[As remarked by several commentators, you need $S$ to be orientable. I added this hypothesis and fixed some of the notation. - Sam]

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    $\begingroup$ Is the equality sign supposed to mean "is homeomorphic to"? $\endgroup$
    – YCor
    Jul 27, 2021 at 22:42
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    $\begingroup$ $M$ could be a nontrivial line bundle over $S$. Maybe you want to assume that both $S$ and $M$ are orientable--and also that $S$ is not the sphere. $\endgroup$ Jul 28, 2021 at 0:27
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    $\begingroup$ As @TomGoodwillie's comment says, you also need to assume that the surface $S$ is orientable: there are orientable non-trivial line bundles over non-orientable surfaces. The simplest such example is a line bundle over the Klein bottle; another way to see it is as the mapping cyclinder of the double cover of the Klein bottle by the torus. $\endgroup$
    – HJRW
    Jul 28, 2021 at 12:03
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    $\begingroup$ When $M$ admits a complete hyperbolic metric, this is true, but you need the Tameness Theorem for that. In general, I don't know. $\endgroup$ Jul 28, 2021 at 16:36
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    $\begingroup$ On the other hand, if $S$ is required to be orientable (as suggested by Tom Goodwillie and HJRW), and $M$ admits a complete metric with non-negative sectional curvature, then the claim is true by the Soul Theorem. $\endgroup$ Jul 28, 2021 at 22:06

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I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$

Suppose that $M$ is a non-compact, connected, oriented three-manifold without boundary, with universal cover homeomorphic to $\RR^3$. Suppose that $F$ is a compact, connected, oriented surface with genus at least one. Suppose that the fundamental groups of $M$ and $F$ are isomorphic. Is $M$ homeomorphic to $F \times \RR$?

The answer is "no". We can build a counterexample using the paper Some examples of exotic non-compact three-manifolds by Scott and Tucker. Here I closely follow their wording and notation, starting at the bottom of page 488.

Consider their example manifold $M_5$. This is a three-manifold with boundary a closed surface $F$. The universal cover of $M_5$ is homeomorphic to $\RR^2 \times \RR_{\geq 0}$, and the inclusion of $F$ into $M_5$ is a homotopy equivalence. However, $M_5$ is not "almost compact" (that is, it is not tame), and $M_5$ is not homeomorphic to $F \times \RR_{\geq 0}$.

Now double $M_5$ across its boundary to obtain $M = D(M_5)$. This does not change the fundamental group. Note that $M$ has two ends, separated by (the image of) $F$. By Scott-Tucker, neither end is tame. Thus $M$ is not homeomorphic to $F \times \RR$. However, the universal cover of the double is (in this case) the double of the universal cover; thus it is $\RR^2 \times \RR$, as desired.

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    $\begingroup$ You may want to add an explanation why the universal cover is $R^3$. (This comes from the fact that the universal cover of $M_5$ is a half-space.) $\endgroup$ Jul 30, 2021 at 13:40
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    $\begingroup$ Great! I didn't know the reference, thank you. $\endgroup$ Jul 30, 2021 at 15:01
  • $\begingroup$ @MoisheKohan - good point - added. $\endgroup$
    – Sam Nead
    Jul 30, 2021 at 15:10

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