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I think the following two statements are true, and most likely are in the literature. If so, could someone point me to some references? If not, counterexamples?

  1. Let $Y$ be a closed oriented connected 3-manifold, and $[\theta]\in H^1(Y;\mathbb{Z})$ be a primitive class. Then a minimal genus embedded surface representing the Poincar'e dual of $[\theta]$ may be chosen to be connected. (Here, the embedded surface has "minimal genus" if $\chi_-(\Sigma):=\max (-\chi(\Sigma),0)$ is minimal among embedded surfaces representing the same homology class.)
  2. In the preceding context, suppose $\theta=df$ where $f$ is a circle-valued Morse function with no local extrema. Then one of the regular level surfaces of $f$ is a minimal genus embedded surface representing the Poincar'e dual of $[\theta]$.

Many thanks in advance!

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  • $\begingroup$ (1) yes. If the fiber of the map $Y \to S^1$ is not connected, you simply tube the fibers together. Regarding (2), I am not certain how you can interpret a cohomology class as the derivative of a function to the circle. $\endgroup$
    – Ryan Budney
    Jun 18, 2021 at 3:20
  • $\begingroup$ @Ryan Budney: Thanks! For (2): Think of the circle as $\mathbb{R}/\mathbb{Z}$, then given $f: Y\to \mathbb{R}/\mathbb{Z}$, $df$ is a closed 1-form. $\endgroup$
    – user48975
    Jun 18, 2021 at 6:45
  • $\begingroup$ By "no extrema" I suppose you mean no critical points that are local extrema? Because it's unclear what extrema means for circle-valued functions. Off the top of my head I don't see how to come to conclusion (2) for an arbitrary circle-valued Morse function, but one can certainly come to the conclusion in the homotopy-class of $f$. Is that not enough for your purposes? $\endgroup$
    – Ryan Budney
    Jun 18, 2021 at 6:55
  • $\begingroup$ Yes, I should have said "local extrema". I don't know what "the conclusion in the homotopy class of $f$" means, but I need that the fiber of $f$ with minimal $\chi_-$ (among all fibers of $f$) has the minimal $\chi_-$ among embedded surfaces representing the same homology class. $\endgroup$
    – user48975
    Jun 18, 2021 at 9:21
  • $\begingroup$ i.e. if you are comfortable working with a map simply homotopic to $f$, then yes the conclusion holds. The idea is to apply step (1) to a fiber, and then go in reverse to construct a new map $\tilde f$, homotopic to $f$, whose fiber is exactly your tubed-together manifold from step (1). $\endgroup$
    – Ryan Budney
    Jun 18, 2021 at 18:33

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