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I'm trying to understand a step in a proof, where one starts with a non-compact manifold $V$ containing a trapped (2-sided, closed) surface $\Sigma$ that's non-separating. In order to complete the proof, one needs to somehow double that manifold such that in the end one has a (still non-compact) manifold where the surface now separates it. They do this by cutting the manifold $V$ along $\Sigma$, thus obtaining a (still connected) manifold $\tilde{V}$ with two separate boundaries diffeomorphic to $\Sigma$, and then gluing $\mathbb{Z}$ such copies of $\tilde{V}$ end-to-end along the boundaries. Then each ex-boundary ($\Sigma$) separates the new manifold.

My question is: why do I need $\mathbb{Z}$ such copies instead of just two? Can it happen that $\tilde{V}$ turns out to be compact, although $V$ wasn't?

Thanks guys.

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    $\begingroup$ If you used two copies the surface wouldn’t separate the resulting manifold. $\endgroup$
    – HJRW
    Commented Jan 4, 2020 at 21:00
  • $\begingroup$ I mean 2 copies glued together along just one of the boundaries instead of both? $\endgroup$
    – aceituna
    Commented Jan 4, 2020 at 22:46
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    $\begingroup$ Then you get a manifold with boundary (I assume you do not want it). Anyway, could you tell what do you need to prove with this trick (it might help). $\endgroup$
    – Anton Petrunin
    Commented Jan 5, 2020 at 1:35
  • $\begingroup$ Its the second part of the proof for a Variation of the Penrose singularity theorem, switching the condition that the Cauchy surface $V$ contains a trapped surface to $V$ contains a MOTS $\Sigma$ and additionally the generic condition holds on each future and past inextendible null geodesic normal to $\Sigma$. It's Theorem 3.2 from the paper "Topological censorship from the initial data point of view" from Eichmair, Galloway and Pollack. $\endgroup$
    – aceituna
    Commented Jan 5, 2020 at 9:54
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    $\begingroup$ What does MOTS mean? Anyway, if I understood it correctly, the answer to your second, more specific, question is “no” — if $V$ is non-compact then neither is $\tilde V$, however you construct it. $\endgroup$
    – HJRW
    Commented Jan 5, 2020 at 12:09

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