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Let $M^3$ be a smooth closed orientable manifold. Does there exist a non negative integer $g_0$ such that every closed orientable embedded surface $\Sigma \subset M$ of genus $g \geq g_0$ represents the trivial class in the second homology group $H_2(M; \mathbb{Z})$?

My thoughts: if every surface separates, then we are done. If this is not the case, let $\Sigma$ be one nonseparating surface of genus $g$. Then, attaching small handles to $\Sigma$ would produce nonseparating surfaces of every genus $\geq g$.

(sorry if it is a silly question…)

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    $\begingroup$ I think your argument is fine. If you can find a surface of genus $g$ that is not nullhomologous, then you can stabilize it to increase the genus to whatever you want without changing the homology class. It's more interesting to try to minimize the genus of a surface representing a given homology class (c.f. the Thurston norm). $\endgroup$
    – Andy Putman
    Commented Sep 23, 2022 at 2:27
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    $\begingroup$ Another more interesting question might be to ask if there are incompressible surfaces of arbitrarily large genus that represent non-trivial classes in $H_2$. (Note that your high-genus examples will all be compressible.) I believe this is always true for fibred hyperbolic manifolds $M$ with $b_2(M)>1$, again using the theory of the Thurston norm -- specifically the fact that fibred classes form an open subset of the unit norm ball. $\endgroup$
    – HJRW
    Commented Sep 23, 2022 at 9:58
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    $\begingroup$ Even the hypothesis on $b_2$ is not necessary. Here is one reference: Knots with infinitely many incompressible Seifert surfaces by Robin Wilson. I assume that these remain distinct and incompressible after longitudinal filling. And if they don't, then I bet there is another family with this extra property. :) $\endgroup$
    – Sam Nead
    Commented Sep 23, 2022 at 10:58

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